Answer: B. (−7, −2)
SOLVINGS
y^2 + x^2 = 53 ….. (Equation I)
y − x = 5 ….. (Equation II)
Make y the subject of equation II
y = x + 5 ….. (Equation III)
Substitute (x + 5) for the value of y in Equation I
∴ (x + 5)^2 + x^2 = 53
x^2 + 10x + 25 + x^2 = 53
2 x^2 + 10x + 25 = 53
2 x^2 + 10x + 25 – 53 = 0
2 x^2 + 10x – 28 = 0 ….. (Equation IV)
Factor 2 out of equation IV;
2 (x^2 + 5x - 14) = 0 ….. (Equation V)
Divide both sides of equation V by 0
∴ x^2 + 5x – 14 = 0
x^2 + (7x – 2x) – 14 = 0
x^2 + 7x – 2x – 14 = 0 ….. (Equation VI)
Divide Equation VI into two groups (x^2 + 7x and – 2x – 14)
x + 7 is a common factor in both groups
Factor out x + 7 in Equation VI
∴ x(x + 7) – 2(x +7) = 0
(x – 2) (x +7) = 0
∴ x – 2 = 0 OR x + 7 = 0
x = 2 OR x = -7
x = 2, -7
If x = 2
Substitute 2 for the value of x in equation II
y − x = 5
y – 2 = 5
y = 5 +2
y = 7(x =2, y =7) … Not included in the options
If x = -7
Substitute -7 for the value of x in equation II
y − x = 5
y – (-7) = 5
y + 7 = 5
y = 5 – 7
y = -2(x = -7, y = -2) …. This tally with option B
