8.4 Solve Equations with Fraction or Decimal Coefficients

Learning objectives

Learning objectives

• Solve equations with fraction coefficients

• Solve equations with decimal coefficients

Solve Equations with Fraction Coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation $\frac{1}{8}\,x+\frac{1}{2}=\frac{1}{4}.$

This method worked fine, but many students don’t 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 of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions. Let’s solve the same equation again, but this time use the method that clears the fractions.

Example 8.37

Solution

1. Find the least common denominator. The denominators $8$, $2$, and $4$ have least common denominator $8$.
2. Clear the fractions. Multiply both sides by $8$ using the Multiplication Property of Equality: $8\left(\frac{1}{8}x+\frac{1}{2}\right)=8\left(\frac{1}{4}\right)$, so $x+4=2$.
3. Apply the Subtraction Property of Equality. Subtract $4$ from both sides: $x+4-4=2-4$, which gives $x=-2$.
4. Check by substitution and state the result. Substitute $x=-2$ into the original equation: $\frac{1}{8}(-2)+\frac{1}{2}=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}$, so the solution is $x=-2$.

Takeaway: Multiplying every side by the least common denominator removes all fractions without changing the solution set.

Notice in Example 8.37 that 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.

Example 8.38

Solution

1. Find the least common denominator. The denominators $2$, $4$, and $3$ have least common denominator $12$.
2. Clear the fractions. Multiply both sides by $12$: $12(7)=12\left(\frac{1}{2}x+\frac{3}{4}x-\frac{2}{3}x\right)$, which becomes $84=6x+9x-8x$.
3. Combine like terms. The right side is $(6+9-8)x=7x$, so $84=7x$.
4. Apply the Division Property of Equality. Divide both sides by $7$: $\frac{84}{7}=\frac{7x}{7}$, giving $x=12$.
5. Check by substitution. Substitute $x=12$: $\frac{1}{2}(12)+\frac{3}{4}(12)-\frac{2}{3}(12)=6+9-8=7$.
6. Check by substitution and state the result. Substituting $x=12$ gives $\frac{1}{2}(12)+\frac{3}{4}(12)-\frac{2}{3}(12)=7$, so the solution is $x=12$.

Takeaway: Clearing every denominator with the least common denominator produces integer coefficients that are easier to combine.

In the next example, we’ll have variables and fractions on both sides of the equation.

Example 8.39

Solution

1. Find the least common denominator. The denominators $3$, $6$, and $2$ have least common denominator $6$.
2. Clear the fractions. Multiply both sides by $6$: $6\left(x+\frac{1}{3}\right)=6\left(\frac{1}{6}x-\frac{1}{2}\right)$, which gives $6x+2=x-3$.
3. Collect the variable terms. Subtract $x$ from both sides: $6x-x+2=x-x-3$, so $5x+2=-3$.
4. Remove the constant. Subtract $2$ from both sides: $5x+2-2=-3-2$, giving $5x=-5$.
5. Apply the Division Property of Equality. Divide both sides by $5$: $\frac{5x}{5}=\frac{-5}{5}$, so $x=-1$.
6. Check by substitution and state the result. Substituting $x=-1$ gives $-1+\frac{1}{3}=-\frac{2}{3}$ on the left and $\frac{1}{6}(-1)-\frac{1}{2}=-\frac{1}{6}-\frac{1}{2}=-\frac{2}{3}$ on the right, so the solution is $x=-1$.

Takeaway: When fractions occur on both sides, clearing them first makes variable collection and constant removal much easier.

In Example 8.40, we’ll start by using the Distributive Property. This step will clear the fractions right away!

Example 8.40

Solution

1. Distribute the fractional factor. Expand $1=\frac{1}{2}(4x+2)$ to obtain $1=2x+1$.
2. Remove the constant. Subtract $1$ from both sides: $1-1=2x+1-1$, which gives $0=2x$.
3. Apply the Division Property of Equality. Divide both sides by $2$: $\frac{0}{2}=\frac{2x}{2}$, so $x=0$.
4. Check by substitution and state the result. Substitute $x=0$ into the original equation: $1=\frac{1}{2}(4(0)+2)=\frac{1}{2}(2)=1$, so the solution is $x=0$.

Takeaway: Distributing a fraction can clear it immediately when the factor divides each term inside the parentheses.

Many times, there will still be fractions, even after distributing.

Example 8.41

Solution

1. Distribute on both sides. Expand $\frac{1}{2}(y-5)=\frac{1}{4}(y-1)$ to obtain $\frac{1}{2}y-\frac{5}{2}=\frac{1}{4}y-\frac{1}{4}$.
2. Clear the remaining denominators. Multiply both sides by $4$: $4\left(\frac{1}{2}y-\frac{5}{2}\right)=4\left(\frac{1}{4}y-\frac{1}{4}\right)$, which gives $2y-10=y-1$.
3. Collect the variable terms. Subtract $y$ from both sides: $2y-y-10=y-1-y$, so $y-10=-1$.
4. Remove the constant. Add $10$ to both sides: $y-10+10=-1+10$, giving $y=9$.
5. Check by substitution and state the result. Substitute $y=9$: $\frac{1}{2}(9-5)=2$ and $\frac{1}{4}(9-1)=2$, so the solution is $y=9$.

Takeaway: Distributing first may reduce the denominators that still need to be cleared before solving.

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, $0.3=\frac{3}{10}$ and $0.17=\frac{17}{100}.$ So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator.

Example 8.42

Solution

1. Choose a multiplier to clear the decimal. Since $0.8=\frac{8}{10}$, multiply both sides by $10$: $10(0.8x-5)=10(7)$.
2. Distribute and simplify. This gives $8x-50=70$.
3. Remove the constant. Add $50$ to both sides: $8x-50+50=70+50$, so $8x=120$.
4. Apply the Division Property of Equality. Divide both sides by $8$: $\frac{8x}{8}=\frac{120}{8}$, giving $x=15$.
5. Check by substitution and state the result. Substitute $x=15$ into the original equation: $0.8(15)-5=12-5=7$, so the solution is $x=15$.

Takeaway: Multiplying by a power of ten clears decimal coefficients while preserving equality on both sides.

Example 8.43

Solution

1. Choose a multiplier that clears every decimal. The greatest number of decimal places is two, so multiply both sides by $100$ using the Multiplication Property of Equality.
2. Clear the decimals. Multiplying each term by $100$ gives $6x+2=25x-150$, because $0.06(100)=6$, $0.02(100)=2$, $0.25(100)=25$, and $-1.5(100)=-150$.
3. Collect the variable terms. Subtract $6x$ from both sides: $6x-6x+2=25x-6x-150$, which gives $2=19x-150$.
4. Remove the constant. Add $150$ to both sides: $2+150=19x-150+150$, so $152=19x$.
5. Apply the Division Property of Equality. Divide both sides by $19$: $\frac{152}{19}=\frac{19x}{19}$, and $152\div19=8$, so $x=8$.
6. Check by substitution and state the result. Substitute $x=8$: $0.06(8)+0.02=0.48+0.02=0.50$ and $0.25(8)-1.5=2-1.5=0.50$, so the solution is $x=8$.

Takeaway: Clearing decimals by multiplying every term by the LCD $100$ turns the equation into $6x+2=25x-150$, giving $x=8$.

The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation.

Example 8.44

Solution

1. Distribute the decimal factor. Expand $0.25x+0.05(x+3)=2.85$ to obtain $0.25x+0.05x+0.15=2.85$.
2. Combine like terms. The variable terms give $0.30x$, so the equation becomes $0.30x+0.15=2.85$.
3. Clear the decimals. Multiply both sides by $100$: $100(0.30x+0.15)=100(2.85)$, which gives $30x+15=285$.
4. Remove the constant. Subtract $15$ from both sides: $30x+15-15=285-15$, so $30x=270$.
5. Apply the Division Property of Equality. Divide both sides by $30$: $\frac{30x}{30}=\frac{270}{30}$, giving $x=9$.
6. Check by substitution and state the result. Substituting $x=9$ into the original equation gives $0.25(9)+0.05(9+3)=2.25+0.6=2.85$, so the solution is $x=9$.

Takeaway: Distributing before clearing decimals makes the coefficient arithmetic easier to see and check.

Section 8.4 Exercises

Section 8.4 Exercises

Solve equations with fraction coefficients

In the following exercises, solve the equation by clearing the fractions.

Solve Equations with Decimal Coefficients

In the following exercises, solve the equation by clearing the decimals.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

1. Solve: $\frac{1}{4}\,x+\frac{1}{2}=\frac{5}{8}.$

2. Solve: $\frac{1}{6}\,y-\frac{1}{3}=\frac{1}{6}.$

3. Solve: $6=\frac{1}{2}\,v+\frac{2}{5}\,v-\frac{3}{4}\,v.$

1. Solve: $-1=\frac{1}{2}\,u+\frac{1}{4}\,u-\frac{2}{3}\,u.$

2. Solve: $a+\frac{3}{4}=\frac{3}{8}\,a-\frac{1}{2}.$

3. Solve: $c+\frac{3}{4}=\frac{1}{2}\,c-\frac{1}{4}.$

1. Solve: $-11=\frac{1}{2}(6p+2).$

2. Solve: $8=\frac{1}{3}(9q+6).$

3. Solve: $\frac{1}{5}(n+3)=\frac{1}{4}(n+2).$

1. Solve: $\frac{1}{2}(m-3)=\frac{1}{4}(m-7).$

2. Solve: $0.6x-1=11.$

3. Solve: $1.2x-3=9.$

1. Solve: $0.14h+0.12=0.35h-2.4.$

2. Solve: $0.65k-0.1=0.4k-0.35.$

3. Solve: $0.25n+0.05(n+5)=2.95.$

1. Solve: $0.10d+0.05(d-5)=2.15.$

2. $\frac{1}{4}\,x-\frac{1}{2}=-\frac{3}{4}$

3. $\frac{5}{6}\,y-\frac{2}{3}=-\frac{3}{2}$

1. $\frac{1}{2}\,a+\frac{3}{8}=\frac{3}{4}$

2. $2=\frac{1}{3}\,x-\frac{1}{2}\,x+\frac{2}{3}\,x$

3. $\frac{1}{4}\,m-\frac{4}{5}\,m+\frac{1}{2}\,m=-1$

1. $x+\frac{1}{2}=\frac{2}{3}\,x-\frac{1}{2}$

2. $\frac{1}{3}\,w+\frac{5}{4}=w-\frac{1}{4}$

3. $\frac{1}{2}\,x-\frac{1}{4}=\frac{1}{12}\,x+\frac{1}{6}$

1. $\frac{1}{3}\,b+\frac{1}{5}=\frac{2}{5}\,b-\frac{3}{5}$

2. $1=\frac{1}{6}(12x-6)$

3. $\frac{1}{4}(p-7)=\frac{1}{3}(p+5)$

1. $\frac{1}{2}(x+4)=\frac{3}{4}$

2. $0.6y+3=9$

3. $3.6j-2=5.2$

1. $0.4x+0.6=0.5x-1.2$

2. $0.23x+1.47=0.37x-1.05$

3. $0.48x+1.56=0.58x-0.64$

1. $0.9x-1.25=0.75x+1.75$

2. $0.05n+0.10(n+8)=2.15$

3. $0.10d+0.25(d+5)=4.05$

1. $0.05(q-5)+0.25q=3.05$

2. Coins Taylor has $\text{\textdollar 2.00}$ in dimes and pennies. The number of pennies is $2$ more than the number of dimes. Solve the equation $0.10d+0.01(d+2)=2$ for $d,$ the number of dimes.

3. Explain how to find the least common denominator of $\frac{3}{8},\,\frac{1}{6},\,\mathrm{and}\,\frac{2}{3}.$

1. If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?