Question 17

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Jambmaths question: 

Solve the equation ${{m}^{2}}+{{n}^{2}}=29,\text{  }m+n=7$

Option A: 

(2,3) and (3,5)

Option B: 

(2,5) and (5,2)

Option C: 

(5,2) and (5,3)

Option D: 

(5,3) and (3,5)

Jamb Maths Solution: 

$\begin{align}  & \text{Note:        }{{m}^{2}}+{{n}^{2}}={{(m+n)}^{2}}-2mn \\ & {{m}^{2}}+{{n}^{2}}={{(m+n)}^{2}}-2mn=29----(i) \\ & m+n=7---(ii) \\ & \text{Substitute }7\text{ for }m+n\text{ in equation }(i) \\ & {{7}^{2}}-2mn=29 \\ & mn=10------(iii) \\ & \text{From }(ii),\text{ }m=7-n \\ & \text{substitute }(7-n)\text{ for }m\text{  in (iii)} \\ & (7-n)n=10 \\ & {{n}^{2}}-7n+10=0 \\ & (n-2)(n-5)=0 \\ & n=2\text{ or }n=5 \\ & \text{when }n=2,\text{ }m=7-2=5 \\ & \text{when }n=5,\text{ }m=7-5=2 \\ & (m,n)=(2,5) \\ & (m,n)=(5,2) \\\end{align}$Option B

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