Question

solve by cross multiply ​

Answer 1

Answer:

Step-by-step explanation:

$a_1x+b_1y+c1=0\\\\a_2x+b_2y+c_2=0\\\\\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{-y}{a_1c_2-a_2c_1}=\dfrac{1}{a_1b_2-a_2b_1} \\\\ax+by=a^2=>ax+by-a^2=0\\\\bx-ay=ab \rightarrow bx-ay-ab=0\\\\\dfrac{x}{-ab^2-a^3}=\dfrac{-y}{-a^2b+a^2b}=\dfrac{1}{-a^2-b^2}\\ \\y=0\\\\x=\dfrac{-ab^2-a^3}{-(a^2+b^2)}\\\\x=\dfrac{-a(a^2+b^2)}{-(a^2+b^2)}\\\\x=a$