Problem of the Week

Updated at Oct 3, 2022 1:58 PM

This week we have another equation problem:

How would you solve the equation \(\frac{4(2+t)}{2+{t}^{2}}=\frac{20}{11}\)?

Let's start!



\[\frac{4(2+t)}{2+{t}^{2}}=\frac{20}{11}\]

1
Multiply both sides by \(2+{t}^{2}\).
\[4(2+t)=\frac{20}{11}(2+{t}^{2})\]

2
Simplify  \(\frac{20}{11}(2+{t}^{2})\)  to  \(\frac{20(2+{t}^{2})}{11}\).
\[4(2+t)=\frac{20(2+{t}^{2})}{11}\]

3
Multiply both sides by \(11\).
\[44(2+t)=20(2+{t}^{2})\]

4
Expand.
\[88+44t=40+20{t}^{2}\]

5
Move all terms to one side.
\[88+44t-40-20{t}^{2}=0\]

6
Simplify  \(88+44t-40-20{t}^{2}\)  to  \(48+44t-20{t}^{2}\).
\[48+44t-20{t}^{2}=0\]

7
Factor out the common term \(4\).
\[4(12+11t-5{t}^{2})=0\]

8
Factor out the negative sign.
\[4\times -(5{t}^{2}-11t-12)=0\]

9
Divide both sides by \(4\).
\[-5{t}^{2}+11t+12=0\]

10
Multiply both sides by \(-1\).
\[5{t}^{2}-11t-12=0\]

11
Split the second term in \(5{t}^{2}-11t-12\) into two terms.
\[5{t}^{2}+4t-15t-12=0\]

12
Factor out common terms in the first two terms, then in the last two terms.
\[t(5t+4)-3(5t+4)=0\]

13
Factor out the common term \(5t+4\).
\[(5t+4)(t-3)=0\]

14
Solve for \(t\).
\[t=-\frac{4}{5},3\]

Done

Decimal Form: -0.8, 3
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