Solve the equation: m² + n² = 29, m + n = 7
A. (5, 3) and (3, 5)
B. (2, 5) and (5, 2)
C. (5, 2) and (5, 3)
D. (2, 3) and (3, 5)
QUICK ANSWER…
B
DETAILS…
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
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 /
