Problem of the Week

Updated at Aug 31, 2020 5:40 PM

This week's problem comes from the equation category.

How would you solve the equation \(\frac{5}{4t}\times \frac{5}{t-3}=\frac{25}{16}\)?

Let's begin!



\[\frac{5}{4t}\times \frac{5}{t-3}=\frac{25}{16}\]

1
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{5\times 5}{4t(t-3)}=\frac{25}{16}\]

2
Simplify  \(5\times 5\)  to  \(25\).
\[\frac{25}{4t(t-3)}=\frac{25}{16}\]

3
Multiply both sides by \(4t(t-3)\).
\[25=\frac{25}{16}\times 4t(t-3)\]

4
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[25=\frac{25\times 4t(t-3)}{16}\]

5
Simplify  \(25\times 4t(t-3)\)  to  \(100t(t-3)\).
\[25=\frac{100t(t-3)}{16}\]

6
Simplify  \(\frac{100t(t-3)}{16}\)  to  \(\frac{25t(t-3)}{4}\).
\[25=\frac{25t(t-3)}{4}\]

7
Multiply both sides by \(4\).
\[100=25t(t-3)\]

8
Expand.
\[100=25{t}^{2}-75t\]

9
Move all terms to one side.
\[100-25{t}^{2}+75t=0\]

10
Factor out the common term \(25\).
\[25(4-{t}^{2}+3t)=0\]

11
Factor out the negative sign.
\[25\times -({t}^{2}-3t-4)=0\]

12
Divide both sides by \(25\).
\[-{t}^{2}+3t+4=0\]

13
Multiply both sides by \(-1\).
\[{t}^{2}-3t-4=0\]

14
Factor \({t}^{2}-3t-4\).
\[(t-4)(t+1)=0\]

15
Solve for \(t\).
\[t=4,-1\]

Done
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