Problem of the Week

Updated at Jul 19, 2021 4:12 PM

How can we solve the equation \({(\frac{5}{n})}^{2}\times \frac{3-n}{2}=25\)?

Below is the solution.



\[{(\frac{5}{n})}^{2}\times \frac{3-n}{2}=25\]

1
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[\frac{{5}^{2}}{{n}^{2}}\times \frac{3-n}{2}=25\]

2
Simplify  \({5}^{2}\)  to  \(25\).
\[\frac{25}{{n}^{2}}\times \frac{3-n}{2}=25\]

3
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{25(3-n)}{{n}^{2}\times 2}=25\]

4
Regroup terms.
\[\frac{25(3-n)}{2{n}^{2}}=25\]

5
Multiply both sides by \(2{n}^{2}\).
\[25(3-n)=25\times 2{n}^{2}\]

6
Simplify  \(25\times 2{n}^{2}\)  to  \(50{n}^{2}\).
\[25(3-n)=50{n}^{2}\]

7
Divide both sides by \(25\).
\[3-n=2{n}^{2}\]

8
Move all terms to one side.
\[3-n-2{n}^{2}=0\]

9
Multiply both sides by \(-1\).
\[2{n}^{2}+n-3=0\]

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

11
Factor out common terms in the first two terms, then in the last two terms.
\[n(2n+3)-(2n+3)=0\]

12
Factor out the common term \(2n+3\).
\[(2n+3)(n-1)=0\]

13
Solve for \(n\).
\[n=-\frac{3}{2},1\]

Done

Decimal Form: -1.5, 1
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