Question

x-y=a+b,ax+by=a2-b2 in cross multiplication method

Answer 1

GIVEN :

The equations are x-y=a+b, $ax+by=a^2-b^2$ in cross multiplication method

TO FIND :

The values of x and y from the given equations.

SOLUTION :

Given equations are x-y=a+b, $ax+by=a^2-b^2$

It can be written as

$x-y-(a+b)=0\hfill (1)$

$ax+by-(a^2-b^2)=0\hfill (2)$

By using the cross multiplication method:

The formula for the equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is given by

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

Here $a_1=1$ $b_1=-1$ and $c_1=-(a+b)$

$a_2=a$ $b_2=b$ and $c_2=-(a^2-b^2)$

Substitute the values in formula we have

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

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

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

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

Equating $\frac{x}{a^2+ab}=\frac{1}{b+a}$

$\frac{x}{a(a+b)}=\frac{1}{b+a}$

$x=\frac{a(a+b)}{b+a}$

⇒ x=a

Equating $\frac{y}{-ba-b^2}=\frac{1}{b+a}$

$\frac{y}{-b(a+b)}=\frac{1}{b+a}$

$y=\frac{-b(a+b)}{b+a}$

⇒ y=-b

∴ the values of x and y from the given equations by using Cross Multiplication method are x=a and y=-b