Topics
2. Solving Equations
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United States of America
Grades 9-12

2.01 Linear and absolute value equations and inequalities

Lesson
Practice

Introduction

We solved linear equations and inequalities as well as absolute value equations and inequalities in Algebra 1, and here we will review the concept of using inverse operations to isolate the variable and find solutions.

Solving linear equations and inequalities

There is more than one way to solve an equation or inequality, but it's important to remember that the goal of finding the solution to an equation or inequality is isolating the variable on one side of the equation using inverse operations.

A linear equation in one variable can have one, none, or an infinite number of solutions, as shown:

  • x=a, where a is a number (a unique solution)

  • a=a, where a is a number (infinitely many solutions)

  • a=b, where a and b are different numbers (no solutions)

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The solution to an equation may be graphed on a number line.

For example, consider the equation x+2=6

The solution, x=4, may be graphed as shown on the number line.

A linear inequality has an infinite number of solutions, with its solution written in inequality notation or interval notation.

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For example, consider the inequality x + 2 > 6

The solution set may be graphed on a number line.

Inequality notation: x>4

Interval notation: (4, \infty)

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For example, consider the inequality x + 2 \geq 6

The solution set may be graphed on a number line.

Inequality notation: x \geq 4

Interval notation: [4, \infty)

When solving an absolute value equation or inequality, isolate the absolute value expression, then create two equations or inequalities and solve them separately.

Examples

Example 1

Solve the following equation. Round your solution to the nearest tenth. 4 + \dfrac{2}{3}\left(9a - \dfrac{3}{2}\right) = 6\left(3 - \dfrac{2}{3}a\right)

Worked Solution
Apply the idea
\displaystyle 4 + \dfrac{2}{3}\left(9a - \dfrac{3}{2}\right)\displaystyle =\displaystyle 6\left(3 - \dfrac{2}{3}a\right)
\displaystyle 4 + 6a - 1\displaystyle =\displaystyle 18 - 4aDistributive property
\displaystyle 3 + 6a\displaystyle =\displaystyle 18 - 4aCombine like terms
\displaystyle 3 + 10a\displaystyle =\displaystyle 18Add 4a to both sides
\displaystyle 10a\displaystyle =\displaystyle 15Subract 3 from both sides
\displaystyle a\displaystyle =\displaystyle 1.5Divide by 10 on both sides
Reflect and check

Check a=1.5 by substituting it into the original equation:

\displaystyle 4 + \dfrac{2}{3}\left(9a - \dfrac{3}{2}\right)\displaystyle =\displaystyle 6\left(3 - \dfrac{2}{3}a\right)
\displaystyle 4 + \dfrac{2}{3}\left(9(1.5) - \dfrac{3}{2}\right)\displaystyle =\displaystyle 6\left(3 - \dfrac{2}{3}(1.5)\right)
\displaystyle 4 + \dfrac{2}{3}\left(12\right)\displaystyle =\displaystyle 6\left(2\right)
\displaystyle 4 + 8\displaystyle =\displaystyle 12

Since the equation is a true statement, we can confirm that a=1.5 is the solution.

Example 2

Solve the following inequality. Round your solution to the nearest tenth. -4.5x + 2.6x -\dfrac{0.4(x +1)}{3} \leq 0.7-(3.4x+10)

Worked Solution
Apply the idea
\displaystyle -4.5x + 2.6x -\dfrac{0.4(x +1)}{3}\displaystyle \leq\displaystyle 0.7-(3.4x+10)Original inequality
\displaystyle -1.9x -\dfrac{0.4(x +1)}{3}\displaystyle \leq\displaystyle 0.7-(3.4x+10)Combine like terms
\displaystyle -1.9x -\dfrac{0.4(x+1)}{3}\displaystyle \leq\displaystyle 0.7-3.4x-10Distributive property
\displaystyle -1.9x -\dfrac{0.4(x+1)}{3}\displaystyle \leq\displaystyle -3.4x-9.3Combine like terms
\displaystyle -\dfrac{0.4(x+1)}{3}\displaystyle \leq\displaystyle -1.5x-9.3Add 1.9x to both sides
\displaystyle 0.4(x+1)\displaystyle \geq\displaystyle 4.5x+27.9Multiply by -3 on both sides
\displaystyle 0.4x+0.4\displaystyle \geq\displaystyle 4.5x+27.9Distributive property
\displaystyle 0.4x\displaystyle \geq\displaystyle 4.5x+27.5Subtract 0.4 from both sides
\displaystyle -4.1x\displaystyle \geq\displaystyle 27.5Subtract 4.5x from both sides
\displaystyle x\displaystyle \leq\displaystyle -6.707317Divide by -4.1 on both sides

The solution to the inequality is approximately x \leq -6.7

Reflect and check

The solution to the inequality is all values of x that are less than or equal to -6.7. The solution on a number line is

-10-9-8-7-6-5-4-3-2-1012345

Example 3

Two buses leave Kansas City, Oklahoma at 1 o'clock in the afternoon. One bus is traveling east and averaging a driving speed of 35 miles per hour while the other was driving west and averaging a driving speed of 60 miles per hour. Write and solve an equation to determine what time the buses will be 300 miles apart.

Worked Solution
Create a strategy

The unknown in this problem is time. Since the information provided is given around miles per hour, we can state that the unknown variable is x, hours.

Write an equation that equals 300, the total distance between the two buses. Since the units of 300 is miles, we know that the units of the terms in the equation should have miles as the units.

One bus is driving 35 miles per hour, so the expression 35x represents the total miles the bus drives in x hours.

The other bus is driving 60 miles per hour, so the expression 60x represents the total miles the bus drives in x hours.

Apply the idea

The equation that represents the amount of time the buses will be driving in order to be 300 miles apart is 35x+60x=300.

\displaystyle 35x+60x\displaystyle =\displaystyle 300Model equation
\displaystyle 95x\displaystyle =\displaystyle 300Combine like terms
\displaystyle x\displaystyle =\displaystyle 3.15789Divide by 95 on both sides

In 3.2 hours, the buses will be 300 miles apart. To determine the time of day that this occurs, we should convert 0.2 hours into minutes by multiplying 0.2 \text{ hours} \times \dfrac{60 \text{ minutes}}{1 \text{ hour}} = 12 \text{ minutes}.

3 hours and 12 minutes after 1 o'clock in the afternoon is 4:12PM.

Reflect and check

We can determine if the solution to the problem makes sense by calculating how far each bus actually traveled in 3.2 hours.

The bus driving 35 miles per hour would have driven \dfrac{35 \text{ miles}}{1 \text{ hour}} \times 3.2 \text{ hours} = 112 \text{ miles.}

The bus driving 60 miles per hour would have driven \dfrac{60 \text{ miles}}{1 \text{ hour}} \times 3.2 \text{ hours} = 192 \text{ miles}.

In total, this is 304 miles.

Idea summary

Solve equations and inequalities by performing inverse operations. We can always check the solution to an equation or inequality by substituting a solution into the problem and evaluating to determine if it is mathematically true or if it makes sense in context.

Solving absolute value equations and inequalities

Recall that absolute value equations and inequalities are solved using similar properties as equations and inequalities. The absolute value of a number is the number's distance from zero, leading to different types of solutions.

An absolute value equation in one variable can have one, two, or no solutions.

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The solution to an absolute value equation may be graphed on a number line.

For example, consider the equation |x|=4

The solutions, x=-4 and x=4, may be graphed as shown on the number line.

An absolute value inequality in one variable may have many solutions. The solution set of an absolute value inequality may be written as a compound inequality or in interval notation.

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For example, consider the inequality |x| > 4

The solution set may be graphed on a number line.

Inequality notation: x<-4 or x>4

Interval notation: (-\infty, -4) \cup (4, \infty)

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For example, consider the inequality |x| \leq 4

The solution set may be graphed on a number line.

Inequality notation: -4 \leq x \leq 4

Interval notation: [-4,4]

Examples

Example 4

Solve the equation. Round your solution to the nearest tenth. \dfrac{1}{2}-2|\dfrac{3}{5}x+4|=-6

Worked Solution
Create a strategy

Isolate the absolute value expression, then create two equations and solve them separately.

Apply the idea
\displaystyle \dfrac{1}{2}-2|\dfrac{3}{5}x+4|\displaystyle =\displaystyle -6Given equation
\displaystyle -2|\dfrac{3}{5}x+4|\displaystyle =\displaystyle -6.5Subtract \dfrac{1}{2} from both sides
\displaystyle |\dfrac{3}{5}x+4|\displaystyle =\displaystyle 3.25Divide by -2 on both sides

The equations to solve are \dfrac{3}{5}x+4=3.25 and \dfrac{3}{5}x+4=-3.25

First,

\displaystyle \dfrac{3}{5}x+4\displaystyle =\displaystyle 3.25Equation with positive 3.25
\displaystyle \dfrac{3}{5}x\displaystyle =\displaystyle -0.75Subtract 4 from both sides
\displaystyle x\displaystyle =\displaystyle -1.25Multiply by \dfrac{5}{3} on both sides

Next,

\displaystyle \dfrac{3}{5}x+4\displaystyle =\displaystyle -3.25Equation with negative 3.25
\displaystyle \dfrac{3}{5}x\displaystyle =\displaystyle -7.25Subtract 4 from both sides
\displaystyle x\displaystyle =\displaystyle -12.0833Multiply by \dfrac{5}{3} on both sides

The solutions to the absolute value equations are approximately -1.3 and -12.1.

Reflect and check

If the absolute value expression was equal to a negative value, we would not move forward with solving the equation because the equation would have no solutions. The absolute value of an expression can never be equal to a negative value.

Example 5

Solve the inequality. Round your solution to the nearest tenth. \dfrac{3.5s + |s - 4.25|}{2} \leq 1.75s + 8.85

Worked Solution
Apply the idea
\displaystyle \dfrac{3.5s + |s - 4.25|}{2}\displaystyle \leq\displaystyle 1.75s + 8.85Given inequality
\displaystyle 3.5s + |s - 4.25|\displaystyle \leq\displaystyle 3.5s + 17.7Multiply by 2 on both sides
\displaystyle |s - 4.25|\displaystyle \leq\displaystyle 17.7Subtract 3.5s on both sides

Since |s-4.25| is less than or equal to 17.7, we can write the absolute value inequality as the compound inequality s-4.25 \leq 17.7 or s-4.25 \geq -17.7 and solve.

First,

\displaystyle s - 4.25\displaystyle \leq\displaystyle 17.7Inequality with positive 17.7
\displaystyle s\displaystyle \leq\displaystyle 21.95Add 4.25 on both sides

Next,

\displaystyle s - 4.25\displaystyle \geq\displaystyle -17.7Inequality with negative 17.7
\displaystyle s\displaystyle \geq\displaystyle -13.45Add 4.25 on both sides

The solution to the inequality is approximately -13.5 \leq s \leq 22.0 .

Reflect and check

The solution to the inequality is all values of s that are between -13.5 and 22, inclusive. The solution on a number line is

-20-15-10-50510152025
Idea summary

An absolute value equation is an equation where at least one expression contains an absolute value, for example:

|ax+b| = k

When k>0, an absolute value equation has two solutions. When k=0, an absolute value equation has one solution. When k<0, an absolute value equation has no solutions.

For absolute value inequalities with an algebraic expression p(x) and k>0,

  • \left|p(x)\right|<k is within k units of 0 and can be written as -k<p(x)<k
  • \left|p(x)\right|> k is more than k units of 0 and can be written as p(x)<-k or p(x)> k

Outcomes

A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems.

A.CED.A.4

Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.

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