Question

ax+by=(a-b) , bx - ay =(a+b) simultaneous linear equations using cross multiplication

Answer 1

Answer:

$x=1\,,\,y=-1$

Step-by-step explanation:

Given: $ax+by=a-b\,,\,bx-ay=a+b$

To solve: the given linear equations

Solution:

Consider the equations:

$A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0$

By method of cross multiplication:

$\frac{x}{B_1C_2-B_2C_1}=\frac{y}{C_1A_2-C_2A_1}=\frac{1}{A_1B_2-A_2B_1}$

For equations: $ax+by=a-b\,,\,bx-ay=a+b$

$ax+by-(a-b)=0\\bx-ay-(a+b)=0$

Take $A_1=a\,,\,B_1=b\,,\,C_1=-(a-b)\,,\,A_2=b\,,\,B_2=-a\,,\,C_2=-(a+b)$

So,

$\frac{x}{-b(a+b)-a(a-b)}=\frac{y}{-b(a-b)+a(a+b)}=\frac{1}{-a^2-b^2}\\\frac{x}{-ab-b^2-a^2+ab}=\frac{y}{-ab+b^2+a^2+ab}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}\,,\,\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\x=1\,,\,y=-1$