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 2+5x4(x+2)=932\frac{2+\frac{5}{x}}{4(x+2)}=\frac{9}{32}?

Here are the steps:



2+5x4(x+2)=932\frac{2+\frac{5}{x}}{4(x+2)}=\frac{9}{32}

1
Multiply both sides by 4(x+2)4(x+2).
2+5x=932×4(x+2)2+\frac{5}{x}=\frac{9}{32}\times 4(x+2)

2
Use this rule: ab×cd=acbd\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}.
2+5x=9×4(x+2)322+\frac{5}{x}=\frac{9\times 4(x+2)}{32}

3
Simplify  9×4(x+2)9\times 4(x+2)  to  36(x+2)36(x+2).
2+5x=36(x+2)322+\frac{5}{x}=\frac{36(x+2)}{32}

4
Simplify  36(x+2)32\frac{36(x+2)}{32}  to  9(x+2)8\frac{9(x+2)}{8}.
2+5x=9(x+2)82+\frac{5}{x}=\frac{9(x+2)}{8}

5
Multiply both sides by the Least Common Denominator: 8x8x.
16x+40=9x(x+2)16x+40=9x(x+2)

6
Simplify.
16x+40=9x2+18x16x+40=9{x}^{2}+18x

7
Move all terms to one side.
16x+409x218x=016x+40-9{x}^{2}-18x=0

8
Simplify  16x+409x218x16x+40-9{x}^{2}-18x  to  2x+409x2-2x+40-9{x}^{2}.
2x+409x2=0-2x+40-9{x}^{2}=0

9
Multiply both sides by 1-1.
9x2+2x40=09{x}^{2}+2x-40=0

10
Split the second term in 9x2+2x409{x}^{2}+2x-40 into two terms.
9x2+20x18x40=09{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)=0x(9x+20)-2(9x+20)=0

12
Factor out the common term 9x+209x+20.
(9x+20)(x2)=0(9x+20)(x-2)=0

13
Solve for xx.
x=209,2x=-\frac{20}{9},2

Done

Decimal Form: -2.222222, 2