ORCCA Linear Equations and Inequalities Chapter Review

Section 2.6 Linear Equations and Inequalities Chapter Review

Subsection 2.6.1 Solving Multistep Linear Equations

In Section 2.1 we covered the steps to solve a linear equation and the differences between simplifying expressions, evaluating expressions and solving equations.

Example 2.6.1.

Solve for $a$ in $4-(3-a)=-2-2(2a+1)\text{.}$

Explanation

To solve this equation, we will simplify each side of the equation, manipulate it so that all variable terms are on one side and all constant terms are on the other, and then solve for $a\text{:}$

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-3+a\amp=-2-4a-2\\ 1+a\amp=-4-4a\\ 1+a\addright{4a}\amp=-4-4a\addright{4a}\\ 1+5a\amp=-4\\ 1+5a\subtractright{1}\amp=-4\subtractright{1}\\ 5a\amp=-5\\ \divideunder{5a}{5}\amp=\divideunder{-5}{5}\\ a\amp=-1 \end{align*}

Checking the solution $-1$ in the original equation, we get:

\begin{align*} 4-(3-a)\amp=-2-2(2a+1)\\ 4-(3-(\substitute{-1}))\amp\stackrel{?}{=}-2-2(2(\substitute{-1})+1)\\ 4-(4)\amp\stackrel{?}{=}-2-2(-1)\\ 0\amp\stackrel{\checkmark}{=}0 \end{align*}

Therefore the solution to the equation is $-1$ and the solution set is $\{-1\}\text{.}$

Subsection 2.6.2 Solving Multistep Linear Inequalities

In Section 2.2 we covered how solving inequalities is very much like how we solve equations, except that if we multiply or divide by a negative we switch the inequality sign.

Example 2.6.2.

Solve for $x$ in $-2-2(2x+1)\gt4-(3-x)\text{.}$ Write the solution set in both set-builder notation and interval notation.

Explanation

\begin{align*} -2-2(2x+1)\amp\gt4-(3-x)\\ -2-4x-2\amp\gt4-3+x\\ -4x-4\amp\gt x+1\\ -4x-4\subtractright{x}\amp\gt x+1\subtractright{x}\\ -5x-4\amp\gt1\\ -5x-4\addright{4}\amp\gt1\addright{4}\\ -5x\amp\gt5\\ \divideunder{-5x}{-5}\amp\lt\divideunder{5}{-5}\\ x\amp\lt-1 \end{align*}

Note that when we divided both sides of the inequality by $-5\text{,}$ we had to switch the direction of the inequality symbol.

The solution set in set-builder notation is $\{x\mid x\lt-1\}\text{.}$ The solution set in interval notation is $(-\infty,-1)\text{.}$

Subsection 2.6.3 Linear Equations and Inequalities with Fractions

In Section 2.3 we covered how to eliminate denominators in an equation with the LCD to help solve the equation.

Example 2.6.3.

Solve for $x$ in $\frac{1}{4}x+\frac{2}{3}=\frac{1}{6}\text{.}$

Explanation

We'll solve by multiplying each side of the equation by $12\text{:}$

\begin{align*} \frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\ \multiplyleft{12}\left(\frac{1}{4}x+\frac{2}{3}\right)\amp=\multiplyleft{12}\frac{1}{6}\\ 12\cdot\left(\frac{1}{4}x\right)+12\cdot\left(\frac{2}{3}\right)\amp=12\cdot\frac{1}{6}\\ 3x+8\amp=2\\ 3x\amp=-6\\ \divideunder{3x}{3}\amp=\divideunder{-6}{3}\\ x\amp=-2 \end{align*}

Checking the solution:

\begin{align*} \frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\ \frac{1}{4}(\substitute{-2})+\frac{2}{3}\amp\stackrel{?}{=}\frac{1}{6}\\ -\frac{2}{4}+\frac{2}{3}\amp\stackrel{?}{=}\frac{1}{6}\\ -\frac{6}{12}+\frac{8}{12}\amp\stackrel{?}{=}\frac{1}{6}\\ \frac{2}{12}\amp\stackrel{?}{=}\frac{1}{6}\\ \frac{1}{6}\amp\stackrel{\checkmark}{=}\frac{1}{6} \end{align*}

The solution is therefore $-2\text{.}$ We write the solution set s $\{-2\}\text{.}$

Subsection 2.6.4 Special Solution Sets

In Section 2.4 we covered linear equations that have no solutions and also linear equations that have infinitely many solutions. When solving linear inequalities, it's also possible that no solution exists or that all real numbers are solutions.

Example 2.6.4.

Solve for $x$ in the equation $3x=3x+4\text{.}$
Solve for $t$ in the inequality $4t+5\gt 4t+2\text{.}$

Explanation

To solve this equation, we need to move all terms containing $x$ to one side of the equals sign:

\begin{align*} 3x\amp=3x+4\\ 3x\subtractright{3x}\amp=3x+4\subtractright{3x}\\ 0\amp=4 \end{align*}

This equation has no solution. We write the solution set as $\emptyset\text{,}$ which is the symbol for the empty set.
To solve for $t\text{,}$ we will first subtract $4t$ from each side to get all terms containing $t$ on one side:

\begin{align*} 4t+5\amp\gt 4t+2\\ 4t+5\subtractright{4t}\amp\gt 4t+2\subtractright{4t}\\ 5\amp\gt 2 \end{align*}

All values of the variable $t$ make the inequality true. The solution set is all real numbers, which we can write as $\{t\mid t\text{ is a real number}\}$ in set notation, or $(-\infty,\infty)$ in interval notation.

Subsection 2.6.5 Isolating a Linear Variable

In Section 2.5 we covered how to solve an equation when there are multiple variables in the equation.

Example 2.6.5.

Solve for $x$ in $y=mx+b\text{.}$

Explanation

\begin{align*} y\amp=mx+b\\ y\subtractright{b}\amp=mx+b\subtractright{b}\\ y-b\amp=mx\\ \divideunder{y-b}{m}\amp=\divideunder{mx}{m}\\ \frac{y-b}{m}\amp=x \end{align*}

Exercises 2.6.6 Exercises

1.

Solve $\displaystyle{ {3\!\left(x+7\right)-7}={35} }\text{.}$
Evaluate $\displaystyle{{3\!\left(x+7\right)-7}}$ when $x=7\text{.}$
Simplify $\displaystyle{{3\!\left(x+7\right)-7}}\text{.}$

2.

Solve $\displaystyle{ {5\!\left(y-4\right)+7}={22} }\text{.}$
Evaluate $\displaystyle{{5\!\left(y-4\right)+7}}$ when $y=7\text{.}$
Simplify $\displaystyle{{5\!\left(y-4\right)+7}}\text{.}$

3.

Solve the equation.

${-16}={-7b-8-b}$

4.

Solve the equation.

${14}={-3A-2-A}$

5.

Solve the equation.

$\displaystyle{ {5+9\!\left(B-3\right)}={-7-\left(8-2B\right)} }$

6.

Solve the equation.

$\displaystyle{ {3+10\!\left(m-8\right)}={-74-\left(3-2m\right)} }$

7.

Solve the equation.

$\displaystyle{ -6-7n+3 = -n+9-6n }$

8.

Solve the equation.

$\displaystyle{ -9-9q+2 = -q+7-8q }$

9.

Solve the equation.

$\displaystyle{ {21}={\frac{x}{5}+\frac{x}{2}} }$

10.

Solve the equation.

$\displaystyle{ {22}={\frac{r}{3}+\frac{r}{8}} }$

11.

Solve the equation.

$\displaystyle{ {\frac{t-1}{6}}={\frac{t+4}{8}} }$

12.

Solve the equation.

$\displaystyle{ {\frac{b-5}{4}}={\frac{b+1}{6}} }$

13.

Solve this inequality.

$\displaystyle{ {3-\left(y+6\right)} \lt {-11} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

14.

Solve this inequality.

$\displaystyle{ {4-\left(y+9\right)} \lt {0} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

15.

Solve this inequality.

$\displaystyle{ 8(k-7) \leq 8(k-3) }$

16.

Solve this inequality.

$\displaystyle{ 10(k-5) \leq 10(k-1) }$

17.

Solve this inequality.

$\displaystyle{ {5+9\!\left(x-7\right)} \lt {-34-\left(4-4x\right)} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

18.

Solve this inequality.

$\displaystyle{ {1+8\!\left(x-10\right)} \lt {-87-\left(4-2x\right)} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

19.

Solve this inequality.

$\displaystyle{ {-{\frac{1}{4}}t} > {{\frac{2}{3}}t-22} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

20.

Solve this inequality.

$\displaystyle{ {-{\frac{5}{6}}t} > {{\frac{4}{3}}t-26} }$

In set-builder notation, the solution set is .

In interval notation, the solution set is .

21.

Solve this linear equation for $x\text{.}$

\begin{equation*} Ax+By=C \end{equation*}

22.

Solve this linear equation for $y\text{.}$

\begin{equation*} Ax+By=C \end{equation*}

23.

Solve this linear equation for $B\text{.}$

\begin{equation*} C=a-\frac{7B}{x} \end{equation*}

24.

Solve this linear equation for $m\text{.}$

\begin{equation*} r=c-\frac{8m}{t} \end{equation*}

25.

Matthew has ${\$87}$ in his piggy bank. He plans to purchase some Pokemon cards, which costs ${\$2.85}$ each. He plans to save ${\$52.80}$ to purchase another toy. At most how many Pokemon cards can he purchase?

Write an equation to solve this problem.

Matthew can purchase at most Pokemon cards.

26.

Chris has ${\$89}$ in his piggy bank. He plans to purchase some Pokemon cards, which costs ${\$2.25}$ each. He plans to save ${\$71}$ to purchase another toy. At most how many Pokemon cards can he purchase?

Write an equation to solve this problem.

Chris can purchase at most Pokemon cards.

27.

Use a linear equation to solve the word problem.

Daniel has ${\$95.00}$ in his piggy bank, and he spends ${\$3.50}$ every day. Sydney has ${\$40.00}$ in her piggy bank, and she saves ${\$1.50}$ every day.

If they continue to spend and save money this way, how many days later would they have the same amount of money in their piggy banks?

days later, Daniel and Sydney will have the same amount of money in their piggy banks.

28.

Use a linear equation to solve the word problem.

Sean has ${\$100.00}$ in his piggy bank, and he spends ${\$2.50}$ every day. Kurt has ${\$8.00}$ in his piggy bank, and he saves ${\$1.50}$ every day.

If they continue to spend and save money this way, how many days later would they have the same amount of money in their piggy banks?

days later, Sean and Kurt will have the same amount of money in their piggy banks.

29.

Use a linear equation to solve the word problem.

Massage Heaven and Massage You are competitors. Massage Heaven has $4200$ registered customers, and it gets approximately $700$ newly registered customers every month. Massage You has $9450$ registered customers, and it gets approximately $350$ newly registered customers every month. How many months would it take Massage Heaven to catch up with Massage You in the number of registered customers?

These two companies would have approximately the same number of registered customers months later.

30.

Use a linear equation to solve the word problem.

Two truck rental companies have different rates. V-Haul has a base charge of ${\$60.00}\text{,}$ plus ${\$0.35}$ per mile. W-Haul has a base charge of ${\$45.40}\text{,}$ plus ${\$0.45}$ per mile. For how many miles would these two companies charge the same amount?

If a driver drives miles, those two companies would charge the same amount of money.

31.

A rectangle’s perimeter is ${146\ {\rm ft}}\text{.}$ Its length is ${2\ {\rm ft}}$ shorter than four times its width. Use an equation to find the rectangle’s length and width.

Its width is .

Its length is .

32.

A rectangle’s perimeter is ${216\ {\rm ft}}\text{.}$ Its length is ${4\ {\rm ft}}$ longer than three times its width. Use an equation to find the rectangle’s length and width.

Its width is .

Its length is .