Question

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

`x/a = y/b`

ax + by = a² + b²

Answer 1

Given: `x/a = y/b`, ax + by = a² + b²

Step-wise calculation:

1. From `x/a = y/b`, cross-multiply:

bx = ay 

⇒ bx – ay = 0

Write both equations in standard form:

(1) bx – ay + 0 = 0 

(2) ax + by – (a² + b²) = 0

2. Use the cross‑multiplication formula for a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0: 

`x/(b_1c_2 - b_2c_1) = y/(c_1a_2 - c_2a_1)`

= `1/(a_1b_2 - a_2b_1)`

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

3. Compute the needed combinations:

b₁c₂ – b₂c₁ = (–a) × (–(a² + b²)) – b × 0 

= a(a² + b²) 

= a³ + ab²

c₁a₂ – c₂a₁ = 0 × a – (–(a² + b²)) × b 

= b(a² + b²) 

= a²b + b³ 

a₁b₂ – a₂b₁ = b × b – a × (–a) 

= b² + a² 

= a² + b²

4. Thus `x/(a^3 + a b^2) = y/(a^2b + b^3)` 

= `1/(a^2 + b^2)`

Therefore `x = (a^3 + ab^2)/(a^2 + b^2)`

= `a xx (a^2 + b^2)/(a^2 + b^2)`

= a 

`y = (a^2b + b^3)/(a^2 + b^2)`

= `b xx (a^2 + b^2)/(a^2 + b^2)`

= b

Provided a² + b² ≠ 0 i.e., a and b are not both zero, the solution is x = a, y = b.