A pair of linear equations which has a unique solution x = 2, y = –3 is a) x – 4y –14 = 0
5x – y – 13 = 0
b) 2x – y = 1
3x + 2y = 0
c) x + y = –1
2x – 3y = –5
d) 2x + 5y = –11
4x + 10y = –22

a)x - 4y -14 =0

5x - y - 13 =0

Step-by-step explanation:

Using substitution method to solve equation above gives:

x - 4y - 14 =0 eq1

5x - y -13 =0 eqn2

From eqn1, making x the subject formula:

x - 4y - 14 =0

x - 4y =14

x = 14+4y ...eqn3

From eqn2, substitute value of x and solve for y:

5(14+4y)-y-13 =0

70+20y-y-13 =0

70+19y-13 =0

70 - 13 = -19y

57 = -19y

divide both sides by -19

-57/19 = -19y/-19, then

y = -3

From eqn1:

x = 14 + 4y

substitute the value of y in the expression above

x = 14 + 4(-3)

x = 14 + (-12)

x = 14 - 12

x = 2

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°