# 2.3: Solving Linear Equations- Part 1

- Page ID
- 18335

Learning Objectives

- Identify linear equations with one variable and verify their solutions.
- Use the properties of equality to solve basic linear equations.
- Use multiple steps to solve linear equations by isolating the variable.
- Solve linear equations where the coefficients are fractions or decimals.

## Linear Equations with One Variable and Their Solutions

Learning how to solve various algebraic equations is one of the main goals in algebra. This section introduces the basic techniques used for solving linear equations with one variable.

An equation is a statement indicating that two algebraic expressions are equal. A linear equation with one variable, \(x\), is an equation that can be written in the general form \(ax+b=0\), where \(a\) and \(b\) are real numbers and \(a≠0\). Here are some examples of linear equations, all of which are solved in this section:

\(x+3=-5\qquad 2x-5=15\qquad\frac{5}{3}x+2=-8\)

A solution to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation \(2x+3=13\) is \(x\), and the solution is \(x=5\). To verify this, substitute the value \(5\) for \(x\) and check that you obtain a true statement.

\(\begin{aligned} 2x+13&=13 \\ 2(\color{Cerulean}{5}\color{black}{)+3}&=13 \\ 10+3&=13 \\ 13&=13\quad\color{Cerulean}{\checkmark} \end{aligned}\)

Alternatively, when an equation is equal to a constant, we can verify a solution by substituting the value for the variable and show that the result is equal to that constant. In this sense, we say that solutions satisfy the equation.

Example \(\PageIndex{1}\)

Is \(x=3\) a solution to \(−2x−3=−9\)?

**Solution**:

\(-2x-3=-2(\color{Cerulean}{3}\color{black}{)-3=-6-3=-9}\quad\color{Cerulean}{\checkmark}\)

**Answer**:

Yes, it is a solution, because \(x=3\) satisfies the equation.

Example \(\PageIndex{2}\)

Is \(a=−\frac{1}{2}\) a solution to \(−10a+5=25\)?

**Solution**:

\(-10a+5=-10\color{black}{\left( \color{Cerulean}{-\frac{1}{2}} \right) +5=5+5=10\neq 25}\quad\color{red}{x}\)

**Answer**:

No, it is not a solution, because \(a=−\frac{1}{2}\) does not satisfy the equation.

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, then substitute the appropriate values. By making use of parentheses we avoid some common errors when working the order of operations.

Example \(\PageIndex{3}\)

Is \(y=−3\) a solution to \(2y−5=−y−14\)?

**Solution**:

\(\begin{aligned} 2y-5&=-y-14 \\ 2(\:\:)-5&=-(\:\:)-14 &\color{Cerulean}{Replace\:variables\:with\:parentheses.} \\ 2(\color{Cerulean}{-3}\color{black}{)-5}&=-(\color{Cerulean}{-3}\color{black}{)-14}&\color{Cerulean}{Substitute\:the\:appropriate\:values.} \\ -6-5&=3-14 &\color{Cerulean}{Simplify.} \\ -11&=-11\quad\color{Cerulean}{\checkmark}&\color{Cerulean}{True}\end{aligned}\)

**Answer**:

Yes, it is a solution, because \(y=−3\) produces a true statement.

Exercise \(\PageIndex{1}\)

Is \(x=−3\) a solution to \(−2x+5=−1\)?

**Answer**-
No

## Solving Basic Linear Equations

We begin by defining equivalent equations as equations with the same solution set. Consider the following two linear equations and check to see if the solution is \(x=7\).

\(\begin{array}{c|c} {\begin{aligned} 3x-5&=16 \\ 3(\color{Cerulean}{7}\color{black}{)-5}&=16 \\ 21-5&=16 \\ 16&=16\quad\color{Cerulean}{\checkmark} \end{aligned}}&{\begin{aligned} 3x&=21 \\ 3(\color{Cerulean}{7}\color{black}{)}&=21 \\ 21&=21\quad\color{Cerulean}{\checkmark} \end{aligned}} \end{array}\)

Here we can see that the two linear equations \(3x−5=16\) and \(3x=21\) are equivalent because they share the same solution set, namely, \(\{7\}\). The goal is to develop a systematic process to find equivalent equations until the variable is isolated:

\(\left.\begin{aligned} 3x-5&=16 \\ 3x&=21 \\x&=7 \end{aligned}\right\} \quad\color{Cerulean}{Equivalent\:equations}\)

To do this, use the properties of equality. Given algebraic expressions A and B, where c is a real number, we have the following:

\(\begin{array}{ll}{\textbf{Addition Property of Equality}:} & { If\:A=B,\: then\: A\color{Cerulean}{+c}\color{black}{=B}\color{Cerulean}{+c}} \\ {\textbf{Subtraction Property of Equality:}}&{If\:A=B,\:then\:A\color{Cerulean}{-c}\color{black}{=B}\color{Cerulean}{-c}} \\ {\textbf{Multiplication Property of Equality:}}&{If\:A=B,\:then\:\color{Cerulean}{c}\color{black}{A=}\color{Cerulean}{c}\color{black}{B}} \\ {\textbf{Division Property of Equality:}}&{If\:A=B,\:then\:\color{black}{\frac{A}{\color{Cerulean}{c}}=\frac{B}{\color{Cerulean}{c}}}\:c\neq 0}\end{array}\)

Note

Multiplying or dividing both sides of an equation by \(0\) is carefully avoided. Dividing by \(0\) is undefined and multiplying both sides by \(0\) results in the equation \(0 = 0\).

To summarize, equality is retained and you obtain an equivalent equation if you add, subtract, multiply, or divide * both sides *of an equation by any nonzero real number. The technique for solving linear equations involves applying these properties in order to

*on one side of the equation. If the linear equation has a constant term, then we add to or subtract it from both sides of the equation to obtain an equivalent equation where the variable term is isolated.*

**isolate the variable**Example \(\PageIndex{4}\)

Solve:

\(x+3=−5\).

**Solution**:

To isolate the variable \(x\) on the left side, subtract \(3\) from both sides.

\(\begin{aligned} x+3&=-5 \\ x+3\color{Cerulean}{-3}&=-5\color{Cerulean}{-3}&\color{Cerulean}{Subtract\:3\:from\:both\:sides.} \\ x+0&=-8 \\ x&=-8 \end{aligned}\)

**Answer**:

The solution is \(x=−8\).

To check that this is true, substitute \(−8\) into the original equation and simplify to see that it is satisfied: \(x+3=-8+3=-5\quad\checkmark\).

In the previous example, after subtracting \(3\) from both sides, you get \(x+0=−8\). By the additive identity property of real numbers, this is equivalent to \(x=−8\). This step is often left out in the presentation of the solution.

If the * variable term* of the equation (including the coefficient) is isolated, then apply the multiplication or division property of equality to obtain an equivalent equation with the

*isolated. In other words, our goal is to obtain an equivalent equation with \(x\) or \(1x\) isolated on one side of the equal sign.*

**variable**Example \(\PageIndex{5}\)

Solve:

\(−5x=−35\).

**Solution**:

The coefficient of \(x\) is \(–5\), so divide both sides by \(−5\).

\(\begin{aligned} -5x&=-35 \\ \frac{-5x}{\color{Cerulean}{-5}}&=\frac{-35}{\color{Cerulean}{-5}} &\color{Cerulean}{Divide\:both\:sides\:by\:-5.} \\1x&=7 &\color{Cerulean}{Here\: \frac{-5}{-5}=1\:and\:\frac{-35}{-5}=+7.} \\ x&=7 \end{aligned}\)

**Answer**:

The solution is \(x=7\). Perform the check mentally by substituting \(7\) for \(x\) in the original equation.

In the previous example, after dividing both sides by \(−5, x\) is left with a coefficient of \(1\), because \(\frac{−5}{−5}=1\). In fact, when we say “isolate the variable,” we mean to change the coefficient of the variable to \(1\), because \(1x=7\) is equivalent to \(x=7\). This step is often left out of the instructional examples even though its omission is sometimes a source of confusion.

Another important property is the symmetric property: for any algebraic expressions \(A\) and \(B\),

\[\text{If }A=B,\text{ then }B=A\]

The equation \(2=x\) is equivalent to \(x=2\). It does not matter on which side we choose to isolate the variable.

Example \(\PageIndex{6}\)

Solve:

\(2=5+x\).

**Solution**:

Isolate the variable \(x\) by subtracting \(5\) from both sides of the equation.

\(\begin{aligned} 2&=5+x \\ 2\color{Cerulean}{-5}&=5+x\color{Cerulean}{-5} &\color{Cerulean}{Subtract\:5\:from\:both\:sides.} \\ -3&=x \\ x&=-3 &\color{Cerulean}{Apply\:the\:symmetric\:property.} \end{aligned}\)

**Answer**:

The solution is \(−3\), and checking the solution shows that \(2 = 5 − 3\).

Exercise \(\PageIndex{2}\)

Solve:

\(6=x-4\)

**Answer**-
\(x=10\)

## Isolating the Variable in Two Steps

A linear equation of the form \(ax+b=c\) takes two steps to solve. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking solutions in the following examples is left to the reader.

Example \(\PageIndex{7}\)

Solve:

\(2x-5=15\).

**Solution**:

\(\begin{aligned} 2x-5&=15 \\ 2x-5\color{Cerulean}{+5}&=15\color{Cerulean}{+5} &\color{Cerulean}{Add\:5\:to\:both\:sides.} \\ 2x&=20 \\ \frac{2x}{\color{Cerulean}{2}}&=\frac{20}{\color{Cerulean}{2}} &\color{Cerulean}{Divide\:both\:sides\:by\:2.} \\ x&=10 \end{aligned}\)

**Answer**:

The solution is \(10\).

Example \(\PageIndex{8}\)

Solve:

\(−3x−2=9\).

**Solution**:

\(\begin{aligned} -3x-2&=9 \\ -3x-2\color{Cerulean}{+2}&=9\color{Cerulean}{+2} &\color{Cerulean}{Add\:2\:to\:both\:sides.} \\ -3x&=11 \\ \frac{-3x}{\color{Cerulean}{-3}}&=\frac{11}{\color{Cerulean}{-3}} &\color{Cerulean}{Divide\:both\:sides\:by\:-3.} \\ x&=-\frac{11}{3} \end{aligned}\)

**Answer**:

The solution is \(-\frac{11}{3}\).

Example \(\PageIndex{9}\)

Solve:

\(6−5y=−14\).

**Solution**:

When no sign precedes the term, it is understood to be positive. In other words, think of this as \(+6−5y=−14\). Begin by subtracting \(6\) from both sides of the equal sign.

\(\begin{aligned} 6-5y&=-14 \\ 6-5y\color{Cerulean}{-6}&=-14\color{Cerulean}{-6} &\color{Cerulean}{Subtract\:6\:from\:both\:sides.} \\ -5y&=-20 \\ \frac{-5y}{\color{Cerulean}{-5}}&=\frac{-20}{\color{Cerulean}{-5}}&\color{Cerulean}{Divide\:both\:sides\:by\:-5.} \\ y&=4 \end{aligned}\)

**Answer**:

The solution is \(4\).

Example \(\PageIndex{10}\)

Solve:

\(3x+\frac{1}{2}=\frac{2}{3}\).

**Solution**:

\(\begin{aligned} 3x+\frac{1}{2}&=\frac{2}{3} \\ 3x+\frac{1}{2}\color{Cerulean}{-\frac{1}{2}}&=\frac{2}{3}\color{Cerulean}{-\frac{1}{2}}&\color{Cerulean}{Subtract\:\frac{1}{2}\:from\:both\:sides.} \\ 3x&=\color{black}{\frac{2\cdot\color{OliveGreen}{2}}{3\cdot\color{OliveGreen}{2}}-\frac{1\cdot\color{OliveGreen}{3}}{2\cdot\color{OliveGreen}{3}}} &\color{Cerulean}{Obtain\:equivalent\:fractions\:with} \\ &&\color{Cerulean}{a\:common\:denominator.} \\ 3x&=\frac{4}{6}-\frac{3}{6} \\ 3x&=\frac{1}{6} \\ \frac{3x}{\color{Cerulean}{3}}&=\frac{6}{\color{Cerulean}{3}} &\color{Cerulean}{Divide\:both\:sides\:by\:3.} \\ x&=\frac{1}{6}\div\color{Cerulean}{3}\color{black}{=\frac{1}{6}\cdot}\color{Cerulean}{\frac{1}{3}}\color{black}{=\frac{1}{18}} \end{aligned}\)

**Answer**:

The solution is \(\frac{1}{18}\).

Example \(\PageIndex{11}\)

Solve:

\(3−y=1\).

**Solution**:

\(\begin{aligned} 3-y&=1 \\ 3-y\color{Cerulean}{-3}&=1\color{Cerulean}{-3} \\ -y&=-2 \end{aligned}\)

Recall that \(−y\) is equivalent to \(−1y\); divide both sides of the equation by \(−1\).

\(\begin{aligned} -y&=-2\\ \frac{-1y}{\color{Cerulean}{-1}}&=\frac{-2}{\color{Cerulean}{-1}} \\ y&=2 \end{aligned}\)

Alternatively, multiply both sides of \(−y=−2\) by \(−1\) and achieve the same result:

\(\begin{aligned} -y&=-2 \\ \color{Cerulean}{(-1)}\color{black}{(-y)}&=\color{Cerulean}{(-1)}\color{black}{(-2)} \\ y&=2 \end{aligned}\)

**Answer**:

The solution is \(2\).

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation. First, apply the addition or subtraction property of equality to isolate the variable term and then apply the multiplication or division property of equality to isolate the variable on one side of the equation.

Exercise \(\PageIndex{3}\)

Solve:

\(−7x+6=27\).

**Answer**-
\(x=−3\)

## Multiplying by the Reciprocal

To solve an equation like \(\frac{3}{4}x=1\), we can isolate the variable by dividing both sides by the coefficient. For example

\(\begin{aligned} \frac{3}{4}x&=1 \\ \frac{\frac{3}{4}x}{\color{Cerulean}{\frac{3}{4}}}&=\frac{1}{\color{Cerulean}{\frac{3}{4}}}\\x&=1\cdot\color{Cerulean}{\frac{4}{3}} \\ x&=\frac{4}{3} \end{aligned}\)

On the left side of the equal sign, the fraction cancels. On the right side, we have a complex fraction and multiply by the reciprocal of the coefficient. You can save a step by recognizing this and start by multiplying both sides of the equation by the reciprocal of the coefficient.

\(\begin{aligned} \frac{3}{4}x&=1 \\ \color{Cerulean}{\frac{4}{3}}\color{black}{\cdot\frac{3}{4}x}&=\color{Cerulean}{\frac{4}{3}}\color{black}{\cdot 1} \\ x&=\frac{4}{3} \end{aligned}\)

Recall that the product of reciprocals is \(1\), in this case \(\frac{4}{3}⋅\frac{3}{4}=1\), leaving the variable isolated.

Example \(\PageIndex{12}\)

Solve:

\(\frac{5}{3}x+2=−8\).

**Solution**:

Isolate the variable term using the addition property of equality and then multiply both sides of the equation by the reciprocal of the coefficient \(\frac{5}{3}\).

\(\begin{aligned} \frac{5}{3}x+2&=-8 \\ \frac{5}{3}x+2\color{Cerulean}{-2}&=-8\color{Cerulean}{-2} &\color{Cerulean}{Subtract\:2\:from\:both\:sides.} \\ \frac{5}{3}x&=-10 \\ \color{Cerulean}{\frac{3}{5}}\color{black}{\cdot\frac{5}{3}x}&=\color{Cerulean}{\frac{3}{5}}\color{black}{\cdot (-10)}&\color{Cerulean}{Multiply\:both\:sides\:by\:\frac{3}{5}.}\\1x&=3\cdot (-2) \\ x&=-6 \end{aligned}\)

**Answer**:

The solution is \(-6\).

Example \(\PageIndex{13}\)

Solve:

\(−\frac{4}{5}x−5=15\).

**Solution**:

\(\begin{aligned} -\frac{4}{5}x-5&=15 \\ -\frac{4}{5}x-5\color{Cerulean}{+5}&=15\color{Cerulean}{+5} \\ -\frac{4}{5}x&=20 \end{aligned}\)

The reciprocal of \(−\frac{4}{5}\) is \(−\frac{5}{4}\) because \((−\frac{5}{4})(−\frac{4}{5})=+\frac{20}{20}=1\). Therefore, to isolate the variable \(x\), multiply both sides by \(−\frac{5}{4}\).

\(\begin{aligned} \color{Cerulean}{-\frac{5}{4}}\color{black}{\cdot\left(-\frac{4}{5}x \right)}&=\color{Cerulean}{-\frac{5}{4}}\color{black}{\cdot (20)} \\ 1x&=-5\cdot 5 \\ x&=-25 \end{aligned}\)

**Answer**:

The solution is \(-25\).

Exercise \(\PageIndex{4}\)

Solve:

\(\frac{2}{3}x−9=−4\).

**Answer**-
\(x=\frac{15}{2}\)

## Key Takeaways

- Linear equations with one variable can be written in the form \(ax+b=0\), where \(a\) and \(b\) are real numbers and \(a≠0\).
- To “solve a linear equation” means to find a numerical value that can replace the variable and produce a true statement.
- The properties of equality provide tools for isolating the variable and solving equations.
- To solve a linear equation, first isolate the
*variable term*by adding the opposite of the constant term to both sides of the equation. Then isolate the*variable*by dividing both sides of the equation by its coefficient. - After isolating a variable term with a fraction coefficient, solve by multiplying both sides by the reciprocal of the coefficient.

Exercise \(\PageIndex{5}\) Solutions to Linear Equations

Is the given value a solution to the linear equation?

- \(x−6=20; x=26\)
- \(y+7=−6; y=−13\)
- \(−x+5=17; x=12\)
- \(−2y=44; y=11\)
- \(4x=−24; x=−6\)
- \(5x−1=34; x=−7\)
- \(−2a−7=−7; a=0\)
- \(−\frac{1}{3}x−4=−5; x=−3\)
- \(−\frac{1}{2}x+\frac{2}{3}=−\frac{1}{4}; x=\frac{11}{6}\)
- \(−8x−33=3x; x=3\)
- \(3y−5=−2y−15; y=−2\)
- \(3(2x+1)=−4x−3; x=−\frac{1}{2}\)
- \(\frac{1}{2}y−\frac{1}{3}=\frac{1}{3}y+\frac{1}{6}; y=3\)
- \(−\frac{4}{3}y+\frac{1}{9}=−\frac{2}{3}y−\frac{1}{9}; y=\frac{1}{3}\)

**Answer**-
1. Yes

3. No

5. Yes

7. Yes

9. Yes

11. Yes

13. Yes

Exercise \(\PageIndex{6}\) Solving Basic Linear Equations

Solve.

- \(x+3=13\)
- \(y−4=22\)
- \(−6+x=12\)
- \(9+y=−4\)
- \(x−\frac{1}{2}=\frac{1}{3}\)
- \(x+\frac{2}{3}=−\frac{1}{5}\)
- \(x+2\frac{1}{2}=3\frac{1}{3}\)
- \(−37+x=−37\)
- \(4x=−44\)
- \(-9x=63\)
- \(−y=13\)
- \(−x=−10\)
- \(−9x=0\)
- \(−3a=−33\)
- \(27=18y\)
- \(14=−7x\)
- \(31. 5.6a=−39.2\)
- \(−1.2y=3.72\)
- \(\frac{1}{3}x=−\frac{1}{2}\)
- \(−\frac{t}{12}=\frac{1}{4}\)
- \(−\frac{7}{3}x=\frac{1}{2}\)
- \(\frac{x}{5}=−3\)
- \(\frac{4}{9}y=−\frac{2}{3}\)
- \(−\frac{5}{8}y=−\frac{5}{2}\)

**Answer**-
1. \(10\)

3. \(18\)

5. \(\frac{5}{6}\)

7. \(\frac{5}{6}\)

9. \(−11\)

11. \(−13\)

13. \(0\)

15. \(\frac{3}{2}\)

17. \(−7\)

19. \(−\frac{3}{2}\)

21. \(−\frac{3}{14}\)

23. \(−\frac{3}{2}\)

Exercise \(\PageIndex{7}\) Solving Linear Equations

Solve.

- \(5x+7=32\)
- \(4x−3=21\)
- \(3a−7=23\)
- \(12y+1=1\)
- \(21x−7=0\)
- \(−3y+2=−13\)
- \(−5x+9=8\)
- \(22x−55=−22\)
- \(4.5x−2.3=6.7\)
- \(1.4−3.2x=3\)
- \(9.6−1.4y=−10.28\)
- \(4.2y−3.71=8.89\)
- \(3−2y=−11\)
- \(−4−7a=24\)
- \(−10=2x−5\)
- \(24=6−12y\)
- \(\frac{5}{6}x−\frac{1}{2}=\frac{2}{3}\)
- \(\frac{1}{2}x+\frac{1}{3}=\frac{2}{5}\)
- \(4a−\frac{2}{3}=−\frac{1}{6}\)
- \(\frac{3}{5}x−\frac{1}{2}=\frac{1}{10}\)
- \(−\frac{4}{5}y+\frac{1}{3}=\frac{1}{15}\)
- \(−\frac{9}{16}x+\frac{4}{3}=\frac{4}{3}\)
- \(−x+5=14\)
- \(−y−7=−12\)
- \(75−a=200\)
- \(15=5−x\)
- \(−8=4−2x\)
- \(33−x=33\)
- \(18=6−y\)
- \(−12=−2x+3\)
- \(−3=3.36−1.2a\)
- \(0=−3.1a+32.55\)
- \(\frac{1}{4}=−\frac{3}{8}+10x\)
- \(70=50−\frac{1}{2}y\)

**Answer**-
1. \(5\)

3. \(10\)

5. \(\frac{1}{3}\)

7. \(\frac{1}{5}\)

9. \(2\)

11. \(14.2\)

13. \(7\)

15. \(−\frac{5}{2}\)

17. \(\frac{7}{5}\)

19. \(\frac{1}{8}\)

21. \(\frac{1}{3}\)

23. \(−9\)

25. \(−125\)

27. \(6\)

29. \(−12\)

31. \(5.3\)

33. \(\frac{1}{16}\)

Exercise \(\PageIndex{8}\) Solving Linear Equations

Translate the following sentences into linear equations and then solve.

- The sum of \(2x\) and \(5\) is equal to \(15\).
- The sum of \(−3x\) and \(7\) is equal to \(14\).
- The difference of \(5x\) and \(6\) is equal to \(4\).
- Twelve times \(x\) is equal to \(36\).
- A number \(n\) divided by \(8\) is \(5\).
- Six subtracted from two times a number \(x\) is \(12\).
- Four added to three times a number \(n\) is \(25\).
- Three-fourths of a number \(x\) is \(9\).
- Negative two-thirds times a number \(x\) is equal to \(20\).
- One-half of a number \(x\) plus \(3\) is equal to \(10\).

**Answer**-
1. \(2x+5=15; x=5\)

3. \(5x−6=4; x=2\)

5. \(\frac{n}{8}=5; n=40\)

7. \(3n+4=25; n=7\)

9. \(−\frac{2}{3}x=20; x=−30\)

Exercise \(\PageIndex{9}\) Solving Linear Equations

Find a linear equation of the form \(ax+b=0\) with the given solution, where \(a\) and \(b\) are integers. (Answers may vary.)

- \(x=2\)
- \(x=−3\)
- \(x=-\frac{1}{2}\)
- \(x=\frac{2}{3}\)

**Answer**-
1. \(x−2=0\)

3. \(2x+1=0\)

Exercise \(\PageIndex{10}\) Discussion Board Topics

- How many steps are needed to solve any equation of the form \(ax+b=c\)? Explain.
- Instead of dividing by \(6\) when \(6x=12\), could you multiply by the reciprocal of \(6\)? Does this always work?

**Answer**-
1. Answers may vary