Question

[ax + by+a] =0
[bx+ay+b] =0 solve the pair of linear equation by cross multiplication method​

Answer 1

Answer:

cross multiplication method

Ax + by = a-b becomes

(ax + by)/(a - b) = 1

bx - ay = a + b becomes

(bx-ay)/(a+b) = 1

So now we can equate the two:

(ax + by)/(a-b) = (bx - ay)/(a+b)

So by cross multiplication:

(ax + by)(a + b) = (bx - ay)(a - b)

a^2x + abx + aby + b^2y = abx - b^2x - a^2y + aby

a^2x + b^2y = -b^2x - a^2y

a^2x + b^2x = -a^2y - b^2y

(a^2 + b^2)x = -(a^2 + b^2)y

x = -y

Substituting into the first equation:

-ay + by = a-b

-(a - b)y = a-b

y = -1

So x = 1

To check: a(1) + b(-1) = a-b, b(1) - a(-1) = a + b

Step-by-step explanation:

Answer 2

Answer:

ax + by + a = 0

bx + ay + b=0

by cross multiply

0(ax+by+a)=0(bx+ay+b)

ax+by+a=bx+ay+b

• plz mark me as braniest