Collect, Combine Like Terms

Simplify.

x + x Solution

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'Like terms' have the same variable and the same degree (exponent).

= 2x

x + x + x + x + x Solution

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'Like terms' have the same variable and the same degree (exponent).

= 5x

2x + x + 3x Solution

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'Like terms' have the same variable and the same degree (exponent).

= 6x

x + 2x - 2x Solution

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'Like terms' have the same variable and the same degree (exponent).

= 3x - 2x = x

x + y + x + x + y Solution

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'Like terms' have the same variable and the same degree (exponent).

= 3x + 2y

x + 2y + x - x - y Solution

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'Like terms' have the same variable and the same degree (exponent).

= 2y - y + x + x - x

= y + x

2a + b + 3c - a + b - 2c Solution

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'Like terms' have the same variable and the same degree (exponent).

= 2a - a + b + b + 3c - 2c

= a + 2b + c

x + x + 1 Solution

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'Like terms' have the same variable and the same degree (exponent).

= 2x + 1

2 + x + 1 + x Solution

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'Like terms' have the same variable and the same degree (exponent).

= 2x + 3

1 + 2y + x + 5 + x - y + 10 Solution

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'Like terms' have the same variable and the same degree (exponent).

= 1 + 5 + 10 + 2y - y + x + x

= 16 + y + 2x

Representations of Powers: Pre-Product Rule

Simplify leaving your answer in the most reduced exponent form.

2 × 2 × 2 Solution

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= 2³

3 × 3 × 3 × 3 × 3 Solution

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= 3⁵

b × b × b Solution

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= b³

c × c × c × c × c Solution

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= c⁵

3 × a × a Solution

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= 3a²

2y × 2y Solution

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= (2)(2) y²

= 4y²

3a × 4a × a Solution

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= 12a³

–b × 2a Solution

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= -2ab

–5b × 2ab Solution

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= -10ab²

–2a × –ab × 3c Solution

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= 6a²bc

Exponents: Multiplication Uses Product Rule

Write each with a single, positive exponent, showing your work without using a calculator.

$3^2 × 3^5$ Solution

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= 3^{2 + 5} = 3⁷

$2^2 × 2^3 × 2^1 × 2$ Solution

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= 2^{2 + 3 + 1 + 1} = 2⁷

$5^2 × 5^2$ Solution

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= 5^{2 + 2} = 5⁴

$(-10)^5 × (-10)^7$ Solution

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= (-10)^{5 + 7}
= (-10)¹²

$\left(\dfrac{2}{3}\right)^4 × \left(\dfrac{2}{3}\right)^3$ Solution

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$\left(\dfrac{2}{3}\right)$

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Hint Unavailable

$\quad \left(\dfrac{2}{3}\right)^{4 + 3} \\ \\ \quad = \left(\dfrac{2}{3}\right)^7$

$\left(\dfrac{1}{2}\right)^1 × \left(\dfrac{1}{2}\right)^2 × \left(\dfrac{1}{2}\right)^3$ Solution

Hint Clear Info

$\left(\dfrac{1}{2}\right)$

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Hint Unavailable

$\quad \left(\dfrac{1}{2}\right)^{1 + 2 + 3} \\ \\ \quad = \left(\dfrac{1}{2}\right)^6$

Exponents: Division is Quotient Rule

Simplify each with a single, positive exponent, showing your work without using a calculator.

$7^7 \div 7^3$ Solution

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= 7^{7 - 3}

= 7⁴

$(-2)^8 \div (-2)^3$ Solution

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= (-2)^{8 - 3}

= (-2)⁵

$(-5)^6 \div (-5)^4$ Solution

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= (-5)^{6 - 4}

= (-5)²

$3^9 \div 3^4 \div 3^1 \div 3 \div 3^2$ Solution

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= 3⁹ ÷ 3⁴ ÷ 3¹ ÷ 3¹ ÷ 3²

= 3^{9 - 4 - 1 - 1 - 2}

= 3¹

= 3

$\dfrac{(5^4)}{5^2}$ Solution

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= 5⁴ ÷ 5²

= 5^{4 - 2}

= 5²

Exponents: Power of a Power

Simplify each with a single, positive exponent, showing your work without using a calculator.

$\left(3^2\right)^4$ Solution

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= 3^{(2 × 4)}

= 3⁸

$\left(2^2\right)^3$ Solution

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= 2^{(2 × 3)}

= 2⁶

$\left(\left(2^2\right)^3\right)^2$ Solution

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= 2^{(2 × 3 × 2)}

= 2¹²

EXPONENTS: GENERAL MIX

Simplify each with integers, or fractions, or positive exponents.

3 - 3² + 3³ Solution

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Hint Unavailable

3 - 3² + 3³
= 3 - 9 + 27
= 21

$\left( \dfrac{1}{3} × \dfrac{2}{1} \right)^3 \div \left( \dfrac{3}{2} \right)^2$ Solution

Hint Clear Info

$\left(\dfrac{2}{3}\right)$

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$\begin{align} & = \left( \dfrac{1}{3} × \dfrac{2}{1} \right)^3 \div \left( \dfrac{3}{2} \right)^2 \\ \\ & = \left( \dfrac{1 × 2}{3 × 1} \right)^3 × \left( \dfrac{2}{3} \right)^2 \\ \\ & = \left( \dfrac{2}{3} \right)^3 × \left( \dfrac{2}{3} \right)^2 \\ \\ & = \left( \dfrac{2}{3} \right)^{3+2} \\ \\ & = \left( \dfrac{2}{3} \right)^{5} \\ \\ \end{align}$

Exponents: Combination of Product and Quotient Rule

Write each with a single, positive exponent, showing your work without using a calculator. Simplify fully.

$\dfrac{(8^4)(8^2)}{(8^1)(8^3)}$ Solution

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= (8⁴)(8²) ÷ (8¹)(8³)

= 8^{4 + 2} ÷ 8^{1 + 3}

= 8⁶ ÷ 8⁴

= 8^{6 - 4}

= 8²

= (2³)²

= 2³⁽²⁾

= 2⁶

$\dfrac{(2.1)^3(2.1)^1}{(2.1)^2(2.1)}$ Solution

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Hint Unavailable

= (2.1)^{3 + 1 - (2 + 1)}

= (2.1)^{4 - 3}

= (2.1)¹

= 2.1

Understanding Powers of Ten with Scientific Notation

Convert the following scientific notation numbers.

1.634 × 10^-9 Solution

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Hint Unavailable

2.50 × 10⁴ Solution

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Hint Unavailable

Understanding Powers of Ten with Scientific Notation

Convert to Scientific Notation

0.0000519 Solution

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× 10

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Hint Unavailable

50,525 Solution

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× 10

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Hint Unavailable

Distributive Property

Simplify

2(6)(3) = 2(6) × 2(3) Solution

1. True
2. False

This does not use the distributive property because everything is multiplied here: 2(6)(3)

The distributive property is used when different terms are separated by addition or subtraction.

4(x + 2) = 4(4) × 4(2) Solution

1. True
2. False

4(x + 2) does use the distributive property.

2(3x + 1) Solution

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$\begin{align} & 2(3x + 1) \\ \\ & = 2(3x) + 2(1) \\ \\ & = 6x + 2 \\ \\ \end{align}$

4(1 + 2x – 2) Solution

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= 4(1) + 4(2x) + 4(-2)

= 4 + 8x - 8

= 4 - 8 + 8x

= -4 + 8x

-x(1 - x) Solution

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= -x(1) - x(-x)

= -x + x²

x(3y + 2x) Solution

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= x(3y) + x(2x)

= 3xy + 2x²

Polynomial Expressions up to 2 Variables: Addition, Subtraction

Simplify by distributing and collecting like terms.

$1 + (x + 1) + 1$ Solution

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Hint Unavailable

$\begin{align} & 1 + (x + 1) + 1 \\ \\ & = 1 + x + 1 + 1 \\ \\ & = x + 1 + 1 + 1 \\ \\ & = x + 3 \\ \\ \end{align}$

$(x + 2) + (2x + 4)$ Solution

Hint Clear Info

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Hint Unavailable

= x + 2 + 2x + 4

= x + 2x + 2 + 4

= 3x + 6

$2 - (5x + 1)$ Solution

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Hint Unavailable

= 2 - 1(5x + 1)

= 2 - 1(5x) - 1(1)

= 2 - 5x - 1

= 1 - 5x

$(8x + 6) - 2x - 3$ Solution

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Hint Unavailable

= 8x + 6 - 2x - 3

= 8x - 2x + 6 - 3

= 6x + 3

$(2x + 4) + (3x + 2) - (3x - 3)$ Solution

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= 2x + 4 + 3x + 2 - 3x + 3

= 2x + 3x - 3x + 4 + 2 + 3

= 2x + 9

$(6x - 3x) - (4y + 2y)$ Solution

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= 3x - 6y

$(2x + 3y) + (x \ – y)$ Solution

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= 3x + 2y

$2x + 3y + 4xy \ – 3x + 8yx – 2x$ Solution

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= 2x - 3x - 2x + 3y + 4xy + 8xy

= -3x + 3y + 12xy

$2x^2y + 3x^2y$ Solution

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= 5x²y

$5x^2y \ – x^2y + 2xy^2$ Solution

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= 5x²y - 1x²y + 2xy²

= 4x²y + 2xy²

$(4x^2y + 1xy^2) + (3x^2y + 5xy^2)$ Solution

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= 4x²y + 1xy² + 3x²y + 5xy²

= 4x²y + 3x²y + 1xy² + 5xy²

= 7x²y + 6xy²

$(5x^2y \ – 5xy^2) + (–3x^2y + 7xy^2)$ Solution

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Hint Unavailable

= 5x²y - 5xy² - 3x²y + 7xy²

= 5x²y - 3x²y - 5xy² + 7xy²

= 2x²y + 2xy²

$(5x^2y \ – 2xy^2) \ – \ (1x^2y \ – 4xy^2)$ Solution

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$\begin{align} & (5x^2y \ – 2xy^2) \ – \ (1x^2y \ – 4xy^2) \\ \\ & = 5x^2y \ – 2xy^2 – 1x^2y + 4xy^2 \\ \\ & = 5x^2y \ – 1x^2y – 2xy^2 + 4xy^2 \\ \\ & = 4x^2y + 2xy^2 \\ \\ \end{align}$

Polynomials up to 2 Variables: Multiplication

Simplify by distributing and collecting like terms. Order the terms with decreasing exponents for example, x² + 2x - 3.

$x(2x)$ Solution

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$\begin{align} & x(2x) \\ \\ & = 2x^2 \\ \\ \end{align}$

$x(x + 1)$ Solution

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= x(x) + x(1)

= x² + x

$–x(3x \ – 3)$ Solution

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= -x(3x) - x(-3)

= -3x² + 3x

$2x(3 + 4x) + 3x(x \ – 2)$ Solution

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Hint Unavailable

= 2x(3) + 2x(4x) + 3x(x) + 3x(-2)

= 6x + 8x² + 3x² - 6x

= 11x²

$2y^2(5) \ – 5y^2 + 4y + 3$ Solution

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Hint Unavailable

= (2)(5)y² - 5y² + 4y + 3

= 10y² - 5y² + 4y + 3

= 5y² + 4y + 3

$2x(3x^2 + 1) \ – x(3 + 2x)$ Solution

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= (2x)(3x²) + (2x)(1) - (x)(3) - (x)(2x)

= (2)(3)x^{1 + 2} + 2x - 3x - 2x^{1 + 1}

= 6x³ - 2x² - x

* $\quad 2y(4y + 3y^2 – 4) + 1$ Solution

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= (2y)(4y) + (2y)(3y²) + (2y)(-4) + 1

= 6y³ + 8y² - 8y + 1

* $\quad –2y^2 (3y \ – 5) + 3y^3 – 2y^2 + 4y + 3$ Solution

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= -6y³ + 10y² + 3y³ - 2y² + 4y + 3

= -3y³ + 8y² + 4y + 3

Linear Equations

Solve the following equation (by removing the decimals to make it easier for yourself). Solution $1.4 - 0.5x = -0.9x + 3$

x =

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Multiply everything by 10 to eliminate the decimals. $\begin{align} 1.4 - 0.5x & = -0.9x + 3 \\ \\ 10(1.4 - 0.5x) & = 10(-0.9x + 3) \\ \\ 10(1.4) - 10(0.5x) & = 10(-0.9x) + 10(3) \\ \\ 14 - 5x & = -9x + 30 \\ \\ 14\ \underline{ - 14} - 5x & = -9x + 30\ \underline{ - 14} \\ \\ -5x & = -9x + 30 - 14 \\ \\ -5x & = -9x + 16 \\ \\ -5x\ \underline{ + 9x} & = -9x\ \underline{ + 9x} + 16 \\ \\ -5x + 9x & = 16 \\ \\ 4x & = 16 \\ \\ x & = \frac{16}{4} \\ \\ x & = 4 \\ \\ \end{align}$

Cross Multiplication

Cross multiplication is possible in both of the following cases. Solution $\dfrac{5}{4} × \dfrac{3}{2x} \quad\quad\quad\quad\quad\quad\quad \dfrac{2}{7} = \dfrac{3}{4x}$

1. True
2. False

Cross Multiplication

Solve.

$\quad \dfrac{x}{2} = \dfrac{3}{1}$ Solution

x =

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$\begin{align} \dfrac{x}{2} & = \dfrac{3}{1} \\ \\ 1(x) & = 2(3) \\ \\ x & = 6 \\ \\ \end{align}$

$\quad \dfrac{2}{3} = \dfrac{2x}{9}$ Solution

x =

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$\begin{align} \dfrac{2}{3} & = \dfrac{2x}{9} \\ \\ 9(2) & = 3(2x) \\ \\ 18 & = 6x \\ \\ 6x & = 18 \\ \\ x & = \frac{18}{6} \\ \\ x & = 3 \\ \\ \end{align}$

$\quad \dfrac{15}{6x} = \dfrac{2}{4}$ Solution

x =

Hint Clear Info

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$\begin{align} \dfrac{15}{6x} & = \dfrac{2}{4} \\ \\ \dfrac{15 (4)}{} & = \dfrac{2 (6x)}{} \\ \\ 60 & = 12x \\ \\ 12x & = 60 \\ \\ x & = \dfrac{60}{12} \\ \\ x & = 5 \\ \\ \end{align}$

$\quad \dfrac{3}{18} = \dfrac{4}{3x}$ Solution

x =

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$\begin{align} \dfrac{3}{18} & = \dfrac{4}{3x} \\ \\ \dfrac{3 (3x)}{} & = \dfrac{4 (18)}{} \\ \\ 9x & = 72 \\ \\ x & = \dfrac{72}{9} \\ \\ x & = 8 \\ \\ \end{align}$

$\quad \dfrac{-7}{12} = \dfrac{x}{3}$ Solution

Hint Clear Info

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$\begin{align} \dfrac{-7}{12} & = \dfrac{x}{3} \\ \\ \dfrac{-7 (3)}{} & = \dfrac{x (12)}{} \\ \\ -21 & = 12x \\ \\ 12x & = -21 \\ \\ \dfrac{\cancel{12}x}{\cancel{12}} & = \dfrac{-21}{12} \\ \\ 1x & = \dfrac{-21 \div 3}{12 \div 3} \\ \\ x & = \dfrac{-7}{4} \\ \\ \end{align}$

$\quad \dfrac{27}{6} = -9x$ Solution

Hint Clear Info

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$\begin{align} \dfrac{27}{6} & = -9x \\ \\ -9x & = \dfrac{27}{6} \\ \\ \dfrac{-9x}{1} & = \dfrac{27}{6} \\ \\ \dfrac{-9x (6)}{} & = \dfrac{27 (1)}{} \\ \\ -54x & = 27 \\ \\ \dfrac{\cancel{-54}x}{\cancel{-54}} & = \dfrac{27}{-54} \\ \\ x & = \dfrac{27}{-54} \\ \\ x & = \dfrac{27 \div 27}{-54 \div 27} \\ \\ x & = \dfrac{1}{-2} \\ \\ x & = \dfrac{-1}{2} \\ \\ \end{align}$

Algebra with Fractions and Fractional Coefficients Using Cross Multiplication

Solve

$\dfrac{2x\ +\ 2}{3} = \dfrac{4}{1}$ Solution

x =

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$\begin{align} \dfrac{2x\ +\ 2}{3} & = \dfrac{4}{1} \\ \\ 1(2x + 2) & = 3(4) \\ \\ 2x + 2 & = 12 \\ \\ 2x & = 12 - 2 \\ \\ 2x & = 10 \\ \\ x & = \frac{10}{2} \\ \\ x & = 5 \\ \\ \end{align}$

$\dfrac{x\ -\ 1}{6} = \dfrac{x\ +\ 3}{2}$ Solution

x =

Hint Clear Info

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2 options: cross multiply, or multiply each by the lowest common multiple.
(Cross multiplication shown here:) $\begin{align} \dfrac{x\ -\ 1}{6} & = \dfrac{x\ +\ 3}{2} \\ \\ 2(x - 1) & = 6(x + 3) \\ \\ 2x - 2 & = 6x + 18 \\ \\ 2x - 6x & = 18 + 2 \\ \\ -4x & = 20 \\ \\ x & = \dfrac{20}{-4} \\ \\ x & = -5 \\ \\ \end{align}$

$\dfrac{1}{4}(8x\ +\ 16) = \dfrac{1}{2}(2x\ +\ 4)$ Solution

x =

Hint Clear Info

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x = -2

Solving Equations with Fractions

Solve. Solution $8\left(\dfrac{2x}{4}\right) - 8\left(\dfrac{x}{8}\right) = 8\left(\dfrac{3}{2}\right)$

x =

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Get rid of the fractions, by reducing like this, $\begin{align} 8\left(\dfrac{2x}{4}\right) - 8\left(\dfrac{x}{8}\right) & = 8\left(\dfrac{3}{2}\right) \\ \\ ^2\!\cancel{8}\left(\dfrac{2x}{\cancel{4}}\right) - \cancel{8}\left(\dfrac{x}{\cancel{8}}\right) & = \ ^4\!\cancel{8}\left(\dfrac{3}{\cancel{2}}\right) \\ \\ 2(2x) - x & = 4(3) \\ \\ 4x - x & = 12 \\ \\ 3x & = 12 \\ \\ x & = \dfrac{12}{3} \\ \\ x & = 4 \\ \\ \end{align}$

Solving Equations with Fractions

Solve.

$\dfrac{2x}{3} - \dfrac{5}{3} = 1$ Solution

x =

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$\begin{align} \dfrac{2x}{3} - \dfrac{5}{3} & = 1 \\ \\ 3\left(\dfrac{2x}{3}\right) - 3\left(\dfrac{5}{3}\right) & = 3(1) \\ \\ \cancel{3}\left(\dfrac{2x}{\cancel{3}}\right) - \cancel{3}\left(\dfrac{5}{\cancel{3}}\right) & = 3 \\ \\ 2x - 5 & = 3 \\ \\ 2x & = 3 + 5 \\ \\ 2x & = 8 \\ \\ x & = \frac{8}{2} \\ \\ x & = 4 \\ \\ \end{align}$

$\dfrac{4}{3} - \dfrac{x}{3} = 4 - \dfrac{5x}{3}$ Solution

x =

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$\begin{align} \dfrac{4}{3} - \dfrac{x}{3} & = 4 - \dfrac{5x}{3} \\ \\ 3\left(\dfrac{4}{3}\right) - 3\left(\dfrac{x}{3}\right) & = 3\left(4\right) - 3\left(\dfrac{5x}{3}\right) \\ \\ \cancel{3}\left(\dfrac{4}{\cancel{3}}\right) - \cancel{3}\left(\dfrac{x}{\cancel{3}}\right) & = 3\left(4\right) - \cancel{3}\left(\dfrac{5x}{\cancel{3}}\right) \\ \\ 4 - x & = 12 - 5x \\ \\ 4 - x & = 12 - 5x \\ \\ - x + 5x & = 12 - 4 \\ \\ 4x & = 8 \\ \\ \dfrac{\cancel{4}x}{\cancel{4}} & = \dfrac{8}{4} \\ \\ x & = 2 \\ \\ \end{align}$

$\dfrac{10}{9} - \dfrac{2x}{3} = \dfrac{8}{3} - \dfrac{8}{9}$ Solution

x =

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$\begin{align} \dfrac{10}{9} - \dfrac{2x}{3} & = \dfrac{8}{3} - \dfrac{8}{9} \\ \\ 9\left(\dfrac{10}{9}\right) - 9\left(\dfrac{2x}{3}\right) & = 9\left(\dfrac{8}{3}\right) - 9\left(\dfrac{8}{9}\right) \\ \\ \cancel{9}\left(\dfrac{10}{\cancel{9}}\right) - \ ^3\!\cancel{9}\left(\dfrac{2x}{\cancel{3}}\right) & = \ ^3\!\cancel{9}\left(\dfrac{8}{\cancel{3}}\right) - \cancel{9}\left(\dfrac{8}{\cancel{9}}\right) \\ \\ 10 - 3(2x) & = 3(8) - 8 \\ \\ 10 - 6x & = 24 - 8 \\ \\ -6x & = 24 - 8 - 10 \\ \\ -6x & = 6 \\ \\ x & = \dfrac{\cancel{6}}{-\cancel{6}} \\ \\ x & = -\dfrac{1}{1} \\ \\ x & = -1 \\ \\ \end{align}$

$\dfrac{16 - 5x}{6} = 2 - \dfrac{3x}{2}$ Solution

x =

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$\begin{align} \dfrac{16 - 5x}{6} & = 2 - \dfrac{3x}{2} \\ \\ 6\left(\dfrac{16 - 5x}{6}\right) & = 6\left(2\right) - 6\left(\dfrac{3x}{2}\right) \\ \\ \cancel{6}\left(\dfrac{16 - 5x}{\cancel{6}}\right) & = 6\left(2\right) - \ ^3\!\cancel{6}\left(\dfrac{3x}{\cancel{2}}\right) \\ \\ \\ 16 - 5x & = 6(2) - 3(3x) \\ \\ 16 - 5x & = 12 - 9x \\ \\ -5x + 9x & = 12 - 16 \\ \\ 4x & = -4 \\ \\ x & = \dfrac{-4}{4} \\ \\ x & = -1 \\ \\ \end{align}$

$\dfrac{9x + 12}{7} - \dfrac{15x + 10}{14} = \dfrac{17}{14}$ Solution

x =

Hint Clear Info

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$\begin{align} 14\left(\dfrac{9x + 12}{7}\right) - 14\left(\dfrac{15x + 10}{14}\right) & = 14\left(\dfrac{17}{14}\right) \\ \\ ^2\!\cancel{14}\left(\dfrac{9x + 12}{\cancel{7}}\right) - \cancel{14}\left(\dfrac{15x + 10}{\cancel{14}}\right) & = \cancel{14}\left(\dfrac{17}{\cancel{14}}\right) \\ \\ 2(9x + 12) - (15x + 10) & = 17 \\ \\ 2(9x) + 2(12) - 1(15x) - 1(10) & = 17 \\ \\ 18x + 24 - 15x - 10 & = 17 \\ \\ 18x - 15x & = 17 - 24 + 10 \\ \\ 3x & = 3 \\ \\ x & = \dfrac{3}{3} \\ \\ x & = 1 \\ \\ \end{align}$

Rearranging Formulas

Rearrange each formula to isolate for the variable indicated.

$A = \frac{1}{2}b \cdot h$
      (isolate for h) Solution

h =

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$\begin{align} A & = \frac{1}{2}b \cdot h \\ \\ \frac{1}{2}b \cdot h & = A \\ \\ b \cdot h & = 2A \\ \\ h & = \frac{2A}{b} \\ \\ \end{align}$

$P = 2(l + w)$
      (isolate for l)

$A = \pi r^2$
      (isolate for r)

$V = \frac{1}{3}\pi r^2 h$
      (isolate for r)

$a^2 + b^2 = c^2$
      (isolate for b)

$2z = 4(2x - xy)$
      (isolate for x)

$z = \frac{2}{3} y (2w + x)$
      (isolate for x)

$x = \dfrac{nm}{a^2}$
      (isolate for a)

$a = \dfrac{2(b - 5)}{x^2}$
      (isolate for b)

Cross Multiplication with Variables in Denominator (Rational)

Solve. Solution $\dfrac{3x + 2}{x - 2} = \dfrac{1}{3}$

x =

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$\begin{align} \dfrac{3x + 2}{x - 2} & = \dfrac{1}{3} \\ \\ 3(3x + 2) & = 1(x - 2) \\ \\ 3(3x) + 3(2) & = 1(x) + 1(-2) \\ \\ 9x + 6 & = 1x - 2 \\ \\ 9x - 1x & = -2 - 6 \\ \\ 8x & = -8 \\ \\ x & = \dfrac{-8}{8} \\ \\ x & = -1 \\ \\ \end{align}$

Proportions

Part of a baking recipe calls for 600 mL of flour and 5 tablespoons of chocolate chips to make 30 chocolate chip cookies. How much flour and how many tablespoons of chocolate chips would be needed to make 58 cookies? Round your answers to the nearest whole number. Solution

Hint Clear Info

Flour = mL
Chocolate = tablespoons

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Find the amount of flour and chocolate for 1 cookie: $\begin{align} & = \dfrac{600\ mL}{30\ cookies} \\ \\ & = 20\ mL\ flour\ per\ cookie \\ \\ \end{align}$ And ... $\begin{align} & = \dfrac{5\ tablespoons}{30\ cookies} \\ \\ & = 0.166\ tablespoons\ chocolate\ per\ cookie \\ \\ \end{align}$ Use this to calculate the flour and chocolate for 58 cookies by multiplying each: $\begin{align} & = \left(\dfrac{0.166\ tablespoons\ chocolate}{1\ \cancel{cookie}}\right)(58\ \cancel{cookies}) \\ \\ & = 9.66\ tablespoons\ chocolate \\ \\ \end{align}$ Determine the amount of flour: $\begin{align} & = \left(\dfrac{20\ mL\ flour}{1\ \cancel{cookie}}\right)(58\ \cancel{cookies}) \\ \\ & = 1,160\ mL\ flour \\ \\ \end{align}$

Nigel's favourite coffee company says to use 10.0 grams of coffee for every 170 mL of hot water. How much coffee would it take, in grams, to make a big 850 mL jug of coffee?

Organic bananas cost $0.10 per 100 g. If a large bunch of bananas costs $4.00 how much does it weigh?

A cool car can drive 100 km on just 4.0 L of gasoline. If 1.0 L of gasoline costs $1.25 then how much would it cost to drive a distance of 500 km from Montreal to Toronto?

Formulas with Single Degree Variables

Solve the following.

If the area of a soccer field is 8,000m² and the width is 100m, find the length of the field. Solution

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m

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$\begin{align} area & = (length)(width) \\ \\ length & = \dfrac{area}{width} \\ \\ & = \dfrac{8,000m^2}{100m} \\ \\ & = 80\ m \\ \\ \end{align}$

The circumference of a circle is 29.83cm. What is the radius? Solution $C = 2\pi r$

Hint Clear Info

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cm

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$\begin{align} C & = 2\pi r \\ \\ r & = \dfrac{C}{2\pi} \\ \\ & = \dfrac{29.83cm}{2(3.14)} \\ \\ & = 4.75\ cm \\ \\ \end{align}$

Determine the length of a rectangle with a perimeter of 37m and a width of 8.5m. Solution p = 2(length) + 2(width)

Hint Clear Info

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m

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$\begin{align} perimeter & = 2(length) + 2(width) \\ \\ 37m & = 2(length) + 2(8.5m) \\ \\ 37m & = 2(length) + 17m \\ \\ 2(length) & = 37m - 17m \\ \\ 2(length) & = 20m \\ \\ length & = \dfrac{20m}{2} \\ \\ & = 10\ m \\ \\ \end{align}$

If the area of a ginormous triangle is 50m² and the height is 5m, determine the dimension of its base. Solution $a = \dfrac{bh}{2}$

Hint Clear Info

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m

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$\begin{align} a & = \dfrac{bh}{2} \\ \\ 2a & = bh \\ \\ b & = \dfrac{2a}{h} \\ \\ & = \dfrac{2(50m^2)}{5m} \\ \\ & = 20\ m \\ \\ \end{align}$

If a store sells its slow-cooked meat sandwiches for $9.50 each and made $2,156.50 on sandwiches during the whole day, how many sandwiches were sold? Solution

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sandwiches

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$\begin{align} & = \dfrac{$2,156.50}{$9.50} \\ \\ & = 227\ sandwiches \\ \\ \end{align}$

Solving Expressions

Solve the expression below given, y = 2 and x = -2. Solution $2x^3 - 4x^2y^2 - x^3 + 2(3 + x)^2$

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Substitute the 'x' and 'y' values into the expression, $\begin{align} & 2x^3 - 4x^2y^2 - x^3 + 2(3 + x)^2 \\ \\ & = 2(-2)^3 - 4(-2)^2(2)^2 - (-2)^3 + 2(3 + (-2))^2 \\ \\ & = 2(-8) - 4(4)(4) - (-8) + 2(3 - 2)^2 \\ \\ & = 2(-8) - 4(4)(4) + (8) + 2(1)^2 \\ \\ & = -16 - 4(16) + (8) + 2(1) \\ \\ & = -16 - 64 + 8 + 2 \\ \\ & = -16 - 64 + 8 + 2 \\ \\ & = -70 \\ \\ \end{align}$

Squares and Roots

Simplify

$\sqrt{1^2}$ Solution

Hint Clear Info

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A square root cancels out the 'perfect square' number. $\begin{align} & = \sqrt{1^2} \\ \\ & = 1 \\ \\ \end{align}$

$\sqrt{3^2}$ Solution

Hint Clear Info

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$\begin{align} & = \sqrt{3^2} \\ \\ & = 3 \\ \\ \end{align}$

$\sqrt{4}\sqrt{81}$ Solution

Hint Clear Info

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$\begin{align} & = (2)(9) \\ \\ & = 18 \\ \\ \end{align}$

$2\sqrt{5^2}\sqrt{15}$ Solution

Hint Clear Info

$\sqrt{\begin{matrix} \\ \quad \end{matrix}}$

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$\begin{align} & = 2\sqrt{5^2}\sqrt{15} \\ \\ & = 2(5)\sqrt{15} \\ \\ & = 10\sqrt{15} \\ \\ \end{align}$

Squares and Roots: Solve Equations (Guess and Check or Algebra)

Solve for the unknown:

$x = 2\sqrt{3^2}$ Solution

x =

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$\begin{align} x & = 2\sqrt{3^2} \\ \\ x & = 2(3) \\ \\ x & = 6 \\ \\ \end{align}$

$2x = \sqrt{6^2}$ Solution

x =

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$\begin{align} 2x & = 6 \\ \\ x & = 3 \\ \\ \end{align}$

$\sqrt{x^2} = \sqrt{7^2}$ Solution

x =

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$\begin{align} \sqrt{x^2} & = \sqrt{7^2} \\ \\ x & = 7 \\ \\ \end{align}$

$x^2 = 3^2$ Solution

x =

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$\begin{align} \sqrt{x^2} & = \sqrt{3^2} \\ \\ x & = 3 \\ \\ \end{align}$

$\sqrt{x^2} = 2\sqrt{7^2}$ Solution

x =

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$\begin{align} \sqrt{x^2} & = 2\sqrt{7^2} \\ \\ x & = 2(7) \\ \\ x & = 14 \\ \\ \end{align}$

$2\sqrt{x^2} = 2\sqrt{3^2}$ Solution

x =

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$\begin{align} 2\sqrt{x^2} & = 2\sqrt{3^2} \\ \\ 2x & = 2(3) \\ \\ x & = \dfrac{6}{2} \\ \\ x & = 3 \\ \\ \end{align}$

Common Factoring up to Two Degrees

Factor.

10x + 5 Solution

Hint Clear Info

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$\begin{align} & = 10x + 5 \\ \\ & = 5(2x + 1) \\ \\ \end{align}$

25x + 25 Solution

Hint Clear Info

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$\begin{align} & 25x + 25 \\ \\ & = 25(x + 1) \\ \\ \end{align}$

x² + 2x Solution

Hint Clear Info

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$\begin{align} & = x^2 + 2x \\ \\ & = x(x + 2) \\ \\ \end{align}$

2x² - 2x Solution

Hint Clear Info

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$\begin{align} & 2x^2 - 2x \\ \\ & = 2x(x - 1) \\ \\ \end{align}$

8 - 4x - 2x² Solution

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$\begin{align} & 8 - 4x - 2x^2 \\ \\ & = 2(4 - 2x - 2x^2) \\ \\ \end{align}$

-8x³ - 2x² - 2x Solution

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$\begin{align} & = -8x^3 - 2x^2 - 2x \\ \\ & = 2x(-4x^2 - x - 1) \\ \\ \end{align}$

Types of Numbers

Define the following types of numbers: rational, irrational, natural, whole, and integer. Solution

Number Problems

The following equation correctly models the sentence, ten times a number is 25 less than 5 times a number. Solution $10x = 5x - 25$

1. True
2. False

True.

Write an expression for three times more than a number, using 'x' as your variable. Solution

Hint Clear Info

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3x

The correct expression for two consecutive odd integers or two consecutive even integers is always

1. x + 1
2. x
3. x + 2
4. x + 3

An equation for twice a number is equal to four less than three times a number. Solve for x. Show your work. Solution

x =

Hint Clear Info

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$\begin{align} 2x & = 3x - 4 \\ \\ 2x - 3x & = -4 \\ \\ -1x & = -4 \\ \\ 1x & = 4 \\ \\ x & = 4 \\ \\ \end{align}$

Number Problems

Write a let statement defining the variable, make an equation, and solve for the unknown lower number. Show your work.

The sum of two numbers is seventy. One of the numbers is 10 more than 2 times the other number. Solution Video

x =

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Let 'x' represent the first number.
Let (2x + 10) represent the other number. $\begin{align} (x) + (2x + 10) & = 70 \\ \\ x + 2x + 10 & = 70 \\ \\ 3x + 10 & = 70 \\ \\ 3x & = 70 - 10 \\ \\ 3x & = 60 \\ \\ \frac{3x}{3} & = \frac{60}{3} \\ \\ x & = \frac{60}{3} \\ \\ x & = 20 \\ \\ \end{align}$

The sum of two consecutive even integers is 58. Determine the larger of the two numbers. Solution

x =

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Let 'x' represent the first number.
Let (x + 2) represent the other number. $\begin{align} x + (x + 2) & = 58 \\ \\ x + x + 2 & = 58 \\ \\ 2x + 2 & = 58 \\ \\ 2x + 2\ \underline{ - 2} & = 58\ \underline{ - 2} \\ \\ 2x & = 56 \\ \\ \frac{2x}{2} & = \frac{56}{2} \\ \\ x & = \frac{56}{2} \\ \\ & = 28 \\ \\ \\ \\ \\ \\ \text{the other number} & = (x + 2) \\ \\ & = (28 + 2) \\ \\ & = 30 \\ \\ \end{align}$

The sum of three consecutive odd integers is 489. Solution

x =

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Let 'x' represent the first, (x + 2) represent the second, and (x + 4) represent the third number. $\begin{align} x + (x + 2) + (x + 4) & = 489 \\ \\ 3x + 6 & = 489 \\ \\ 3x + 6\ \underline{ - 6} & = 489\ \underline{ - 6} \\ \\ 3x & = 483 \\ \\ \dfrac{3x}{3} & = \dfrac{483}{3} \\ \\ x & = \dfrac{483}{3} \\ \\ & = 161 \\ \\ \text{the second number} & = (x + 2) \\ \\ & = (161 + 2) \\ \\ & = 163 \\ \\ \text{the third number} & = (x + 4) \\ \\ & = (161 + 4) \\ \\ & = 165 \\ \\ \end{align}$

A larger number is 7 more than a smaller number. Three times the larger number plus 4 times the smaller number equals 42. Solution

x =

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Let 'x' be the smaller number and (x + 7) be the larger number. $\begin{align} 3(x + 7) + 4x & = 42 \\ \\ 3x + 3(7) + 4x & = 42 \\ \\ 3x + 21 + 4x & = 42 \\ \\ 3x + 4x + 21\ \underline{ - 21} & = 42\ \underline{ - 21} \\ \\ 7x & = 21 \\ \\ \frac{7x}{7} & = \frac{21}{7} \\ \\ x & = \frac{21}{7} \\ \\ & = 3 \\ \\ \text{the larger number} & = (x + 7) \\ \\ & = (3 + 7) \\ \\ & = 10 \\ \\ \end{align}$

Age Problems

Write a let statement defining the variable, make an equation, and solve for the unknown. Show your work.

Simon's aunt is 6 times his age. How old is Simon if his aunt is 40 years older than him? Solution

Simon:

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years old

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Let 'x' represent Simons age, and (x + 40) his aunt's age. $\begin{align} 6(x) & = x + 40 \\ \\ 6x & = x + 40 \\ \\ 6x\ \underline{ -\ x} & = x\ \underline{ -\ x} + 40 \\ \\ 6x - x & = 40 \\ \\ 5x & = 40 \\ \\ x & = \frac{40}{5} \\ \\ & = 8 \\ \\ \end{align}$ Simon is 8 and his aunt is 48 years old.

Michael is younger than Amanda. Four times Amanda's age plus Michael's age is 84. How old is Amanda if the sum of their ages is 30? Solution

Amanda:

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Let 'x' represent Amanda's age, and (30 - x) Michael's age. $\begin{align} 4(x) + (30 - x) & = 84 \\ \\ 4x + 30 - x & = 84 \\ \\ 4x - x & = 84 - 30 \\ \\ 3x & = 54 \\ \\ x & = \frac{54}{3} \\ \\ & = 18 \\ \\ \end{align}$ Amanda is 18 years old. (Michael is 12).

If Oliver is x years old and Laila is 2 times older than Oliver, determine an expression for both of their ages in 5 years from now.

Age Now	Age in 5 years
x
2x

1. Now = 5x, Later = 10x
2. Now = 6x, Later = 7x
3. Now = x - 5, Later = 2x - 5
4. Now = x + 5, Later = 2x + 5

Money Problems

Determine the correct equation to solve the statement below. The number of ten dollar bills a bank teller has is 8 less than 3 times the number of twenty dollar bills.
If there are 68 bills in total, determine the quantity of twenty dollar bills.

1. 3(x - 8) + 3x = 68
2. 3x + (x + 8) = 68
3. (3x - 8) + x = 68
4. (x - 8) + 3x = 68

Write a let statement defining the variable, make an equation, and solve for the unknown. Show your work. Solution The number of twenty dollar bills a bank teller has is 4 less than 3 times the number of fifty dollar bills.
If there are 44 bills in total, determine the quantity of fifty dollar bills.

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fifty dollar bills

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Let 'x' represent the number of 50 dollar bills. $\begin{align} \text{Number of twenty dollar bills} + \text{Number of fifty dollar bills} & = \text{Total number of bills} \\ \\ (3x - 4) + x & = 44 \\ \\ 3x - 4 + x & = 44 \\ \\ 3x + x - 4 & = 44 \\ \\ 4x & = 44 + 4 \\ \\ 4x & = 48 \\ \\ x & = \frac{48}{4} \\ \\ & = 12 \\ \\ \end{align}$ Therefore there are 12 twenty dollar bills.

Determine the correct equation to solve the statement below. A bank has 88 bills consisting only of five and ten dollar bills.
Determine the number of five dollar bills if the total amount of money is $550.00

1. $5(x) + $10(88 - x) = $550
2. $5(x) + $10(x + 88) = $550
3. $5(x) + $10(88x) = $550
4. $5(x) + $10(x - 88) = $550

Money Problems

Write a let statement defining the variable, make an equation, and solve for the unknown. Show your work.

A person has 12 bills in their wallet consisting only of five and twenty dollar bills. Determine the number of five dollar bills if the total amount of money is $105.00 Solution

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five dollar bills

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Let 'x' represent the number of five dollar bills. Let (12 - x) represent the number of twenty dollar bills. $\begin{align} (\$5)(\text{Number of five dollar bills}) + (\$20)(\text{Number of twenty dollar bills}) & = \text{Total amount of money} \\ \\ (\$5)(x) + (\$20)(12 - x) & = \$105 \\ \\ \$5(x) + \$20(12) - \$20(x) & = \$105 \\ \\ 5(x) + 20(12) - 20(x) & = 105 \\ \\ 5x + 240 - 20x & = 105 \\ \\ 5x - 20x & = 105 - 240 \\ \\ -15x & = -135 \\ \\ 15x & = 135 \\ \\ x & = \frac{135}{15} \\ \\ & = 9 \\ \\ \end{align}$ Therefore there are 9 five dollar bills.

A store sells 'Super Awesome' phones for $500 and 'Pretty Good' phones for $100. If the store sells 95 phones and makes $15,500 in a day, determine the number of Pretty Good phones sold. Solution

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phones

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Let 'x' represent the number of Pretty Good phones. Let (95 - x) represent the number of Super Awesome phones. $\begin{align} (\$100)(\text{Pretty Good phones}) + (\$500)(\text{Super Awesome phones}) & = \text{Total money} \\ \\ (\$100)(x) + (\$500)(95 - x) & = \$15,500 \\ \\ \$100(x) + \$500(95) - \$500(x) & = \$15,500 \\ \\ 100(x) + 500(95) - 500(x) & = 15,500 \\ \\ 100x + 47,500 - 500x & = 15,500 \\ \\ 100x - 500x & = 15,500 - 47,500 \\ \\ -400x & = -32,000 \\ \\ 400x & = 32,000 \\ \\ x & = \frac{32,000}{400} \\ \\ & = 80 \\ \\ \end{align}$ Therefore 80 Pretty Good phones were sold.

Lise has $14 less than 5 times as much money as Sarah. If Lise gives Sarah $15, then Lise will have three times as much money as Sarah. Determine how much money Sarah has in the end. Solution

$

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Let 'x' represent Sarah's money and (5x - 14) Lise's money.
After giving $15, Lise has (5x - 14 - 15) and Sarah has (x + 15). $\begin{align} 5x - 14 - 15 & = 3(x + 15) \\ \\ 5x - 29 & = 3(x) + 3(15) \\ \\ 5x - 29 & = 3x + 45 \\ \\ 5x - 3x & = 45 + 29 \\ \\ 2x & = 74 \\ \\ x & = \frac{74}{2} \\ \\ & = 37 \\ \\ \\ \\ \\ \\ x + 15 & = 37 + 15 = 52 \\ \\ \end{align}$ Sarah has $52 after Lise gives her money.

Geometry Problems

A triangle has side lengths of 2x + 4, 3x - 2, and x - 10. If the total perimeter is 5x + 4, write and equation to determine the lengths of the sides of the triangle.

1. (2x + 4) + (3x - 2) = (5x + 4) + (x - 10)
2. (2x + 4) + (3x - 2) + (x - 10) = 5x + 4
3. (2x + 4)(3x - 2)(x - 10) = 5x + 4
4. $\dfrac{(2x + 4)(3x - 2)(x - 10)}{2} = 5x + 4$

Determine the lengths of the sides of the triangle based on the statement below. Solution A triangle has side lengths of x - 4, 2x - 8, and 3x - 4.
The total perimeter is x + 14.

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$\begin{align} (x - 4) + (2x - 8) + (3x - 4) & = x + 14 \\ \\ x - 4 + 2x - 8 + 3x - 4 & = x + 14 \\ \\ x + 2x + 3x - 4 - 8 - 4 & = 14 \\ \\ 6x - 16 & = 14 \\ \\ 6x & = 14 + 16 \\ \\ 6x & = 30 \\ \\ x & = \frac{30}{6} \\ \\ & = 5 \\ \\ \\ \\ \text{Side 1} & = x - 4 = 5 - 4 = \underline{1} \\ \\ \text{Side 2} & = 2x - 8 = 2(5) - 8 = \underline{2} \\ \\ \text{Side 3} & = 3x - 4 = 3(5) - 4 = \underline{11} \\ \\ \end{align}$

The length of a rectangle is 1 less than twice the width. If the perimeter of the rectangle is 70, find the length of the sides. Write a let statement defining the variable, make an equation, and solve for the unknown. Show your work. Solution

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Length: Width:

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Let 'x' represent the width. Let (2x - 1) represent the length. $\begin{align} 2(Length) + 2(Width) & = Perimeter \\ \\ 2(2x - 1) + 2(x) & = 70 \\ \\ 2(2x) - 2(1) + 2x & = 70 \\ \\ 4x - 2 + 2x & = 70 \\ \\ 6x - 2 & = 70 \\ \\ 6x & = 70 + 2 \\ \\ 6x & = 72 \\ \\ x & = \frac{72}{6} \\ \\ & = 12 \\ \\ \\ \\ Length & = (2x - 1) = 2(12) - 1 = 24 - 1 = 23 \\ \\ \end{align}$ The length is 23 and the width is 12.

The area of a circle is 4 less than twice the area of a square. The area of the circle is 2 cm² more than the area of the triangle. If the total area of the shapes combined is 30 cm², determine the area of the square. Write a let statement defining the variable, make an equation, and solve for the unknown. Show your work. Solution

area =

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cm²

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Let 'x' represent the area of the square. Let (2x - 4) represent the area of the circle.
Let (2x - 4- 2) represent the area of the triangle. $\begin{align} x + (2x - 4) + (2x - 4 - 2) & = 30 \\ \\ x + 2x - 4 + 2x - 6 & = 30 \\ \\ x + 2x + 2x - 10 & = 30 \\ \\ 5x - 10 & = 30 \\ \\ 5x & = 30 + 10 \\ \\ 5x & = 40 \\ \\ x & = \frac{40}{5} \\ \\ & = 8 \\ \\ \end{align}$ ∴ the area of the square is 8cm²

Distance, Speed, Time Problems

Write a let statement defining the variable, make an equation, and solve for the unknown. Show your work.

The high speed trains in Japan have an average speed of 240 km/hr and the high speed trains in North America have an average speed of 140 km/hr. If the train in North America leaves the station 1.0 hour before the train in Japan leaves its station, determine the time it takes the Japanese train to catch up by travelling the same distance. Solution

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hours

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Let 'x' represent the time the Japanese train takes.
Let x + 1.0 represent the time the North American train takes. $\begin{align} \text{Distance of Japanese train} & = \text{Distance of North American train} \\ \\ Speed_{Japan}\ ×\ Time & = Speed_{North\ America}\ ×\ Time \\ \\ 240 × (x) & = 140 × (x + 1) \\ \\ 240x & = 140(x) + 140(1) \\ \\ 240x & = 140x + 140 \\ \\ 240x - 140x & = 140 \\ \\ 100x & = 140 \\ \\ x & = \frac{140}{100} \\ \\ & = 1.4\ hours \\ \\ \end{align}$

Town A is 100km apart from town B. If a car leaves town A travelling at 100km/hr at the same time that a bicycle leaves town B travelling at 15 km/hr, determine the distance travelled by the bicycle when they meet. Solution

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km

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Let 'x' represent the distance travelled by the bicycle. Let (100 - x) represent the distance travelled by the car. $\begin{align} \text{Time of car} & = \text{Time of bicycle} \\ \\ \frac{Distance}{Speed} & = \frac{Distance}{Speed} \\ \\ \frac{100 - x}{100} & = \frac{x}{15} \\ \\ 15(100 - x) & = 100(x) \\ \\ 15(100) - 15(x) & = 100x \\ \\ 1500 - 15x & = 100x \\ \\ 1500 & = 100x + 15x \\ \\ 1500 & = 115x \\ \\ 115x & = 1500 \\ \\ x & = \frac{1500}{115} \\ \\ & = 13.04\ km \\ \\ \end{align}$

Josh is competing in an NYC race in which he can bike 10km/hr, and run 20km/hr. If each event is the same distance, and the total time is 1.5 hours, find the time it takes Josh to run. Solution

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hour

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Let 'x' represent the time to run, and let (1.5 - x) represent the time to bike. $\begin{align} \text{Run Distance} & = \text{Bike Distance} \\ \\ Speed_{Run}\ ×\ Time_{Run} & = Speed_{Bike}\ ×\ Time_{Bike} \\ \\ (10) × (x) & = (20) × (1.5 - x) \\ \\ 10x & = 20(1.5) - 20(x) \\ \\ 10x & = 30 - 20x \\ \\ 10x + 20x & = 30 \\ \\ 30x & = 30 \\ \\ x & = \frac{30}{30} \\ \\ & = 1 \\ \\ \end{align}$ ∴ Josh runs the event in 1.0 hours.

Direct and Partial Variation

Which of the following is consistent with a direct variation relation? Solution

1. y-intercept at (0, 3)
2. equation y = 3x - 2
3. a mechanic charges an hourly rate plus a constant, fixed cost.
4. point at (0, 0)

A direct variation is in the form y = mx. There is no fixed cost.

A partial variation is in the form y = mx + b.

The direct variation always crosses through the y-axis at the origin (0, 0).

Which of the following equations is not a partial variation? Solution

1. y = x + 1
2. y = 2x - 2
3. y = -¾x + 2
4. y = 3x

A direct variation is in the form y = mx.

A partial variation is in the form y = mx + b.

Determining Rates of Different Scenarios

Determine the rate for each of the following scenarios

A car can travel 550km on a 40L tank of gas. Find the rate of fuel consumption per Liter of gas. Solution

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km/L

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$\begin{align} Rate & = \frac{\text{Change in dependent variable}}{\text{Change in independent variable}} \\ \\ & = \frac{550\ km}{40\ L} \\ \\ & = 13.75 \tfrac{km}{L} \\ \\ \end{align}$

A mechanical orange picker can collect 2000 oranges in 4 minutes. Find the rate of collection in seconds. Solution

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oranges/second

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$\begin{align} & = \frac{2000\ oranges}{240\ seconds} \\ \\ & = 8.3 \tfrac{oranges}{second} \\ \\ \end{align}$

A painter can paint 336m² of a building in 48 minutes. Find the rate of painting, in minutes. Solution

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m²/min

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$\begin{align} & = \frac{336\ m^2}{48\ min} \\ \\ & = 7 \tfrac{m^2}{min} \\ \\ \end{align}$

A dog can eat 58 kernels of spilled popcorn in 26.5 seconds. Find the rate of consumption. Solution

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kernels/sec

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$\begin{align} & = \frac{58}{26.5} \\ \\ & = 2.2 \tfrac{kernels}{second} \\ \\ \end{align}$

Calculating Slope

Given the equations...

Calculate the slope between the points P₁(3, -2) and P₂(4, 8). Solution

m =

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Givens:
x₁ = 3
y₁ = -2
x₂ = 4
y₂ = 8 $\begin{align} m & = \frac{y_2\ -\ y_1}{x_2\ -\ x_1} \\ \\ & = \frac{8 - (-2)}{4 - 3} \\ \\ & = \frac{8 + 2}{4 - 3} \\ \\ & = \frac{10}{1} \\ \\ & = 10 \\ \\ \end{align}$ Note: be careful when subtracting negative 2, it becomes positive!

Calculate the slope between the points P₁(-4, 5) and P₂(2, -1). Solution

m =

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Givens:
x₁ = -4
y₁ = 5
x₂ = 2
y₂ = -1 $\begin{align} Slope = m & = \frac{\Delta y}{\Delta x} \\ \\ m & = \frac{y_2\ -\ y_1}{x_2\ -\ x_1} \\ \\ & = \frac{-1 - 5}{2 - (-4)} \\ \\ & = \frac{-6}{6} \\ \\ & = -1 \\ \\ \end{align}$ Note: be careful when subtracting negative 4, it becomes positive!

Rank the following slopes from lowest to highest. Solution

Slope	Rise (cm)	Run (cm)
I	3	4
II	4	6
III	5	7
IV	2	8

1. I, II, III, IV
2. III, I, IV, II
3. IV, II, III, I
4. I, III, II, IV

$Slope = m = \frac{\Delta y}{\Delta x} = \frac{y_2\ -\ y_1}{x_2\ -\ x_1}$ Calculate the slope for each and order from lowest to highest.

Calculate the slope when the change in x is -6 with a first difference of 3. Solution

The first difference is the $\Delta y$.
$\Delta y = 3$
$\Delta x = -6$ $\begin{align} Slope = m & = \frac{\Delta y}{\Delta x} \\ \\ & = \frac{3}{-6} \\ \\ & = \frac{1}{-2} \\ \\ & = -\frac{1}{2} \\ \\ \end{align}$

A ladder is leaning up against a wall 4m above the ground. If the bottom of the ladder is 2m from the base of the wall, determine the slope of the ladder. Solution

m =

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$\begin{align} Slope = m & = \frac{\Delta y}{\Delta x} \\ \\ & = \frac{4}{2} \\ \\ & = 2 \\ \\ \end{align}$

Rate of Change

A cyclist records their distance travelled as they ride their bicycle.

Determine the distance travelled from 1 to 4 hours. Solution

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km

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We don't add all the distances. The cyclist starts at 10km and goes to 40km. The total distance travelled between 10km and 40km is: $\begin{align} \text{Distance travelled} & = \text{Change in y} \\ \\ & = \Delta y \\ \\ & = y_2 - y_1 \\ \\ & = 40 - 10 \\ \\ & = 30\ km \\ \\ \end{align}$ $\Delta$ means "change".

Calculate the rate of change (slope) of the distance over time from 1 to 4 hours. Solution

m =

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km/hr

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Givens:
x₁ = 1
y₁ = 10
x₂ = 4
y₂ = 40 $\begin{align} Slope & = \frac{\Delta y}{\Delta x} \\ \\ & = \frac{y_2 - y_1}{x_2 - x_1} \\ \\ & = \frac{40 - 10}{4 - 1} \\ \\ & = \frac{30\ km}{3\ hr} \\ \\ & = 10 \tfrac{km}{hr} \\ \\ \end{align}$

Writing Linear Equations

A line has a slope of 2 and a y-intercept of 3. The equation of a line in slope-y-intercept form is given below.

Determine a coordinate on this line. Solution

1. (0, 3)
2. (1, 6)
3. (2, 8)
4. (3, 8)

The y-intercept given, and the coordinate is (0, 3)

Remember that the y-intercept always occurs where x = 0

Write an equation for this line. Solution

y =

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y = mx + b

m is slope, b is y-intercept. y = 2x + 3

Writing Linear Equations

Write an equation for a line with a slope of $-\dfrac{1}{5}$ that passes through the point (4, 2). Show your work and reduce fully. Solution Video

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$-\,\dfrac{}{\quad\quad}x \ + \ \dfrac{}{\quad\quad}$

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Givens:
m = $-\dfrac{1}{5}$
x = 4
y = 2


Need to calculate b! $\begin{align} y & = mx + b \\ \\ y & = -\dfrac{1}{5}x + b \\ \\ 2 & = -\dfrac{1}{5}(4) + b \\ \\ 2 & = -\dfrac{4}{5} + b \\ \\ b & = 2 + \dfrac{4}{5} \\ \\ b & = \dfrac{10}{5} + \dfrac{4}{5} \\ \\ b & = \dfrac{14}{5} \\ \\ \\ \\ & \underline{y = -\dfrac{1}{5}x + \dfrac{14}{5}} \\ \\ \end{align}$

Write an equation for a line with a slope of $\dfrac{3}{2}$ that passes through the point (-1, -3). Show your work. Solution

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$\dfrac{}{\quad\quad}x \ - \ \dfrac{}{\quad\quad}$

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Need to calculate b! $\begin{align} y & = \dfrac{3}{2}x + b \\ \\ -3 & = \dfrac{3}{2}(-1) + b \\ \\ -3 & = \dfrac{-3}{2} + b \\ \\ b & = -3 - \dfrac{-3}{2} \\ \\ b & = -3 + \dfrac{3}{2} \\ \\ b & = -\dfrac{6}{2} + \dfrac{3}{2} \\ \\ b & = -\dfrac{3}{2} \\ \\ \\ \\ \\ \\ y & = \dfrac{3}{2}x - \dfrac{3}{2} \\ \\ \end{align}$

Ax + By + C = 0 and y = mx + b

The following equations are the same, where the first one is in standard form, and the second one is in slope y-intercept form. Solution y = ¼x - 3
-x + 4y + 12 = 0

1. True
2. False

See how to convert from slope y-intercept into standard form: $\begin{align} y & = \dfrac{1}{4}x - 3 \\ \\ (4)y & = (\cancel{4})\dfrac{1}{\cancel{4}}x - (4)3 \\ \\ 4y & = 1x - 12 \\ \\ 4y - x + 12 & = 0 \\ \\ - x + 4y + 12 & = 0 \\ \\ \end{align}$

Ax + By + C = 0 and y = mx + b

For the following points M(2, -2) and N(5, 7)

Calculate the slope (m). Solution

m =

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Determine the different x and y values, and plug into the equation for slope (m): $\begin{align} m & = \dfrac{y_2 - y_1}{x_2 - x_1} \\ \\ & = \dfrac{7 - (-2)}{5 - 2} \\ \\ & = \dfrac{9}{3} \\ \\ & = 3 \\ \\ \end{align}$

Calculate the y-intercept (b). Solution

b =

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Use the slope calculate from before (m = 3), and make an equation like this: $\begin{align} y & = mx + b \\ \\ y & = 3x + b \\ \\ \end{align}$ Then choose either of the points M(2, -2) or N(5, 7) to plug into the equation to find 'b': $\begin{align} y & = mx + b \\ \\ y & = 3x + b \\ \\ (7) & = 3(5) + b \\ \\ 7 & = 15 + b \\ \\ 7 - 15 & = b \\ \\ b & = -8 \\ \\ \end{align}$

Determine the equation. Solution

y =

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Put the slope (m) and y-intercept (b) from before into an equation: $\begin{align} y & = (m)x + (b) \\ \\ y & = 3x - 8 \\ \\ \end{align}$

Ax + By + C = 0 and y = mx + b

Rearrange the equation in standard form below, into slope y-intercept form: y = mx + b. Show your work. Solution $3x - 2y + 1 = 0$

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$\dfrac{}{\quad\quad}x \ + \ \dfrac{}{\quad\quad}$

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$\begin{align} 3x - 2y + 1 & = 0 \\ \\ - 2y + 1 & = -3x \\ \\ - 2y & = -3x - 1\\ \\ 2y & = 3x + 1\\ \\ y & = \dfrac{3}{2}x + \dfrac{1}{2}\\ \\ \end{align}$

Relations

According to the following table, the average score on a test depends on the age of the student. Solution

Age	Score
Grade 5	76%
Grade 6	77%
Grade 7	75%
Grade 8	79%
Grade 9	76%

1. True
2. False

There is no correlation (neither positive, nor negative). To determine this, see that the change in 'y' (score), for a given change in 'x' is inconsistent.

Relationship Between Two Variables

Complete the following table relating the side length of a square to its area. Solution

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Side Length (cm)	Area (cm2)
2	4
5	25
7
10
	144

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Side Length (cm)	Area (cm2)
2	4
5	25
7	49
10	100
12	144

The y-intercept (b)

A linear relation always crosses the y-axis when at x = 0.

Determine the y-intercept for the relation below. Solution Video $2y = -\dfrac{4}{9}x - 8$

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The y-intercept occurs at x = 0. $\begin{align} 2y & = -\dfrac{4}{9}x - 8 \\ \\ 2y & = -\dfrac{4}{9}(0) - 8 \\ \\ 2y & = \cancel{-\dfrac{4}{9}(0)} - 8 \\ \\ 2y & = - 8 \\ \\ y & = \frac{-8}{2} \\ \\ y & = -4 \\ \\ \end{align}$ ∴ the y-intercept occurs at the point (0, -4)

Determine the y-intercept in the equation: 5x - 6y = -1 Solution

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$\bigg( \quad , \dfrac{\quad}{\quad} \bigg)$

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Isolate for y.
Determine b. $\begin{align} 5x - 6y & = -1 \\ \\ - 6y & = -1 - 5x \quad\quad\quad \text{multiply each by -1 to change signs to positive} \\ \\ 6y & = 5x + 1 \\ \\ \frac{6}{6}y & = \frac{5}{6}x + \frac{1}{6} \\ \\ y & = \frac{5}{6}x + \frac{1}{6} \\ \\ \end{align}$ Therefore the y-intercept is $\left( 0,\frac{1}{6}\right)$ You can also find the y-intercept by setting x = 0 and solving for y.

Linear Relations: x-intercepts

Which of the following is the correct algebraic method to determine the x-intercept of the relation, y = -3x + 2 Solution

1. (0) = -3x + 2
2. y = -3(0) + 2
3. (0) = -3(0) + 2
4. None of the above

The x-intercept occurs when y = 0. Substitute y = 0 into the equation and solve for x...

Determine the x-intercept of the following function: y = 5 - x Solution

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Solving for the x-intercept (root) and y-intercept ('b')

Given the function y = x - 1, find:

The y-intercept Solution

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The x-intercept Solution

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Deeper Understanding of Slope

Which of the following is an equation of a vertical line? Solution

1. y = x + 1
2. y = 2
3. x = 2
4. x = y + 1

A vertical line crosses the x-axis and has no y-values.

Which of the following is an equation of a horizontal line? Solution

1. x = y
2. y = 1
3. x = 1
4. y = x + 1

Another way to think about it is a line with a slope (M) of 0, and the y-intercept (b) is the whole line.

$\begin{align} y & = mx + b \\ \\ y & = 0x + b \\ \\ y & = \cancel{0x} + b \\ \\ y & = b \\ \\ \\ \\ & ... \\ \\ \text{E.g.)} \quad\quad\quad\quad y & = 1 \\ \\ \end{align}$

Which equation would overlap the x-axis? Solution

1. y = 4
2. y = x
3. x = 1
4. y = 0

The line y = 0 overlaps the x-axis.

Which of the following lines has the steepest slope? Solution

1. y = x + 3
2. y = 2x + 3
3. y = 5x + 3
4. y = 8x + 3

The highest magnitude of |m| = 8...

Which of the following lines has the steepest slope? Solution

1. y = -5x + 1
2. y = -4x + 1
3. y = -2x + 1
4. y = -x + 1

The highest magnitude of |m| = -5...

Rearrange the following equations and determine which has the most shallow (lowest), positive slope. Solution

1. y - 2 = -3x
2. 1 = y - 4x
3. 2y + 6 = x
4. -2x + y + 2 = 0

y = mx + b

A shallow slope has a low magnitude of m.

A positive slope has a positive m. $\begin{align} 2y + 6 & = x \\ \\ 2y & = x - 6 \\ \\ \frac{2}{2}y & = \frac{1}{2}x - \frac{6}{2} \\ \\ \cancel{\frac{2}{2}}y & = \frac{1}{2}x - \frac{\cancel{6}3}{\cancel{2}1} \\ \\ \frac{1}{1}y & = \frac{1}{2}x - \frac{3}{1} \\ \\ y & = \frac{1}{2}x - 3 \\ \\ \end{align}$

Deeper Understanding of Slope

A line has a slope of m = 2 and goes through the point (2, 1).

Determine the equation of the line. Solution

Hint Clear Info

$x \ -$

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} y & = mx + b \\ \\ y & = 2x + b \\ \\ 1 & = 2(2) + b \\ \\ 1 & = 4 + b \\ \\ 1 - 4 & = b \\ \\ -3 & = b \\ \\ b & = -3 \\ \\ \\ \\ \\ \\ y & = 2x - 3 \\ \\ \end{align}$

Write the equation of a line with the same slope, that has a y-intercept 5 points higher than in part 'a'. Solution

Hint Clear Info

$x \ +$

Incorrect Attempts:

CHECK

Hint Unavailable

If the original y-intercept was -3. $\begin{align} y & = 2x - 3 \\ \\ \end{align}$ A y-intercept 5 points higher = -3 + 5, would be cross at +2. $\begin{align} y & = 2x + 2 \\ \\ \end{align}$

Parallel and Perpendicular Lines

The following three lines are parallel. Solution $\begin{align} y & = 6x + 6 \\ \\ y & = 6x + 0 \\ \\ y & = 6x + -6 \end{align}$

1. True
2. False

Lines are parallel when they have the same slope (m), regardless of the y-intercept (b). y = mx + b

Which of the following lines is parallel to: Solution $y = \dfrac{2}{3}x - 4$

1. $y = \dfrac{1}{3}x + 4$
2. $y = 2x - \dfrac{1}{4}$
3. $y = \dfrac{3}{2}x$
4. $y = \dfrac{2}{3}x + 1$

Parallel lines have the same slope, $m = \dfrac{2}{3}$

Note that the perpendicular ┴ slope is the negative reciprocal, $m_{┴} = -\dfrac{3}{2}$