Question

plz solve.. the pair of linear equations ​

Answer 1

Topic :-

Linear Equations

Given :-

$\dfrac{x}{a}+\dfrac{y}{b}=a+b$

$\dfrac{x}{a^2}+\dfrac{y}{b^2}=2$

To Find :-

Values of x and y.

Methodology :-

There are many methods to do it. Here, we will be using Substitution Method.

In Substitution Method, we find value of x or y from first equation then substitute it into second equation to calculate actual values of x and y.

Solution :-

$\dfrac{x}{a}+\dfrac{y}{b}=a+b$

$\dfrac{bx+ay}{ab}=a+b$

$bx+ay=ab(a+b)$

$bx+ay=a^2b+ab^2$

$bx=a^2b+ab^2-ay$

$x=\dfrac{a^2b+ab^2-ay}{b}$

Now, substitute value of x in second equation.

$\dfrac{x}{a^2}+\dfrac{y}{b^2}=2$

$\dfrac{xb^2+ya^2}{a^2b^2}=2$

$xb^2+ya^2=2a^2b^2$

$\dfrac{(a^2b+ab^2-ay)b^2}{b}+ya^2=2a^2b^2$

$(a^2b+ab^2-ay)b+a^2y=2a^2b^2$

$a^2b^2+ab^3-aby+a^2y=2a^2b^2$

$a^2y-aby=a^2b^2-ab^3$

$ay(a-b)=ab^2(a-b)$

$ay=ab^2$

$y=b^2$

Now, substitute value of y in any equation to get value of x.

$\dfrac{x}{a^2}+\dfrac{y}{b^2}=2$

$y=b^2$

$\dfrac{x}{a^2}+\dfrac{b^2}{b^2}=2$

$\dfrac{x}{a^2}+1=2$

$\dfrac{x}{a^2}=1$

$x=a^2$

Answer :-

So, value of

$x=\bold{a^2}\:\:and$

$y=\bold{b^2}$