Chapter 1 Linear Equations

# 1.1 Use a General Strategy to Solve Linear Equations

Learning Objectives

By the end of this section, you will be able to:

• Solve linear equations using a general strategy
• Classify equations
• Solve equations with fraction or decimal coefficients

# Solve Linear Equations Using a General Strategy

Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

## Solution of an Equation

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation.  If the resulting equation is not true, then the number is not a solution of the equation.

Example 1

Determine whether the values are solutions to the equation $5y+3=10y-4$.

a) $y=\frac{3}{5}$

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

$\begin{array}{lrcl} & 5y+3 & = & 10y-4\\ \text{Substitute} {\frac{3}{5}} \text{for y.} & 5({\frac{3}{5}})+3 & \stackrel{?}{=} & 10({\frac{3}{5}})-4 \\ \text{Multiply.} & 3+3 & \stackrel{?}{=} & 6-4\\ \text{Simplify.} & 6 & \neq & 2\\ \end{array}$

Since $y=\frac{3}{5}$ does not result in a true equation, $y=\frac{3}{5}$ is not a solution to the equation $5y+3=10y-4$

b) $y=\frac{7}{5}$

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

$\begin{array}{lrcl} & 5y+3 & = & 10y-4\\ \text{Substitute} {\frac{7}{5}} \text{for y.} & 5({\frac{7}{5}})+3 & \stackrel{?}{=} & 10 ({\frac{7}{5}})-4 \\ \text{Multiply.} & 7+3 & \stackrel{?}{=} & 14-4\\ \text{Simplify.} & 10 & = & 10\\ \end{array}$

Since $y=\frac{7}{5}$ results in a true equation, $y=\frac{7}{5}$ is a solution to the equation $5y+3=10y-4$

Exercise 1

Determine whether the values are solutions to the equation: $9y+2=6y+3$

a) $y=\frac{4}{3}$

b) $y=\frac{1}{3}$

Solution

a) no

b) yes

Determine whether the values are solutions to the equation: $4x-2=2x+1$

a) $x=\frac{3}{2}$

b) $x=-\frac{1}{2}$

Solution

a) yes

b) no

There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.

linear equation is an equation in one variable that can be written as $ax+b=0$, where $a$ and $b$ are real numbers and $a\neq0$.

To solve a linear equation, it is a good idea to have an overall strategy that can be used to solve any linear equation. In the next example, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Example 2

How to Solve a Linear Equation Using a General Strategy

Solve: $7(n-3)-8=-15$.

 Step 1. Simplify each side of the equation as much as possible. Use the Distributive Property. Notice that each side of the equation is now simplified as much as possible. $\begin{array}{rcl} 7(n-3)-8 & = & -15\\ 7n-21 & = & -15 \\ 7n-29 & = & -15\\ \end{array}$ Step 2. Collect all variable terms on one side of the equation. Nothing to do—all n’s are on the left side. Step 3. Collect constant terms on the other side of the equation. To get constants only on the right, add 29 to each side. Simplify. $\begin{array}{rcl} 7n-29{+29} & = & -15{+29}\\ 7n & = & 14\\ \end{array}$ Step 4. Make the coefficient of the variable term equal to 1. Divide each side by 7. Simplify. $\begin{array}{rcl} \frac{7n}{{7}} & = & \frac{14}{{7}}\\ n & = & 2\\ \end{array}$ Step 5. Check the solution. Let $n=2$. Subtract. Check: $\begin{array}{rcl} 7(n-3)-8 & = & -15\\ 7({2}-3)-8 & \stackrel{?}{=} & -15\\ 7(-1)-8 & \stackrel{?}{=} & -15\\ -7-8 & \stackrel{?}{=} & -15\\ -15 & = & -15\\ \end{array}$

Exercise 2

These steps are summarized in the General Strategy for Solving Linear Equations below.

Solve linear equations using a general strategy.
1. Simplify each side of the equation as much as possible.
• Use the Distributive Property to remove any parentheses.
• Combine like terms.
2. Collect all the variable terms on one side of the equation.
• Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms on the other side of the equation.
• Use the Addition or Subtraction Property of Equality.
4. Make the coefficient of the variable term equal to 1.
• Use the Multiplication or Division Property of Equality.
• State the solution to the equation.
5. Check the solution.
• Substitute the solution into the original equation to make sure the result is a true statement.

Example 3

Solve: $\frac{2}{3}(3m-6)=5-m$

$\begin{array}{lrcl} & \frac{2}{3}(3m-6) & = & 5-m\\ \text{Distribute.} & 2m-4 & = & 5-m \\ \text{Add m to both sides to get}\\ \text{the variables on the left.} & 2m{+m}-4 & = & 5-m{+m}\\ \text{Simplify.} & 3m-4 & = & 5\\ \text{Add 4 to both sides to get}\\ \text{the constants on the right.} & 3m-4{+4} & = & 5{+4}\\ \text{Simplify.} & 3m & = & 9\\ \text{Divide both sides by 3.} & \frac{3m}{{3}} & = & 9\\ \text{Simplify.} & m & = & 3\\ \\ \text{Check: Let }m=3. & \frac{2}{3}(3m-6) & = & 5-m\\ & \frac{2}{3}(3\cdot3-6) & \stackrel{?}{=} & 5-3\\ & \frac{2}{3}(9-6) & \stackrel{?}{=} & 2\\ &\frac{2}{3}(3) & = & 2\\ & 2 & = & 2\\ \end{array}$

Exercise 3

We can solve equations by getting all the variable terms to either side of the equal sign. By collecting the variable terms on the side where the coefficient of the variable is larger, we avoid working with some negatives. This will be a good strategy when we solve inequalities later in this chapter. It also helps us prevent errors with negatives.

Example 4

Solve $4(x-1)-2=5(2x+3)+6$.

$\begin{array}{lrcl} & 4(x-1)-2 & = & 5(2x+3)+6\\ \text{Distribute.} & 4x-4-2 & = & 10x+15+6 \\ \text{Combine like terms.} & 4x-6 & = & 10x+21\\ \text{Subtract 4x from both sides to get}\\ \text{the variables on the right, since 10>4.} & 4x{-4x}-6 & = & 10x{-4x}+21\\ \text{Simplify.} & -6 & = & 6x+21\\ \text{Subtract 21 from each side to}\\ \text{get the constants on the left.} & -6{-21} & = & 6x+21{-21}\\ \text{Simplify.} & -27 & = & 6x\\ \text{Divide both sides by 6.} & \frac{-27}{{6}} & = & \frac{6x}{{6}}\\ \text{Simplify.} & -\frac{9}{2} & = & x\\ \\ \text{Check: Let }x=-\frac{9}{2}. & 4(x-1)-2 & = & 5(2x+3)+6\\ & 4(-{\frac{9}{2}}-1)-2 & \stackrel{?}{=} & 5(2({-\frac{9}{2}})+3)+6\\ & 4(-\frac{11}{2})-2 & \stackrel{?}{=} & 5(-9+3)+6\\ & -22-2 & = & 5(-6)+6\\ & -24 & = & -30+6\\ & -24 & = & -24 \end{array}$

Exercise 4

Example 5

Solve $10[3-8(2s-5)]=15(40-5s)$.

$\begin{array}{lrcl} & 10[3-8(2s-5)] & = & 15(40-5s)\\ \text{Simplify the innermost}\\ \text{parentheses first.} & 10[3-16s+40] & = & 15(40-5s) \\ \text{Combine like terms in the}\\ \text{brackets.} & 10[43-16s] & = & 15(40-5s)\\ \text{Distribute.} & 430-160s & = & 600-75s\\ \text{Add 160s to both sides to}\\ \text{get the variables to the right.} & 430-160s{+160s} & = & 600-75s{+160s}\\ \text{Simplify.} & 430 & = & 600+85s\\ \text{Subtract 600 from both sides}\\ \text{to get the constants to the left.} & 430{-600} & = & 600+85s{-600}\\ \text{Simplify.} & -170 & = & 85s\\ \text{Divide both sides by 85.} & \frac{-170}{{85}} & = & \frac{85s}{{85}}\\ \text{Simplify.} & -2 & = & s\\ \\ \text{Check: Let }s=2. & 10[3-8(2s-5)] & = & 15(40-5s)\\ & 10[3-8(2({-2})-5)] & \stackrel{?}{=} & 15(40-5({-2}))\\ & 10[3-8(-4-5)] & \stackrel{?}{=} & 15(40+10)\\ & 10[3-8(-9)] & = & 15(50)\\ & 10(3+72) & = & 750\\ & 10(75) & = & 750\\ & 750 & = & 750\\ \end{array}$

Exercise 5

# Classify Equations

The equation $7x+8=-13$ is true when we replace the variable, x, with the value $-3$ but not true when we replace x with any other value. An equation like this is called a conditional equation. All the equations we have solved so far are conditional equations.

## Conditional Equations

An equation that is true for one value of the variable and false for all other values of the variable is a conditional equation.

Now let’s consider the equation $7y+14=7(y+2)$. Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.

Solve:

 $\begin{array}{lrcl} & 7y+14 & = & 7(y+2)\\ \text{Distribute.} & 7y+14 & = & 7y+14 \\ \text{Subtract 7y from each side to get the y's to one side.} & 7y{-7y}+14 & = & 7y{-7y}+14\\ \text{Simplify—the y's are eliminated.} & 14 & = & 14\\ \end{array}$ But $14=14$ is true.

This means that the equation $7y+14=7(y+2)$ is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable is called an identity.

## Identity

An equation that is true for any value of the variable is called an identity.

The solution of an identity is all real numbers.

What happens when we solve the equation $-8z=-8z+9$?

Solve:

 $\begin{array}{lrcl} & -8z & = & -8z+9\\ \text{Add 8z to both sides to leave the constant alone on the right.} & -8z{+8z} & = & -8z{+8z}+9 \\ \text{Simplify—the z's are eliminated.} & 0 & = & 9\\ \end{array}$ But $0\neq9$.

Solving the equation $-8z=-8z+9$ led to the false statement $0=9$. The equation $-8z=-8z+9$ will not be true for any value of z. It has no solution. An equation that has no solution or that is false for all values of the variable is called a contradiction.

An equation that is false for all values of the variable is called a contradiction.

The next few examples will ask us to classify an equation as a conditional equation, an identity, or a contradiction.

Example 6

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $6(2n-1)+3=2n-8+5(2n+1)$

 $\begin{array}{lrcl} & 6(2n-1)+3 & = & 2n-8+5(2n+1)\\ \text{Distribute.} & 12n-6+3 & = & 2n-8+10n+5\\ \text{Combine like terms.} & 12n-3 & = & 12n-3\\ \text{Subtract 12n from each side to}\\ \text{get the n's to one side.} & 12n{-12n}-3 & = & 12n{-12n}-3\\ \text{Simplify.} & -3 & = & -3\\ \end{array}$ This is a true statement. The equation is an identity. The solution is all real numbers.

Exercise 6

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $4+9(3x-7)=-42x-13+23(3x-2)$.

Solution

identity; all real numbers

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $8(1-3x)+15(2x+7)=2(x+50)+4(x+3)+1$.

Solution

identity; all real numbers

Example 7

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $8+3(a-4)=0$.

$\begin{array}{lrcl} & 8+3(a-4) & = & 0\\ \text{Distribute.} & 8+3a-12 & = & 0\\ \text{Combine like terms.} & 3a-4 & = & 0\\ \text{Add 4 to both sides.} & 3a-4{+4} & = & 0{+4}\\ \text{Simplify.} & 3a & = & 4\\ \text{Divide.} & \frac{3a}{{3}} & = & \frac{4}{{3}}\\ \text{Simplify.} & a & = & \frac{4}{3}\\ \end{array}$

The equation is true when $a=\frac{4}{3}$, so this is a conditional equation.

The solution is $a=\frac{4}{3}$.

Exercise 7

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $11(q+3)-5=19$.

Solution

conditional equation; $q=-\frac{9}{11}$

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $6+14(k-8)=95$.

Solution

conditional equation; $k=\frac{201}{14}$

Example 8

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $5m+3(9+3m)=2(7m-11)$.

$\begin{array}{lrcl} & 5m+3(9+3m) & = & 2(7m-11)\\ \text{Distribute.} & 5m+27+9m & = & 14m-22\\ \text{Combine like terms.} & 14m+27 & = & 14m-22\\ \text{Subtract 14m from both sides.} & 14m+27{-14m} & = & 14m-22{-14m}\\ \text{Simplify.} & 27 & = & -22\\ \end{array}$

But $27\neq-22$. The equation is a contradiction. It has no solution.

Exercise 8

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $12c+5(5+3c)=3(9c-4)$.

Solution

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $4(7d+18)=13(3d-2)-11d$.

Solution

We summarize the methods for classifying equations in the following table.

Type of equation What happens when you solve it? Solution
Conditional Equation True for one or more values of the variables and false for all other values One or more values
Identity True for any value of the variable All real numbers
Contradiction False for all values of the variable No solution

# Solve Equations with Fraction or Decimal Coefficients

We could use the General Strategy to solve the next example. This method would work fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator (LCD) of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called clearing the equation of fractions.

To clear an equation of decimals, we think of all the decimals in their fraction form and then find the LCD of those denominators.

Example 9

How to Solve Equations with Fraction or Decimal Coefficients

Solve: $\frac{1}{12}x+\frac{5}{6}=\frac{3}{4}$

 Step 1. Find the least common denominator of all the fractions and decimals in the equation. What is the LCD of $\frac{1}{12}$, $\frac{5}{6}$, and $\frac{3}{4}$? $\begin{array}{rcl} \frac{1}{12}x+\frac{5}{6} & = & \frac{3}{4}\\ LCD & = & 12\\ \end{array}$ Step 2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals. Multiply both sides of the equation by the LCD, 12. Use the Distributive Property. Simplify—and notice, no more fractions! $\begin{array}{rcl} {12}(\frac{1}{12}x+\frac{5}{6}) & = & {12}(\frac{3}{4})\\ 12\cdot\frac{1}{12}x+12\cdot\frac{5}{6} & = & 12\cdot\frac{3}{4}\\ x+10 & = & 9\\ \end{array}$ Step 3. Solve using the General Strategy for Solving Linear Equations. To isolate the variable term, subtract 10. Simplify. $\begin{array}{rcl} x+10{-10} & = & 9{-10}\\ x & = & -1\\ \\ \text{Check:}\\ \frac{1}{12}x+\frac{5}{6} & = & \frac{3}{4}\\ \frac{1}{12}{(-1)}+\frac{5}{6} & \stackrel{?}{=} & \frac{3}{4}\\ -\frac{1}{12}+\frac{5}{6} & \stackrel{?}{=} & \frac{3}{4}\\ -\frac{1}{12}+\frac{10}{12} & \stackrel{?}{=} & \frac{9}{12}\\ \frac{9}{12} & = & \frac{9}{12}\\ \end{array}$

Exercise 9

Notice in the previous example, once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve. We then used the General Strategy for Solving Linear Equations.

To solve equations with fraction or decimal coefficients:
1. Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
2. Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
This is demonstrated in the next example.

Example 10

Solve. $5=\frac{1}{2}y+\frac{2}{3}y-\frac{3}{4}y$.

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

$\begin{array}{lrcl} \text{Find the LCD of all fractions}\\ \text{in the equation.} & 5 & = & \frac{1}{2}y+\frac{2}{3}y-\frac{3}{4}y\\ \text{The LCD is 12.}\\ \text{Multiply both sides of the equation}\\ \text{by 12.} & {12}(5) & = & {12}\cdot(\frac{1}{2}y+\frac{2}{3}y-\frac{3}{4}y)\\ \text{Distribute.} & 12(5) & = & 12\cdot\frac{1}{2}y+12\cdot\frac{2}{3}y-12\cdot\frac{3}{4}y\\ \text{Simplify—notice, no more fractions.} & 60 & = & 6y+8y-9y\\ \text{Combine like terms.} & 60 & = & 5y\\ \text{Divide by 5.} & \frac{60}{{5}} & = & \frac{5y}{{5}}\\ \text{Simplify.} & 12 & = & y\\ \\ \text{Check: Let }y=12. & 5 & = & \frac{1}{2}y+\frac{2}{3}y-\frac{3}{4}y\\ & 5 & \stackrel{?}{=} & \frac{1}{2}({12})+\frac{2}{3}({12})-\frac{3}{4}({12})\\ & 5 & \stackrel{?}{=} & 6+8-9\\ & 5 & = & 5\\ \end{array}$

Exercise 10

In the next example, we’ll distribute before we clear the fractions.

Example 11

Solve $\frac{1}{2}(y-5)=\frac{1}{4}(y-1)$.

$\begin{array}{lrcl} & \frac{1}{2}(y-5) & = & \frac{1}{4}(y-1)\\ \text{Distribute.} & \frac{1}{2}y-\frac{5}{2} & = & \frac{1}{4}y-\frac{1}{4}\\ \text{Multiply by the LCD, 4.} & {4}(\frac{1}{2}y-\frac{5}{2}) & = & {4}(\frac{1}{4}y-\frac{1}{4})\\ \text{Distribute.} & 4\cdot\frac{1}{2}y-4\cdot\frac{5}{2} & = & 4\cdot\frac{1}{4}y-4\cdot\frac{1}{4}\\ \text{Simplify.} & 2y-10 & = & y-1\\ \text{Collect the variables to the left.} & 2y {-y}-10 & = & y {-y}-1\\ \text{Simplify.} & y-10 & = & -1\\ \text{Collect the constants to the right.} & y-10{+10} & = & -1{+10}\\ \text{Simplify.} & y & = & 9\\ \end{array}$

An alternate way to solve this equation is to clear the fractions without distributing first. If you multiply the factors correctly, this method will be easier.

$\begin{array}{lrcl} & \frac{1}{2}(y-5) & = & \frac{1}{4}(y-1)\\ \text{Multiply by the LCD, 4.} & {4\cdot}\frac{1}{2}(y-5) & = & {4\cdot}\frac{1}{4}(y-1)\\ \text{Multiply 4 times the fractions.} & 2(y-5) & = & 1(y-1)\\ \text{Distribute.} & 2y-10 & = & y-1\\ \text{Collect the variables to the left.} & 2y {-y}-10 & = & y {-y}-1\\ \text{Simplify.} & y-10 & = & -1\\ \text{Collect the constants to the right.} & y-10{+10} & = & -1{+10}\\ \text{Simplify.} & y & = & 9\\ \text{Check: Let }y=9. & \frac{1}{2}(y-5) & = & \frac{1}{4}(y-1)\\ & \frac{1}{2}({9}-5) & \stackrel{?}{=} & \frac{1}{4}({9}-1)\\ \end{array}$

Finish the check on your own.

Exercise 11

When you multiply both sides of an equation by the LCD of the fractions, make sure you multiply each term by the LCD—even if it does not contain a fraction.

Example 12

Solve $\frac{4q+3}{2}+6=\frac{3q+5}{4}$.

$\begin{array}{lrcl} & \frac{4q+3}{2}+6 & = & \frac{3q+5}{4}\\ \text{Multiply by the LCD, 4.} & {4}(\frac{4q+3}{2}+6) & = & 4(\frac{3q+5}{4})\\ \text{Distribute.} & 4\cdot(\frac{4q+3}{2})+4\cdot6 & = & 4\cdot(\frac{3q+5}{4})\\ \text{Simplify.} & 2(4q+3)+24 & = & 3q+5\\ & 8q+6+24 & = & 3q+5\\ &8q+30 & = & 3q+5\\ \text{Collect the variables to the left.} & 8q{-3q}+30 & = & 3q{-3q}+5\\ \text{Simplify.} & 5q+30 & = & 5\\ \text{Collect the constants to the right.} & 5q+30{-30} & = & 5{-30}\\ \text{Simplify.} & 5q & = & -25\\ \text{Divide both sides by 5.} & \frac{5q}{{5}} & = & \frac{-25}{{5}}\\ \text {Simplify.} & q & = & -5\\ \\ \text{Check: Let }q=-5. & \frac{4q+3}{2}+6 & = & \frac{3q+5}{4}\\ & \frac{4({-5})+3}{2}+6 & \stackrel{?}{=} & \frac{3({-5})+5}{4}\\ \end{array}$

Finish the check on your own.

\begin{array}{lrcl} & \frac{4q+3}{2}+6 & = & \frac{3q+5}{4}\\ \text{Multiply by the LCD, 4.} & {4}(\frac{4q+3}{2}+6) & = & 4(\frac{3q+5}{4})\\ \text{Distribute.} & 4\cdot(\frac{4q+3}{2})+4\cdot6 & = & 4\cdot(\frac{3q+5}{4})\\ \text{Simplify.} & 2(4q+3)+24 & = & 3q+5\\ & 8q+6+24 & = & 3q+5\\ & 8q+30 & = & 3q+5\\ \text{Collect the variables to the left.} & 8q{-3q}+30 & = & 3q{-3q}+5\\ \text{Simplify.} & 5q+30 & = & 5\\ \text{Collect the constants to the right.} & 5q+30{-30} & = & 5{-30}\\ \text{Simplify.} & 5q & = & -25\\ \text{Divide both sides by 5.} & \frac{5q}{{5}} & = & \frac{-25}{{5}}\\ \text{Simplify.} & q & = & -5\\ \text{Check: Let }q=-5. & \frac{4q+3}{2}+6 & = & \frac{3q+5}{4}\\ & \frac{4({-5})+3}{2}+6 & \stackrel{?}{=} & \frac{3({-5})+5}{4}\\ & \frac{-20+3}{2}+6 & \stackrel{?}{=} & \frac{-15+5}{4}\\ & \frac{-17}{2}+6 & \stackrel{?}{=} & \frac{-10}{4}\\ & -8.5+6 & \stackrel{?}{=} & -2.5\\ & -2.5 & = & -2.5\\ \end{array}

Exercise 12

Some equations have decimals in them. This kind of equation may occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, $0.7=\frac{7}{10}$ and $0.29=\frac{29}{100}$ So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.

The next example uses an equation that is typical of the ones we will see in the money applications in a later section. Notice that we will clear all decimals by multiplying by the LCD of their fraction form.

Example 13

Solve $0.25x+0.05(x+3)=2.85$.

Look at the decimals and think of the equivalent fractions:

$0.25=\frac{25}{100}$, $0.05=\frac{5}{100}$, $2.85=2\frac{85}{100}$

Notice, the LCD is 100. By multiplying by the LCD, we will clear the decimals from the equation.

$\begin{array}{lrcl} & 0.25x+0.05(x+3) & = & 2.85\\ \text{Distribute first.} & 0.25x+0.05x+0.15 & = & 2.85\\ \text{Combine like terms.} & 0.30x+0.15 & = & 2.85\\ \text{To clear decimals, multiply by 100.} & {100}(0.30x+0.15) & = & {100}(2.85)\\ \text{Distribute.} & 30x+15 & = & 285\\ \text{Subtract 15 from both sides.} & 30x+15{-15} & = & 285{-15}\\ \text{Simplify.} & 30x & = & 270\\ \text{Divide by 30.} & \frac{30x}{{30}} & = & \frac{270}{{30}}\\ \text{Simplify.} & x & = & 9\\ \end{array}$

Check it yourself by substituting $x=9$ into the original equation.

Exercise 13

definition