5.2Linear Equations with Fractional Coefficients

When an equation contains fractions, the key strategy is to clear the fractions first by multiplying both sides by the LCM of the denominators. This converts the equation into one with integer coefficients, which is much easier to solve.

Recall: LCM from Chapter 1, used to find the common denominator.→

Now you try

Solve $\frac{2n + 6}{2} = 4$ .

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\displaystyle \frac{2n + 6}{2}

\displaystyle = 4

\displaystyle 2n + 6

\displaystyle = 8 \quad \scriptsize\textit{multiply both sides by 2}

\displaystyle 2n

\displaystyle = 2

\displaystyle n

\displaystyle = 1

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Equation with fractional coefficients

Solve $\displaystyle \frac{2x - 1}{4} = \frac{x + 5}{6}$ .

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LCM of

4

and

6

12

\displaystyle 3(2x - 1)

\displaystyle = 2(x + 5) \quad \scriptsize\textit{multiply both sides by 12}

\displaystyle 6x - 3

\displaystyle = 2x + 10 \quad \scriptsize\textit{expand}

\displaystyle 4x

\displaystyle = 13

\displaystyle x

\displaystyle = \frac{13}{4}

Now you try

Solve $\frac{z + 1}{3} = \frac{z + 3}{5}$ .

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\displaystyle \frac{z + 1}{3}

\displaystyle = \frac{z + 3}{5}

\displaystyle 5(z + 1)

\displaystyle = 3(z + 3) \quad \scriptsize\textit{cross multiply}

\displaystyle 5z + 5

\displaystyle = 3z + 9

\displaystyle 2z

\displaystyle = 4

\displaystyle z

\displaystyle = 2

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