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3.02 Multi-step equations with rational numbers

Introduction

We've previously looked at  multi-step equations involving integers  . Now we'll look at solving multi-step equations with rational numbers.

Multi-step equations involving decimals

When solving multi-step equations involving decimals, we can follow the same steps as we did when solving equations with integers:

  • Use the distributive property to remove parentheses.
  • Combine like terms to simplify each side of the equals sign.
  • Add or subtract to get the variable term on one side of the equals sign.
  • Multiply or divide to isolate the variable and solve the equation.

Examples

Example 1

Solve the following equation: 0.2(4+x)=-3.8

Worked Solution
Create a strategy

Use the distributive property and properties of equality to remove the parentheses and isolate the variable.

Apply the idea
\displaystyle 0.2(4+x)\displaystyle =\displaystyle -3.8
\displaystyle 0.8+0.2x\displaystyle =\displaystyle -3.8Use distributive property to expand 0.2(4+x)
\displaystyle 0.8+0.2x-0.8\displaystyle =\displaystyle -3.8-0.8Subtract 0.8 to both sides of the equation
\displaystyle 0.2x\displaystyle =\displaystyle -4.6Evaluate the subtraction
\displaystyle \frac{0.2x}{0.2}\displaystyle =\displaystyle \frac{-4.6}{0.2}Divide both sides by 0.2
\displaystyle x\displaystyle =\displaystyle -23Evaluate
Idea summary

To simplify an equation that has decimal terms, follow the same steps that we do when solving equations with integers.

  • Use the distributive property to remove parentheses.
  • Combine like terms to simplify each side of the equals sign.
  • Add or subtract to get the variable term on one side of the equals sign.
  • Multiply or divide to isolate the variable and solve the equation.

Multi-step equations involving fractions

To simplify an equation that has fractional terms, multiply both sides of the equation by the  least common multiple  of all denominators that appear in the equation. This will get rid of all the denominators in the equation.

Once we eliminate the fractional terms, we will follow the same steps that we did when solving equations with integers.

Examples

Example 2

Solve the following equation: \dfrac{9x}{3}+\dfrac{9x}{2}=-5

Worked Solution
Create a strategy

Multiply both sides of the equation by the least common multiple of all of the denominators in the equation.

Apply the idea

We can start by identifying the LCM of 3 and 2 by listing the first few multiples:

2: 4, 6, 8, 10...

3: 6, 9, 12, 15...

The LCM of 2 and 3 is 6.

\displaystyle \dfrac{9x}{3}\times 6 +\dfrac{9x}{2}\times 6\displaystyle =\displaystyle -5\times 6Multiply both sides by the LCM, 6
\displaystyle 18x+27x\displaystyle =\displaystyle -30Evaluate the multiplication on both sides
\displaystyle 45x\displaystyle =\displaystyle -30Evaluate the addition
\displaystyle x\displaystyle =\displaystyle \dfrac{-30}{45}Divide both sides by 45
\displaystyle =\displaystyle \dfrac{-2}{3}Simplify the fraction

Example 3

Solve \dfrac{8}{9}=\dfrac{7}{x}.

Worked Solution
Create a strategy

We can use cross multiplication to rewrite the equation so that we have no fractions.

If \dfrac{a}{b}=\dfrac{c}{d}, then ad=bc.

Apply the idea
\displaystyle 8x\displaystyle =\displaystyle 9\times 7Cross multiply
\displaystyle 8x\displaystyle =\displaystyle 63Evaluate the multiplication
\displaystyle \dfrac{8x}{8}\displaystyle =\displaystyle \dfrac{63}{8}Isolate x by dividing both sides by 8
\displaystyle x\displaystyle =\displaystyle \dfrac{63}{8}Evaluate the division
Idea summary

To simplify an equation that has fractional terms, multiply both sides of the equation by the least common multiple of all denominators that appear in the equation.

We can also use cross multiplication so that we can remove the fractions in the equation before isolating x and solving the equation.

Follow the same steps that we do when solving equations with integers.

Outcomes

8.EE.C.7

Solve linear equations in one variable.

8.EE.C.7.B

Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.

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