Linear equations are equations that only have variables with a power of 1. To solve linear equation problems with fractions, we have to multiply the entire equation by the least common multiple to eliminate the fractions and then use the conventional method of solving first-degree equations.
In this article, we will look at several examples with answers to master this topic and we will also look at some practice problems and practice what we have learned.
ALGEBRA
Relevant for…
Exploring linear equations with fractions examples and practice problems.
ALGEBRA
Relevant for…
Exploring linear equations with fractions examples and practice problems.
Summary of linear equations with fractions
Recall that, for an equation to be of the first degree, all the variables in the equation must have a maximum power of 1. For example, the equations $latex \frac{1}{3}x+2=6$ and $latex \frac{3}{2}x=\frac{1}{2}x+10$ are first degree equations with fractions. We can solve first degree equations with fractions with the following steps:
Step 1: Remove the fractions: We multiply the entire equation by the least common multiple to remove the fractions.
Step 2: Simplify: We remove the parentheses and other grouping signs and combine like terms.
Step 3: Solve for the variable: We use addition and subtraction to move the variable to only one side of the equation.
Step 4: Solve: We use multiplication and division to solve for the variable completely.
Examples with answers of linear equations with fractions
The following examples of linear equations with fractions have their respective solution. Therefore, you can follow the solving process and fully master solving these types of equations.
EXAMPLE 1
Solve the equation $latex \frac{x}{3}+4=6$.
Solution
Step 1: Remove the fractions: We multiply the entire equation by 3:
$latex \frac{x}{3}+4=6$
$latex x+4(3)=6(3)$
$latex x+12=18$
Step 2: Simplify: We have nothing to simplify.
Step 3: Solve for the variable: We subtract 12 from both sides of the equation:
$latex x+12-12=18-12$
$latex x=6$
Step 4: Solve: We have already got the answer:
$latex x=6$
EXAMPLE 2
Find the value of x in the equation $latex \frac{x+1}{2}+2=4$.
Solution
Step 1: Remove the fractions: We multiply the entire equation by 2:
$latex \frac{x+1}{2}+2=4$
$latex x+1+2(2)=4(2)$
$latex x+1+4=8$
Step 2: Simplify: We combine like terms:
$latex x+1+4=8$
$latex x+5=8$
Step 3: Solve for the variable: We subtract 5 from both sides:
$latex x+5-5=8-5$
$latex x=3$
Step 4: Solve: We have already got the answer:
$latex x=3$
EXAMPLE 3
Find the value of x in the equation $latex \frac{2x+5}{3}+2x=7$.
Solution
Step 1: Remove the fractions: We multiply the entire equation by 3:
$latex \frac{2x+5}{3}+2x=7$
$latex 2x+5+2x(3)=7(3)$
$latex 2x+5+6x=21$
Step 2: Simplify: We combine like terms:
$latex 2x+5+6x=21$
$latex 8x+5=21$
Step 3: Solve for the variable: We subtract 5 from both sides:
$latex 8x+5-5=21-5$
$latex 8x=16$
Step 4: Solve: We divide both sides by 8:
$latex \frac{8x}{8}=\frac{16}{8}$
$latex x=2$
Start now: Explore our additional Mathematics resources
EXAMPLE 4
Find the value of x in the equation $latex \frac{3x-4}{4}+6=2x+10$.
Solution
Step 1: Remove the fractions: We multiply the entire equation by 4:
$latex \frac{3x-4}{4}+6=2x+10$
$latex 3x-4+6(4)=2x(4)+10(4)$
$latex 3x-4+24=8x+40$
Step 2: Simplify: We combine like terms:
$latex 3x-4+24=8x+40$
$latex 3x+20=8x+40$
Step 3: Solve for the variable: We subtract 20 and 8x from both sides of the equation:
$latex 3x+20-20=8x+40-20$
$latex 3x=8x+20$
$latex 3x-8x=8x+20-8x$
$latex -5x=20$
Step 4: Solve: We divide both sides by -5:
$latex \frac{-5x}{-5}=\frac{20}{-5}$
$latex x=-4$
EXAMPLE 5
Solve the equation $latex \frac{t+5}{2}+5=\frac{t-6}{3}+10$ for t.
Solution
Step 1: Remove the fractions: We multiply the entire equation by 6:
$latex \frac{t+5}{2}+5=\frac{t-6}{3}+10$
$latex 3(t+5)+5(6)=2(t-6)+10(6)$
$latex 3(t+5)+30=2(t-6)+60$
Step 2: Simplify: We expand the parentheses and combine like terms:
$latex 3(t+5)+30=2(t-6)+60$
$latex 3t+15+30=2t-12+60$
$latex 3t+45=2t+48$
Step 3: Solve for the variable: We subtract 45 and 2 t from both sides:
$latex 3t+45-45=2t+48-45$
$latex 3t=2t+3$
$latex 3t-2t=2t+3-2t$
$latex t=3$
Step 4: Solve: We already got the answer:
$latex t=3$
EXAMPLE 6
Solve the equation $latex \frac{5x-10}{2}+5=2(2x-2)+1$.
Solution
Step 1: Remove the fractions: We multiply the entire equation by 2:
$latex \frac{5x-10}{2}+5=2(2x-2)+1$
$$5x-10+5(2)=2(2)(2x-2)+1(2)$$
$latex 5x-10+10=4(2x-2)+2$
Step 2: Simplify: We remove the parentheses and combine like terms:
$latex 5x-10+10=4(2x-2)+2$
$latex 5x-10+10=8x-8+2$
$latex 5x=8x-6$
Step 3: Solve for the variable: We subtract 8x from both sides of the equation:
$latex 5x-8x=8x-6-8x$
$latex -3x=-6$
Step 4: Solve: We divide both sides by -3:
$latex \frac{-3x}{-3}=\frac{-6}{-3}$
$latex x=2$
EXAMPLE 7
Find the value of z from the equation $latex \frac{2z+1}{3}+\frac{z-1}{2}=\frac{-3z-5}{2}-11$.
Solution
Step 1: Remove the fractions: We multiply the entire equation by 6 to eliminate all fractions:
$latex \frac{2z+1}{3}+\frac{z-1}{2}=\frac{-3z-5}{2}-11$
$$2(2z+1)+3(z-1)=3(-3z-5)-6(11)$$
$$2(2z+1)+3(z-1)=3(-3z-5)-66$$
Step 2: Simplify: We remove the parentheses and combine like terms:
$$2(2z+1)+3(z-1)=3(-3z-5)-66$$
$latex 4z+2+3z-3=-9z-15-66$
$latex 7z-1=-9z-81$
Step 3: Solve for the variable: Add 1 and 9 z from both sides of the equation:
$latex 7z-1+1=-9z-81+1$
$latex 7z=-9z-80$
$latex 7z+9z=-9z-80+9z$
$latex 16z=-80$
Step 4: Solve: We divide both sides by 16:
$latex \frac{16z}{16}=\frac{-80}{16}$
$latex x=-5$
Linear equations with fractions – Practice problems
The following practice problems can be solved to practice the process of solving linear equations with fractions. If you have trouble solving these problems, you can carefully study the solved examples shown above.
See also
Interested in learning more about solving equations? Take a look at these pages:
