Solve using cross multiplication method, ax + by = a^2; bx + ay = b^2

x=a²+ab+b²/a+b, y=-ab/a+b

Step-by-step explanation:

The system of the given equation may b written as:

ax+by-a²=0

bx+ay-b²=0

Here,

a1=a, b1=b, c1 = - a²

a2=b, b2=a and c2 = - b²

By cross multiplication we get

x/b * (-b²) - (-a²) * a = - y/a * (-b²) - (-a²) * b = 1/a*a-b*b

x/-b³+a³ = - y/-ab²+a²b = 1/a²-b²

Now

x/-b³+a³ = 1/a²-b²

x=a³-b³/a²-b²

x = (a-b) (a²+ab+b²) / (a-b) (a+b)

x=a²+ab+b²/a+b

And,

-y/-ab²+a²b = 1/a²-b²

-y=a²b - ab²/a²-b²

y=ab²-a²b/a²-b²

y=ab (b-a) / (a-b) (a+b)

y = - ab (a-b) / (a-b) (a+b)

y = - ab/a+b

Hence x=a²+ab+b²/a+b, y=-ab/a+b ...