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Section 8.1 Introduction to Linear Equations in One Variable

Linear Equations/Keeping Equations in Balance.
Example 8.1.1.

Determine the solution and solution set to the equation \(x+18=32\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

In order to isolate \(x\) we need to subtract \(18\) from both sides of the equation. Here goes.

\begin{align*} x+18\amp=32\\ x+18\subtractright{18}\amp=32\subtractright{18}\\ x\amp=14 \end{align*}

Because the original equation is equivalent to the equation \(x=14\text{,}\) it apparent that the solution to the original equation is \(14\text{.}\) Let's check our solution.

\begin{align*} x+18\amp=32\\ \highlight{14}+18\amp\stackrel{?}{=}32\\ 32\amp\stackrel{\checkmark}{=}32 \end{align*}

Great,our solution checks. The solution to the given equation is \(14\)and the solution set is \({14}\text{.}\)

Example 8.1.2.

Determine the solution and solution set to the equation \(t-54=-62\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

In order to isolate \(t\text{,}\) we need to add \(54\) to both sides of the equation. On with it.

\begin{align*} t-54\amp=-62\\ t-54\addright{54}\amp=-62\addright{54}\\ t\amp=-8 \end{align*}

Because the given equation is equivalent to the equation \(t=-8\text{,}\) it is apparent that the solution to the given equation is \(-8\text{.}\) Before stating our formal conclusion, let's make sure that we're correct.

\begin{align*} t-54\amp=-62\\ \highlight{-8}-52\amp\stackrel{?}{=}-62\\ -62\amp\stackrel{\checkmark}{=}-62 \end{align*}

Our answer checks, so we an go ahead and state that the solution to the given equation is \(-8\) and the solution set is \({-8}\text{.}\)

Example 8.1.3.

Determine the solution and solution set to the equation \(-7x=42\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

In order to isolate \(x\text{,}\) we need to divide both sides of the equation by \(-7\text{.}\) Let's do it.

\begin{align*} -7x\amp=42\\ \divideunder{-7x}{-7}\amp=\divideunder{42}{-7}\\ x\amp=-6 \end{align*}

Because the original equation is equivalent to the equation \(x=-6\text{,}\) the solution to the original equation is \(-6\text{.}\) Let's check our solution.

\begin{align*} 7x\amp=42\\ 7 \cdot \highlight{-6}\amp\stackrel{?}{=}-42\\ -42\amp\stackrel{\checkmark}{=}-42 \end{align*}

The solution to the given equation is -6 and the solution set is \(\{-6\}.\)

Example 8.1.4.

Determine the solution and solution set to the equation \(\frac{7}{5}x=21\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

In order to isolate \(x\text{,}\) we need to multiply both sides of the equation by \(\frac{5}{7}\text{.}\) Let's do that.

\begin{align*} \frac{7}{5}x\amp=21\\ \multiplyleft{\frac{5}{7}}{\frac{7}{5}x}\amp=\multiplyleft{\frac{5}{7}}{21}\\ x\amp=15 \end{align*}

Our proposed solution is \(15\text{.}\) Let's check it.

\begin{align*} \frac{7}{5}x\amp=21\\ \frac{7}{5} \cdot \highlight{\frac{15}{1}}\amp\stackrel{?}{=}21\\ 21\amp\stackrel{\checkmark}{=}21 \end{align*}

The solution to the given equation is \(15\) and the solution set is \({15}\text{.}\)

Example 8.1.5.

Determine the solution and solution set to the equation \(-2y=\frac{6}{5}\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

We could divide both sides of the equation by \(-2\text{.}\) However, because there is a fraction on the right side of the equation, the arithmetic will probably be easier to process if instead we multiplied both sides of the equation by \(-\frac{1}{2}\text{.}\) let's go for the second option.

\begin{align*} -2y\amp=-\frac{6}{5}\\ \multiplyleft{-\frac{1}{2}}{-2y}\amp=\multiplyleft{-\frac{1}{2}}{-\frac{6}{5}}\\ y\amp=-\frac{3}{5} \end{align*}

The solution appears to be \(-\frac{3}{5}\text{.}\) Let's double check.

\begin{align*} -2y\amp=-\frac{6}{5}\\ -\frac{2}{1} \cdot \highlight{-\frac{3}{5}}\amp\stackrel{?}{=}\frac{6}{5}\\ \frac{6}{5}\amp\stackrel{\checkmark}{=}\frac{6}{5} \end{align*}

The solution to the stated equation is \(-\frac{3}{5}\) and the solution set is \(\left\{-\frac{3}{5}\right\}\text{.}\)

Example 8.1.6.

Determine the solution and solution set to the equation \(12=18t\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

We can isolate \(t\) by dividing both sides of the equation by \(18\text{.}\) We need to make sure that we reduce the resultant fraction.

\begin{align*} 12\amp=18t\\ \divideunder{12}{18}\amp=\divideunder{18t}{18}\\ \frac{2}{3}\amp=t \end{align*}

Let's check the apparent solution, \(\frac{2}{3}\text{.}\)

\begin{align*} 12\amp=18t\\ 12\amp\stackrel{?}{=}\frac{18}{1} \cdot \highlight{\frac{2}{3}}\\ 12\amp\stackrel{\checkmark}{=}12 \end{align*}

The solution to the stated equation is \(\frac{2}{3}\) and the solution set is \(\left\{\frac{2}{3}\right\}\text{.}\)

Multi-Step linear Equations in One Variable.
Example 8.1.7.

Determine the solution and solution set to the equation \(3x-7=17\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

We'll begin by adding \(7\) to both sides of the equation and follow that by dividing both sides of the equation by \(3\text{.}\)

\begin{align*} 3x-7\amp=17\\ 3x-7\addright{7}\amp=17\addright{7}\\ 3x\amp=24\\ \divideunder{3x}{3}\amp=\divideunder{24}{3}\\ x\amp=8 \end{align*}

The apparent solution is \(8\text{.}\) Let's check it.

\begin{align*} 3x-7\amp=17\\ 3 \cdot \highlight{8}-7\amp\stackrel{?}{=}17\\ 24-7\amp\stackrel{?}{=}17\\ 17\amp\stackrel{\checkmark}{=}17 \end{align*}

The solution to the stated equation is \(8\) and the solution set is \(\{8\}\text{.}\)

Example 8.1.8.

Determine the solution and solution set to the equation \(13=-\frac{2}{3}y-25\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

We can go ahead and isolate the variable, \(y\text{,}\) on the right side of the equation. We'll begin by adding 25 to both sides of the equation and then we'll multiply both sides of the equation by \(-\frac{3}{2}\text{.}\)

\begin{align*} 13\amp=-\frac{2}{3}y-25\\ 13\addright{25}\amp=-\frac{2}{3}y-25\addright{25}\\ 38\amp=-\frac{2}{3}y\\ \multiplyleft{-\frac{3}{2}}{38}\amp=\multiplyleft{-\frac{3}{2}}{-\frac{2}{3}y}\\ -57\amp=y \end{align*}

Let's check the apparent solution, \(-57\text{.}\)

\begin{align*} 13\amp=-\frac{2}{3}y-25\\ 13\amp\stackrel{?}{=}-\frac{2}{3}Β \cdot \highlight{-\frac{57}{1}}-25\\ 13\amp\stackrel{?}{=}38-25\\ 13\amp\stackrel{\checkmark}{=}13 \end{align*}

The solution to the given equation is \(-57\) and the solution set is \(\{-57\}\text{.}\)

Example 8.1.9.

Determine the solution and solution set to the equation \(12y-8=-8+15y\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

We'll begin by isolation the constant term on the right side of the equation and the variable term on the left side of the equation. We'll then multiply or divide both sides of the equation as necessary.

\begin{align*} 12y-8\amp=-8+15y\\ 12y-8\addright{8}\amp=-8+15y\addright{8}\\ 12y\amp=15y\\ 12y\subtractright{15y}\amp=15y\subtractright{15y}\\ -3y\amp=0\\ \divideunder{-3y}{-3}\amp=\divideunder{0}{-3}\\ y\amp=0 \end{align*}

The proposed solution is \(0\text{.}\) Let's check that.

\begin{align*} 12y-8\amp=-8+15y\\ 12 \cdot \highlight{0}-8\amp\stackrel{?}{=}-8+15 \cdot \highlight{0}\\ 0-8\amp\stackrel{?}{=}-8+0\\ -8\amp\stackrel{\checkmark}{=}-8 \end{align*}

The solution to the stated equation is \(0\) and the solution set is \(\{0\}\text{.}\)

Example 8.1.10.

Determine the solution and solution set to the equation \(x+12=5x+7\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

We begin by isolating the constant on the right side of the equation and the variable expression on the left side of the equation. We then multiply or divide both sides of the equation as necessary.

\begin{align*} x+12\amp=5x+7\\ x+12\subtractright{12}\amp=5x+7\subtractright{12}\\ x\amp=5x-5\\ x\subtractright{5x}\amp=5x-5\subtractright{5x}\\ -4x\amp=-5\\ \divideunder{-4x}{-4}\amp=\divideunder{-5}{-4}\\ x\amp=\frac{5}{4} \end{align*}

Let's check \(\frac{5}{4}\) is the stated equation.

\begin{align*} x+12\amp=5x+7\\ \highlight{\frac{5}{4}}+12\amp\stackrel{?}{=}\frac{5}{1} \cdot \highlight{\frac{5}{4}}+7\\ \frac{5}{4}+\frac{12}{1} \cdot \frac{4}{4}\amp\stackrel{?}{=}\frac{25}{4}+\frac{7}{1} \cdot \frac{4}{4}\\ \frac{5}{4}+\frac{48}{4}\amp\stackrel{?}{=}\frac{25}{4}+\frac{28}{4}\\ \frac{5+48}{4}\amp\stackrel{?}{=}\frac{25+28}{4}\\ \frac{53}{4}\amp\stackrel{\checkmark}{=}\frac{53}{4} \end{align*}

The solution to the original equation is \(\frac{5}{4}\) and the solution set is \(\left\{\frac{5}{4}\right\}\text{.}\)

Example 8.1.11.

Determine the solution and solution set to the equation \(12-3x=5-2x\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

We'll begin by isolating the variable on the left side of the equation and the constant on the right side of the equation. Let's go ahead and do that in a single step.

\begin{align*} 12-3x\amp=5-2x\\ 12-3x\subtractright{12}\addright{2x}\amp=5-2x\subtractright{12}\addright{2x}\\ -x\amp=-7 \end{align*}

We'll that's something new. What do we do with the negative sign in front of \(x\text{?}\) One thing that comes to mind is that \(-x\) is equivalent to \(-1 \cdot x\text{,}\) so we could simply divide both sides of the equation by \(-1\text{.}\) Let's do it.

\begin{align*} \divideunder{-x}{-1}\amp=\divideunder{-7}{-1}\\ x\amp=7 \end{align*}

Let's check \(7\) in the original equation.

\begin{align*} 12-3x\amp=5-2x\\ 12-3 \cdot \highlight{7}\amp\stackrel{?}{=}5-2 \cdot \highlight{7}\\ 12-21\amp\stackrel{?}{=}5-14\\ -9\amp\stackrel{\checkmark}{=}-9 \end{align*}

The solution to the given equation is \(7\) and the solution set is \({7}\text{.}\)

When Strange Thing Turn Up.
Example 8.1.12.

Determine the solution and solution set to the equation \(5-3x=-3x+9\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

Hopefully you can see that if we add \(3x\) to both sides of the equation the variable will disappear from both sides of the equation. Let's see that explicitly.

\begin{align*} 5-3x\amp=-3x+9\\ 5-3x\addright{3x}\amp=-3x+9\addright{3x}\\ 5\amp=9 \end{align*}

The resultant equivalent equation is a contradiction. No value of \(x\) will make five equal to nine! Since an equivalent equation has no solutions, the given equation also has no solutions and its solution set is \(\emptyset\text{.}\)

Example 8.1.13.

Determine the solution and solution set to the equation \(7x+4=4+7x\text{.}\) Make sure that you show the steps of the algebraic process used to solve the equation.

Solution

When we subtract \(7x\) from both sides of the equation, we are left with the identity \(4=4\text{.}\) No matter what the value of \(x\text{,}\) \(4\) is equal to \(4\text{.}\) So every real number is a solution to the stated equation and the solution set is \(\mathbb{R}\text{.}\)

An Option for Dealing with Equations that Contain Fractions.
Example 8.1.14.

Determine the solution to the equation \(\frac{2}{3}x-1=\frac{3}{2}+4x\text{.}\) Show all of the steps in the algebraic process. State both the solution and the solution set.

Solution

We'll begin by clearing the fractions from the equation. The least common denominator of the fractions in the equation is \(6\text{,}\) so we'll begin by multiplying both side of the equation by \(6\text{.}\) Be mindful that we need to distribute the factor of \(6\) regardless of whether or not a given term is a fraction.

\begin{align*} \frac{2}{3}x-1\amp=\frac{3}{2}+4x\\ \multiplyleft{6}\left(\frac{2}{3}x-1\right)\amp=\multiplyleft{6}\left(\frac{3}{2}+4x\right)\\ \frac{6}{1} \cdot \frac{2}{3}x-6 \cdot 1\amp=\frac{6}{1} \cdot \frac{3}{2}+6 \cdot 4x\\ 4x-6\amp=9+24x\\ 4x-6\addright{6}\subtractright{24x}\amp=9+24x\addright{6}\subtractright{24x}\\ -20x\amp=15\\ \divideunder{-20x}{-20}\amp=\divideunder{15}{-20}\\ x\amp=-\frac{3}{4} \end{align*}

We have a collective need to check.

\begin{align*} \frac{2}{3}x-1\amp=\frac{3}{2}+4x\\ \frac{2}{3} \cdot \highlight{-\frac{3}{4}}-1\amp\stackrel{?}{=}\frac{3}{2}+4 \cdot \highlight{-\frac{3}{4}}\\ -\frac{1}{2}-1\amp\stackrel{?}{=}\frac{3}{2}-3\\ -\frac{1}{2}-\frac{2}{2}\amp\stackrel{?}{=}\frac{3}{2}-\frac{6}{2}\\ -\frac{3}{2}\amp\stackrel{\checkmark}{=}-\frac{3}{2} \end{align*}

The solution is \(-\frac{3}{4}\) and the solution set is \(\left\{-\frac{3}{4}\right\}\text{.}\)

Example 8.1.15.

Determine the solution to the equation \(-\frac{x}{3}+\frac{1}{4}=\frac{x}{12}+\frac{2}{3}\text{.}\) Show all of the steps in the algebraic process. State both the solution and the solution set.

Solution

We'll begin by clearing the fractions from the equation. The least common denominator of the fractions in the equation is \(12\text{,}\) so we'll begin by multiplying both side of the equation by \(12\text{.}\)

\begin{align*} -\frac{x}{3}+\frac{1}{4}\amp=\frac{x}{12}+\frac{2}{3}\\ \multiplyleft{12}\left(-\frac{x}{3}+\frac{1}{4}\right)\amp=\multiplyleft{12}\left(\frac{x}{12}+\frac{2}{3}\right)\\ \frac{12}{1} \cdot -\frac{x}{3}+\frac{12}{1} \cdot \frac{1}{4}\amp=\frac{12}{1} \cdot \frac{x}{12}+\frac{12}{1} \cdot \frac{2}{3}\\ -4x+3\amp=x+8\\ -4x+3\subtractright{3}\subtractright{x}\amp=x+8\subtractright{3}\subtractright{x}\\ -5x\amp=5\\ \divideunder{-5x}{-5}\amp=\divideunder{5}{-5}\\ x\amp=-1 \end{align*}

Time to check.

\begin{align*} -\frac{x}{3}+\frac{1}{4}\amp=\frac{x}{12}+\frac{2}{3}\\ -\frac{\highlight{-1}}{3}+\frac{1}{4}\amp\stackrel{?}{=}\frac{\highlight{-1}}{12}+\frac{2}{3}\\ \frac{1}{3}+\frac{1}{4}\amp\stackrel{?}{=}-\frac{1}{12}+\frac{2}{3}\\ \frac{4}{12}+\frac{3}{12}\amp\stackrel{?}{=}-\frac{1}{12}+\frac{8}{12}\\ \frac{7}{12}\amp\stackrel{\checkmark}{=}-\frac{7}{12} \end{align*}

The solution is \(-1\) and the solution set is \(\{-1\}\text{.}\)

Example 8.1.16.

Determine the solution to the equation \(\frac{1}{3}-\frac{2}{3}x=-\frac{x}{5}-\frac{16}{15}\) . Show all of the steps in the algebraic process. State both the solution and the solution set.

Solution

We'll begin by clearing the fractions from the equation. The least common denominator of the fractions in the equation is \(15\text{,}\) so we'll begin by multiplying both side of the equation by \(15\text{.}\)

\begin{align*} \frac{1}{3}-\frac{2}{3}x\amp=-\frac{x}{5}-\frac{16}{15}\\ \multiplyleft{15}\left(\frac{1}{3}-\frac{2}{3}x\right)\amp=\multiplyleft{15}\left(-\frac{x}{5}-\frac{16}{15}\right)\\ \frac{15}{1} \cdot \frac{1}{3}-\frac{15}{1} \cdot \frac{2}{3}x\amp=\frac{15}{1} \cdot -\frac{x}{5}-\frac{15}{1} \cdot \frac{16}{15}\\ 5-10x\amp=-3x-16\\ 5-10x\subtractright{5}\addright{3x}\amp=-3x-16\subtractright{5}\addright{3x}\\ -7x\amp=-21\\ \divideunder{-7x}{-7}\amp=\divideunder{-21}{-7}\\ x\amp=3 \end{align*}

Let's get to checking.

\begin{align*} \frac{1}{3}-\frac{2}{3}x\amp=-\frac{x}{5}-\frac{16}{15}\\ \frac{1}{3}-\frac{2}{3} \cdot \highlight{3}\amp\stackrel{?}{=}-\frac{\highlight{3}}{5}-\frac{16}{15}\\ \frac{1}{3}-2\amp\stackrel{?}{=}-\frac{x}{5}-\frac{16}{15}\\ \frac{5}{15}-\frac{30}{15}\amp\stackrel{?}{=}-\frac{9}{15}-\frac{16}{15}\\ -\frac{25}{15}\amp\stackrel{\checkmark}{=}-\frac{25}{15} \end{align*}

The solution is \(3\) and the solution set is \(\{3\}\text{.}\)

Exercises Exercises

Solve each equation, showing all of the steps in the algebraic process. State both the solution and the solution set.

1.

Solve \(-9x=42\text{.}\)

Solution

All we need to do to isolate \(x\) is divide both sides of the equation by \(-9\text{.}\)

\begin{align*} -9x\amp=42\\ \divideunder{-9x}{-9}\amp=\divideunder{42}{-9}\\ x\amp=-\frac{14}{3} \end{align*}

Let's check the solution.

\begin{align*} -9x\amp=42\\ -\frac{9}{1} \cdot \highlight{-\frac{14}{3}}\amp\stackrel{?}{=}42\\ 42\amp\stackrel{\checkmark}{=}42 \end{align*}

The solution is \(-\frac{14}{3}\) and the solution set is \(\left\{-\frac{14}{3}\right\}\text{.}\)

2.

Solve \(t+11=\frac{33}{2}\text{.}\)

Solution

In order to isolate \(t\text{,}\) we need to subtract \(11\) from both sides of the equation.

\begin{align*} t+11\amp=\frac{33}{2}\\ t+11\subtractright{11}\amp=\frac{33}{2}\subtractright{11}\\ t\amp=\frac{33}{2}-\frac{22}{2}\\ t\amp=\frac{11}{2} \end{align*}

Checking ...

\begin{align*} t+11\amp=\frac{33}{2}\\ \highlight{\frac{11}{2}}+11\amp\stackrel{?}{=}\frac{33}{2}\\ \frac{11}{2}+\frac{22}{2}\amp\stackrel{?}{=}\frac{33}{2}\\ \frac{33}{2}\amp\stackrel{\checkmark}{=}\frac{33}{2} \end{align*}

The solution is \(\frac{11}{2}\) and the solution set is \(\left\{\frac{11}{2}\right\}\text{.}\)

3.

Solve \(\frac{7}{2}y=35\text{.}\)

Solution

We could begin by clearing the fractions from the equation, but I think the equation is simple enough that we're just going to go ahead and isolate \(y\) by multiplying both sides of the equation by \(\frac{2}{7}\text{.}\)

\begin{align*} \frac{7}{2}y\amp=35\\ \multiplyleft{\frac{2}{7}}\frac{7}{2}y\amp=\multiplyleft{\frac{2}{7}}35\\ y\amp=10 \end{align*}

Time to check.

\begin{align*} \frac{7}{2}y\amp=35\\ \frac{7}{2} \cdot \highlight{\frac{10}{1}}\amp\stackrel{?}{=}35\\ 35\amp\stackrel{\checkmark}{=}35 \end{align*}

The solution is \(10\) and the solution set is \(\{10\}\text{.}\)

4.

Solve \(-33=5x+2\text{.}\)

Solution

We can go ahead and isolate \(x\) on the right side of the equation by first subtracting \(2\) from both sides of the equation and then dividing both sides by \(5\text{.}\)

\begin{align*} -33\amp=5x+2\\ -33\subtractright{2}\amp=5x+2\subtractright{2}\\ -35\amp=5x\\ \divideunder{-35}{5}\amp=\divideunder{5x}{5}\\ -7\amp=x \end{align*}

On to the check.

\begin{align*} -33\amp=5x+2\\ -33\amp\stackrel{?}{=}5 \cdot \highlight{-7}+2\\ -33\amp\stackrel{?}{=}-35+2\\ -33\amp\stackrel{\checkmark}{=}-33 \end{align*}

The solution is \(-7\) and the solution set is \(\{-7\}\text{.}\)

5.

Solve \(4-x=-6\)

Solution

Let's get straight to it.

\begin{align*} 4-x\amp=6\\ 4-x\subtractright{4}\amp=6\subtractright{4}\\ -x\amp=2\\ \divideunder{-x}{-1}\amp=\divideunder{2}{-1}\\ x\amp=-2 \end{align*}

Gotta check.

\begin{align*} 4-x\amp=6\\ 4-(\highlight{-2})\amp\stackrel{?}{=}6\\ 4+2\amp\stackrel{?}{=}6\\ 6\amp\stackrel{\checkmark}{=}6 \end{align*}

The solution is \(-2\) and the solution set is \(\{-2\}\text{.}\)

6.

Solve \(4x+3=-12-x\text{.}\)

Solution

Let's moves the terms with a factor of \(x\) to the left side of the equation and the constant terms to the right side of the equation.

\begin{align*} 4x+3\amp=12-x\\ 4x+3\subtractright{3}\addright{x}\amp=12-x\subtractright{3}\addright{x}\\ 5x\amp=9\\ \divideunder{5x}{5}\amp=\divideunder{9}{5}\\ x\amp=\frac{9}{5} \end{align*}

Let's check.

\begin{align*} 4x+3\amp=12-x\\ \frac{4}{1} \cdot \highlight{\frac{9}{5}}+3\amp\stackrel{?}{=}12-\highlight{\frac{9}{5}}\\ \frac{36}{5}+\frac{15}{5}\amp\stackrel{?}{=}\frac{60}{5}-\frac{9}{5}\\ \frac{51}{5}\amp\stackrel{\checkmark}{=}\frac{51}{5} \end{align*}

The solution is \(\frac{9}{5}\) and the solution set is \(\left\{\frac{9}{5}\right\}\text{.}\)

7.

Solve \(3y-7=3y+6\)

Solution

When we try to put both of the \(y\) terms on the left side of the equation they both subtract to \(0\text{.}\)

\begin{align*} 3y-7\amp=3y+6\\ 3y-7\subtractright{3y}\amp=3y+6\subtractright{3y}\\ -7\amp=6 \end{align*}

The final equation is a contradiction which indicates that the original equation has no solutions and that the solution set to the original equation is \(\emptyset\text{.}\)

8.

Solve \(\frac{2}{3}x+\frac{1}{5}=\frac{x}{5}-\frac{4}{15}\text{.}\)

Solution

Let's begin by clearing the fractions from the equation. The least common denominator of the terms in the equation is \(15\text{,}\) so we'll begin by multiplying both sides of the equation by \(15\text{.}\)

\begin{align*} \frac{2}{3}x+\frac{1}{5}\amp=\frac{x}{5}-\frac{4}{15}\\ \multiplyleft{15}\left(\frac{2}{3}x+\frac{1}{5}\right)\amp=\multiplyleft{15}\left(\frac{x}{5}-\frac{4}{15}\right)\\ \frac{15}{1} \cdot \frac{2}{3}x+\frac{15}{1} \cdot \frac{1}{5}\amp=\frac{15}{1} \cdot \frac{x}{5}-\frac{15}{1} \cdot \frac{4}{15}\\ 10x+3\amp=3x-4\\ 10x+3\subtractright{3}\subtractright{3x}\amp=3x-4\subtractright{3}\subtractright{3x}\\ 7x\amp=-7\\ \divideunder{7x}{7}\amp=\divideunder{-7}{7}\\ x\amp=-1 \end{align*}

Whew! Let's check to make sure that we're correct.

\begin{align*} \frac{2}{3}x+\frac{1}{5}\amp=\frac{x}{5}-\frac{4}{15}\\ \frac{2}{3} \cdot \highlight{-1}+\frac{1}{5}\amp\stackrel{?}{=}\frac{\highlight{-1}}{5}-\frac{4}{15}\\ -\frac{10}{15}+\frac{3}{15}\amp\stackrel{?}{=}-\frac{3}{15}-\frac{4}{15}\\ -\frac{7}{15}\amp\stackrel{\checkmark}{=}-\frac{7}{15} \end{align*}

The solution is \(-1\) and the solution set is \(\{-1\}\text{.}\)

9.

Solve \(-2x+8=8-2x\text{.}\)

Solution

Let's move the terms with a factor of \(x\) to the left side of the equation and the constant terms to the right side of the equation.

\begin{align*} -2x+8\amp=8-2x\\ -2x+8\subtractright{8}\addright{2x}\amp=8-2x\subtractright{8}\addright{2x}\\ 0\amp=0 \end{align*}

The last equation is an identity, which tells us that every real number is a solution to the given equation and that the solution set to the given equation is \(\mathbb{R}\text{.}\)

10.

Solve \(\frac{t}{2}-\frac{3}{4}=\frac{13}{6}-\frac{t}{3}\text{.}\)

Solution

We'll begin by eliminating the fractions from the equation. The least common denominator of the fractions in the equation is \(12\text{,}\) so let's multiply both sides of the equation by \(12\text{.}\)

\begin{align*} \frac{t}{2}-\frac{3}{4}\amp=\frac{13}{6}-\frac{t}{3}\\ \multiplyleft{12}\left(\frac{t}{2}-\frac{3}{4}\right)\amp=\multiplyleft{12}\left(\frac{13}{6}-\frac{t}{3}\right)\\ \frac{12}{1} \cdot \frac{t}{2}-\frac{12}{1} \cdot \frac{3}{4}\amp=\frac{12}{1} \cdot \frac{13}{6}-\frac{12}{1} \cdot \frac{t}{3}\\ 6t-9\amp=26-4t\\ 6t-9\addright{9+4t}\amp=26-4t\addright{9+4t}\\ 10t\amp=35\\ \divideunder{10t}{10}\amp=\divideunder{35}{10}\\ t\amp=\frac{7}{2} \end{align*}

Definitely going to check this one (not that I'd even think about not checking any equation :). While we check, let's make use of the fact that diving by \(2\) is equivalent multiplying by \(\frac{1}{2}\) and that diving by \(3\) is equivalent to multiplying by \(\frac{1}{3}\text{.}\)

\begin{align*} \frac{t}{2}-\frac{3}{4}\amp=\frac{13}{6}-\frac{t}{3}\\ \frac{1}{2} \cdot \highlight{\frac{7}{2}}-\frac{3}{4}\amp\stackrel{?}{=}\frac{13}{6}-\frac{1}{3} \cdot \highlight{\frac{7}{2}}\\ \frac{7}{4}-\frac{3}{4}\amp\stackrel{?}{=}\frac{13}{6}-\frac{7}{6}\\ 1\amp\stackrel{\checkmark}{=}1 \end{align*}