Problem of the Week

Updated at Jun 6, 2022 2:02 PM

How would you solve \(\frac{(3-\frac{5}{x})(x+2)}{2}=1\)?

Below is the solution.



\[\frac{(3-\frac{5}{x})(x+2)}{2}=1\]

1
Multiply both sides by \(2\).
\[(3-\frac{5}{x})(x+2)=2\]

2
Expand.
\[3x+6-5-\frac{10}{x}=2\]

3
Simplify  \(3x+6-5-\frac{10}{x}\)  to  \(3x+1-\frac{10}{x}\).
\[3x+1-\frac{10}{x}=2\]

4
Multiply both sides by \(x\).
\[3{x}^{2}+x-10=2x\]

5
Move all terms to one side.
\[3{x}^{2}+x-10-2x=0\]

6
Simplify  \(3{x}^{2}+x-10-2x\)  to  \(3{x}^{2}-x-10\).
\[3{x}^{2}-x-10=0\]

7
Split the second term in \(3{x}^{2}-x-10\) into two terms.
\[3{x}^{2}+5x-6x-10=0\]

8
Factor out common terms in the first two terms, then in the last two terms.
\[x(3x+5)-2(3x+5)=0\]

9
Factor out the common term \(3x+5\).
\[(3x+5)(x-2)=0\]

10
Solve for \(x\).
\[x=-\frac{5}{3},2\]

Done

Decimal Form: -1.666667, 2
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