Question

Solve: 4m – 2n = – 4, 4m + 3n = 16

Answer 1

The given equations are

4m − 2n = – 4        ...(i)

4m + 3n = 16        ...(ii)

Equations (i) and (ii) are in am + bn = c form.

Comparing the given equations with a₁m + b₁n = c₁ and a₂m + b₂n = c₂, we get

a₁ = 4, b₁ = − 2, c₁ = − 4 and

a₂ = 4, b₂ = 3, c₂ = 16

∴ D = `|("a"_1, "b"_1),("a"_2, "b"_2)|`

= `|(4, -2),(4, 3)|`

= (4 × 3) − (– 2 × 4)

= 12 – (– 8)

= 12 + 8

= 20 ≠ 0

D_m = `|("c"_1, "b"_1),("c"_2, "b"_2)|`

= `|(-4, -2),(16, 3)|`

= (–4 × 3) – (– 2 × 16)

= –12 – (– 32)

= –12 + 32

= 20

D_n = `|("a"_1, "c"_1),("a"_2, "c"_2)|`

= `|(4, -4),(4, 16)|`

= (4 × 16) – (– 4 × 4)

= 64 – (– 16)

= 64 + 16

= 80

∴ By Cramer’s rule, we get

m = `("D"_"m")/"D"` and n = `("D"_"n")/"D"`

∴ m = `20/20` and n = `80/20`

∴ m = 1 and n = 4

∴ (m, n) = (1, 4) is the solution of the given equations.

Answer 2

The given equations are

4m − 2n = −4        ...(i)

4m + 3n = 16        ...(ii) 

To eliminate m, subtract Equation 1 from Equation 2:

(4m + 3n) − (4m − 2n) = 16 − (−4)

4m − 4m + 3n + 2n = 16 + 4

5n = 20

n = `20/5`

= 4

Substitute n = 4 into Equation 1:

4m − 2n = −4

4m − 2(4) = −4

4m − 8 = −4

4m = 4

m = `4/4` = 1

m = 1, n = 1