Lesson Objectives
• Demonstrate an understanding of how to Solve a Linear Equation in One Variable
• Demonstrate an understanding of the Multiplication Property of Equality
• Demonstrate an understanding of how to find the LCD for a group of Fractions
• Demonstrate an understanding of multiplication with decimals by powers of 10
• Learn how to Solve Linear Equations in One Variable with Fractions
• Learn how to Solve Linear Equations in One Variable with Decimals

## How to Solve Linear Equations in One Variable with Fractions or Decimals

In the previous lesson, we showed how to solve any linear equation in one variable using a four-step process. In this lesson, we will take things one step further and learn how to solve a linear equation in one variable when fractions or decimals are involved.

### Solving Equations with Fractions

Let's suppose we encountered the following equation: $$\frac{1}{2}x + \frac{3}{4}=\frac{7}{4}$$ We could choose to use our four-step procedure with the fractions involved, however, there is an easier approach. We can clear our equation of fractions by multiplying both sides of the equation by the LCD of all fractions. In this case, we have denominators of 2, 4, and 4. Our LCD will be 4, so we can multiply both sides of our equation by 4 and clear the equation of fractions. Remember, we are allowed to multiply both sides of an equation by the same non-zero number and not change the solution: $$4\left(\frac{1}{2}x + \frac{3}{4}\right)=4 \cdot \frac{7}{4}$$ Let's use our distributive property on the left side: $$\require{cancel}4 \cdot \frac{1}{2}x + 4 \cdot \frac{3}{4}=4 \cdot \frac{7}{4}$$ Now we can do some simplifying and our equation will be cleared of fractions: $$2\cancel{4}\cdot \frac{1}{\cancel{2}}x + \cancel{4}\cdot \frac{3}{\cancel{4}}=\cancel{4}\cdot \frac{7}{\cancel{4}}$$ $$2x + 3=7$$ Now we have an equivalent equation, meaning it has the same solution, and it has been cleared of all fractions. Let's use our four-step process and obtain our solution: $$2x + 3=7$$ Isolate 2x on the left side: $$2x + 3 - 3=7 - 3$$ $$2x=4$$ Isolate the variable x: $$\frac{2}{2}x=\frac{4}{2}$$ $$\frac{\cancel{2}}{\cancel{2}}x=\frac{2\cancel{4}}{\cancel{2}}$$ $$x=2$$ Check: $$\frac{1}{2}x + \frac{3}{4}=\frac{7}{4}$$ $$\frac{1}{2}\cdot 2 + \frac{3}{4}=\frac{7}{4}$$ $$\frac{1}{\cancel{2}}\cdot \cancel{2}+ \frac{3}{4}=\frac{7}{4}$$ $$1 + \frac{3}{4}=\frac{7}{4}$$ $$\frac{4}{4}+ \frac{3}{4}=\frac{7}{4}$$ $$\frac{7}{4}=\frac{7}{4}$$ Let's look at a few examples.
Example 1: Solve each equation. $$\frac{3}{2}x + \frac{8}{3}=\frac{10}{3}+ \frac{5}{2}x$$ Let's begin by clearing the equation of fractions. Our denominators here are 2, 3, 3, and 2. Our LCD will be 6, so we will multiply both sides of the equation by the LCD of 6:
$$6 \left(\frac{3}{2}x + \frac{8}{3}\right)=6 \left( \frac{10}{3}+ \frac{5}{2}x \right)$$ Let's simplify each side. We will begin with the left side of the equation: $$6 \cdot \frac{3}{2}x + 6 \cdot \frac{8}{3}$$ $$3\cancel{6}\cdot \frac{3}{\cancel{2}}x + 2\cancel{6}\cdot \frac{8}{\cancel{3}}$$ Our left side becomes: $$9x + 16$$ Let's now simplify the right side: $$6 \cdot \frac{10}{3}+ 6 \cdot \frac{5}{2}x$$ $$2\cancel{6}\cdot \frac{10}{\cancel{3}}+ 3\cancel{6}\cdot \frac{5}{\cancel{2}}x$$ Our right side becomes: $$20 + 15x$$ Now we can put the two sides together. We have an equation which has been cleared of fractions: $$9x + 16=20 + 15x$$ Let's solve our equation using the four-step process: $$9x + 16 - 16=20 + 15x - 16$$ $$9x=4 + 15x$$ $$9x - 15x=4 + 15x - 15x$$ $$-6x=4$$ Now we can isolate our variable x. Let's divide both sides of the equation by (-6): $$\frac{-6}{-6}x=\frac{4}{-6}$$ $$\frac{\cancel{-6}}{\cancel{-6}}x=\frac{2\cancel{4}}{-3\cancel{6}}$$ We have our solution: $$x=-\frac{2}{3}$$ Check: $$\frac{3}{2}x + \frac{8}{3}=\frac{10}{3}+ \frac{5}{2}x$$ $$\frac{3}{2}\cdot -\frac{2}{3}+ \frac{8}{3}=\frac{10}{3}+ \frac{5}{2}\cdot -\frac{2}{3}$$ Left side: $$\frac{\cancel{3}}{\cancel{2}}\cdot -\frac{\cancel{2}}{\cancel{3}}+ \frac{8}{3}» -1 + \frac{8}{3}$$ Right side: $$\frac{10}{3}+ \frac{5}{\cancel{2}}\cdot -\frac{\cancel{2}}{3}» \frac{10}{3}- \frac{5}{3}$$ $$-1 + \frac{8}{3}=\frac{10}{3}- \frac{5}{3}$$ $$\frac{-3}{3}+ \frac{8}{3}=\frac{10}{3}- \frac{5}{3}$$ $$\frac{5}{3}=\frac{5}{3}$$ Example 2: Solve each equation. $$-\frac{10}{3}\left(-\frac{7}{2}x + 1 \right)=\frac{95}{3}$$ When we encounter an equation with fractions and parentheses are involved, we want to remove parentheses before we clear our fractions: $$-\frac{10}{3}\cdot -\frac{7}{2}x -\frac{10}{3}\cdot 1=\frac{95}{3}$$ $$\frac{35}{3}x -\frac{10}{3}=\frac{95}{3}$$ We can clear the fractions by multiplying both sides of the equation by the LCD, which is 3: $$3 \left(\frac{35}{3}x - \frac{10}{3}\right)=3 \cdot \frac{95}{3}$$ Let's use our distributive property to clean up the left side: $$3 \cdot \frac{35}{3}x - 3 \cdot \frac{10}{3}=3 \cdot \frac{95}{3}$$ Now we can simplify: $$\cancel{3}\cdot \frac{35}{\cancel{3}}x - \cancel{3}\cdot \frac{10}{\cancel{3}}=\cancel{3}\cdot \frac{95}{\cancel{3}}$$ $$35x - 10=95$$ Let's now solve this equation using our four-step process: $$35x - 10=95$$ Isolate the variable term: $$35x - 10 + 10=95 + 10$$ $$35x=105$$ Isolate the variable: $$\frac{35}{35}x=\frac{105}{35}$$ $$\frac{\cancel{35}}{\cancel{35}}x=\frac{3\cancel{105}}{\cancel{35}}$$ $$x=3$$ Check: $$-\frac{10}{3}\left(-\frac{7}{2}x + 1 \right)=\frac{95}{3}$$ $$-\frac{10}{3}\left(-\frac{7}{2}\cdot 3 + 1 \right)=\frac{95}{3}$$ $$-\frac{10}{3}\left(\frac{-21}{2}+ 1 \right)=\frac{95}{3}$$ $$-\frac{10}{3}\left(\frac{-21}{2}+ \frac{2}{2}\right)=\frac{95}{3}$$ $$-\frac{10}{3}\left( -\frac{19}{2}\right)=\frac{95}{3}$$ $$-\frac{5\cancel{10}}{3}\cdot -\frac{19}{\cancel{2}}=\frac{95}{3}$$ $$\frac{95}{3}=\frac{95}{3}$$

### Solving Equations with Decimals

When we encounter an equation with decimals involved, we can clear the decimals by multiplying both sides of the equation by the appropriate power of 10. Recall that when we multiply by 10 or a power of 10, we move the decimal point one place to the right for each zero in the power of 10. To clear the decimals from an equation we look for the largest number of decimal places. We then multiply by an appropriate power of 10 to clear that decimal and all other decimals will be cleared. Let's take a look at an example.
Example 3: Solve each equation.
-16.8 - 5.5x = 8 + 6.9x
The largest number of decimal places (places after the decimal point) is 1. This means we can multiply both sides of the equation by 10 and clear the decimals:
10(-16.8 - 5.5x) = 10(8 + 6.9x)
10 • (-16.8) + 10 • (-5.5x) = 10 • 8 + 10 • 6.9x
-168 + -55x = 80 + 69x
Our equation is cleared of decimals, now we can solve the equation using the four-step process:
Isolate the variable term:
-168 + 168 - 55x = 80 + 168 + 69x
-55x = 248 + 69x
-55x - 69x = 248 + 69x - 69x
-124x = 248
Isolate the variable: $$\frac{-124}{-124}x=\frac{248}{-124}$$ $$\frac{\cancel{-124}}{\cancel{-124}}x=\frac{2 \cancel{248}}{-\cancel{124}}$$ $$x=-2$$ Check:
-16.8 - 5.5x = 8 + 6.9x
-16.8 - 5.5(-2) = 8 + 6.9(-2)
-16.8 + 11 = 8 - 13.8
-5.8 = -5.8

#### Skills Check:

Example #1

Solve each equation. $$-\frac{3}{2}x - \frac{4}{5}x=\frac{529}{60}$$

A
$$x=-\frac{15}{8}$$
B
$$x=\frac{12}{13}$$
C
$$x=-\frac{7}{4}$$
D
$$x=-2$$
E
$$x=-\frac{23}{6}$$

Example #2

Solve each equation. $$-\frac{17}{6}\left(\frac{9}{4}x + 1\right)=\frac{5}{3}x + \frac{617}{32}$$

A
$$x=\frac{3}{2}$$
B
$$x=-\frac{11}{4}$$
C
$$x=-\frac{5}{4}$$
D
$$x=\frac{27}{13}$$
E
$$x=-6$$

Example #3

Solve each equation. $$-2.36 + 0.9x=x - 2.2$$

A
$$x=2$$
B
$$x=6.4$$
C
$$x=-1.6$$
D
$$x=6.7$$
E
$$x=4.23$$