Question

Solve the system of equations by using the method of cross multiplication:

`a/x - b/y = 0, (ab^2)/x + (a^2b)/y = (a^2 + b^2)`, where x ≠ 0 and y ≠ 0

Solve the following system of equations by using the method of cross multiplication:

`a/x - b/y = 0, (ab^2)/x + (a^2b)/y = (a^2 + b^2)`, where x ≠ 0 and y ≠ 0

Answer 1

Substituting `1/x = u` and `1/y = v` in the given equations, we get

au – bv + 0 = 0   ...(i)

ab²u + a²bv – (a² + b²) = 0   ...(ii)

Here, a₁ = a, b₁ = –b, c₁ = 0, a₂ = ab², b₂ = a²b and c₂ = –(a² + b²).

So, by cross-multiplication, we have

`u/(b_1c_2 - b_2c_1) = v/(c_1a_2 - c_2a_1) = 1/(a_1b_2 - a_2b_1)`

⇒ `u/((-b)[-(a^2 + b^2)] - (a^2b)(0)) = v/((0)(ab^2) - (-a^2 - b^2)(a)) = 1/((a)(a^2b) - (ab^2)(-b))`

`⇒u/(b(a^2+b^2)) = v/(a(a^2 + b^2) )= 1/(ab(a^2 + b^2))`

`⇒u= (b(a^2+b^2))/(ab(a^2 + b^2)), v = (a(a^2 + b^2))/(ab(a^2 + b^2))`

⇒ `u = 1/a, v = 1/b`

⇒ `1/x = 1/a, 1/y = 1/b`

⇒ x = a, y = b

Hence, x = a and y = b.