Problem of the Week

Updated at Nov 4, 2019 9:16 AM

For this week we've brought you this equation problem.

How would you solve the equation \(\frac{2+\frac{5}{x}}{4(x+2)}=\frac{9}{32}\)?

Here are the steps:



\[\frac{2+\frac{5}{x}}{4(x+2)}=\frac{9}{32}\]

1
Multiply both sides by \(4(x+2)\).
\[2+\frac{5}{x}=\frac{9}{32}\times 4(x+2)\]

2
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[2+\frac{5}{x}=\frac{9\times 4(x+2)}{32}\]

3
Simplify  \(9\times 4(x+2)\)  to  \(36(x+2)\).
\[2+\frac{5}{x}=\frac{36(x+2)}{32}\]

4
Simplify  \(\frac{36(x+2)}{32}\)  to  \(\frac{9(x+2)}{8}\).
\[2+\frac{5}{x}=\frac{9(x+2)}{8}\]

5
Multiply both sides by the Least Common Denominator: \(8x\).
\[16x+40=9x(x+2)\]

6
Simplify.
\[16x+40=9{x}^{2}+18x\]

7
Move all terms to one side.
\[16x+40-9{x}^{2}-18x=0\]

8
Simplify  \(16x+40-9{x}^{2}-18x\)  to  \(-2x+40-9{x}^{2}\).
\[-2x+40-9{x}^{2}=0\]

9
Multiply both sides by \(-1\).
\[9{x}^{2}+2x-40=0\]

10
Split the second term in \(9{x}^{2}+2x-40\) into two terms.
\[9{x}^{2}+20x-18x-40=0\]

11
Factor out common terms in the first two terms, then in the last two terms.
\[x(9x+20)-2(9x+20)=0\]

12
Factor out the common term \(9x+20\).
\[(9x+20)(x-2)=0\]

13
Solve for \(x\).
\[x=-\frac{20}{9},2\]

Done

Decimal Form: -2.222222, 2
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