Q2 of 71 Page 58

Solve the pair of linear equations x — y = 28 and x — 3y = 0 and if the solution satisfies, y = mx + 5, then find m.

x – y = 28 eq[1] eq[2]

and x – 3y = 0


⇒ x = 3y


Substituting eq[2] in eq[1]


3y – y = 28


⇒ 2y = 28


⇒ y = 14


Now by, x = 3y


⇒ x = 3(14)


⇒ x = 42


∴ in y = mx + 5


putting values of x and y.


14 = 42m + 5


⇒ 14–5 = 42m


⇒


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