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 4(2+t)2+t2=2011\frac{4(2+t)}{2+{t}^{2}}=\frac{20}{11}?

Let's start!



4(2+t)2+t2=2011\frac{4(2+t)}{2+{t}^{2}}=\frac{20}{11}

1
Multiply both sides by 2+t22+{t}^{2}.
4(2+t)=2011(2+t2)4(2+t)=\frac{20}{11}(2+{t}^{2})

2
Simplify  2011(2+t2)\frac{20}{11}(2+{t}^{2})  to  20(2+t2)11\frac{20(2+{t}^{2})}{11}.
4(2+t)=20(2+t2)114(2+t)=\frac{20(2+{t}^{2})}{11}

3
Multiply both sides by 1111.
44(2+t)=20(2+t2)44(2+t)=20(2+{t}^{2})

4
Expand.
88+44t=40+20t288+44t=40+20{t}^{2}

5
Move all terms to one side.
88+44t−40−20t2=088+44t-40-20{t}^{2}=0

6
Simplify  88+44t−40−20t288+44t-40-20{t}^{2}  to  48+44t−20t248+44t-20{t}^{2}.
48+44t−20t2=048+44t-20{t}^{2}=0

7
Factor out the common term 44.
4(12+11t−5t2)=04(12+11t-5{t}^{2})=0

8
Factor out the negative sign.
4×−(5t2−11t−12)=04\times -(5{t}^{2}-11t-12)=0

9
Divide both sides by 44.
−5t2+11t+12=0-5{t}^{2}+11t+12=0

10
Multiply both sides by −1-1.
5t2−11t−12=05{t}^{2}-11t-12=0

11
Split the second term in 5t2−11t−125{t}^{2}-11t-12 into two terms.
5t2+4t−15t−12=05{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)=0t(5t+4)-3(5t+4)=0

13
Factor out the common term 5t+45t+4.
(5t+4)(t−3)=0(5t+4)(t-3)=0

14
Solve for tt.
t=−45,3t=-\frac{4}{5},3

Done

Decimal Form: -0.8, 3
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