You should save time and quickly realize that option B is the most reasonable, this is because m + n = 7, and only option B completely satisfy this.

Besides, this is a quadratic simultaneous equation, we’re going to be solving it by substitution method…

m² + n² = 29 …………………..eq 1

m + n = 7 ……………………….eq 2

from eq 2,

m = 7 – n

substituting m for ‘7 – n’ in eq 1, we have:

(7 – n)² + n² = 29

(7 – n)(7 – n) + n² = 29

49 – 7n – 7n + n² + n² = 29

49 – 14n + 2n² = 29

Rearranging;

2n² – 14n + 49 – 29 = 0

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2n² – 14n + 20 = 0

Dividing through by 2, we have:

n² – 7n + 10 = 0

factorizing this, we have:

n² – 5n – 2n + 10 = 0

n(n – 5) – 2(n – 5) = 0

(n – 5)(n – 2) = 0

(n – 5) = 0, n = 5

(n – 2) = 0, n = 2

n = 5 or 2

to solve for m, recall that m + n = 7,

m = 7 – n

when n = 5, m = 7 – 5 = 2

and when n = 2, m = 7 – 2 = 5

hence, we have the answer, option B.

Now for the right answer to the above question:

Option A is incorrect. 5 + 3 is not = 7
Option B is correct. 5 + 2 = 7
C is incorrect. 5 + 3 is not = 7
D is not the correct answer. 2 + 3 is not = 7

KEY-POINTS…

You may please note these/this:

Under exam conditions (multiple choice), there will be no need to solve this in detail, just inspect the options and realize that only option B reasonably satisfies one of the equations.
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/ culled from 2020 JAMB-UTME mathematics past question 11 /

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