Question

a2/x-b2/y=0,a2b/x+b2a/y=a+b, solve for x by cross multiplication method

Answer 1

this is the required answer

Answer 2

Answer:  The solution is (x, y) = (a², b²).

Step-by-step explanation: The given system of equations to solve is:

$\dfrac{a^2}{x}-\dfrac{b^2}{x}=0~~~~~~~~~~~~~~~~~(i)\\\\\dfrac{a^2b}{x}+\dfrac{b^2a}{y}=a+b~~~~~~~~~(ii)$

Let us consider that

$\dfrac{1}{x}=u,~~~\dfrac{1}{y}=v.$

So, equations (i) and (ii) becomes

$a^2u-b^2v=0~~~~~~~~~~~~~~~~(iii)\\\\a^2bu+b^2av=a+b\\\\\Rightarrow a^2bu+b^2av-(a+b)=0~~~~~~~~~(iv)$

By the method of cross-multiplication, we have from equations (iii) and (iv) that

              u                        v                         1

-b^2                 0                         a^2                      -b^2

b^2a              -(a+b)                   a^2b                      b^2a

Therefore, we have

$\dfrac{u}{b^2\times(a+b)-0\times b^2a}=\dfrac{v}{0\times a^2b+(a+b)\times a^2}=\dfrac{1}{a^2\times b^2a+a^2b\times b^2}\\\\\\\Rightarrow \dfrac{u}{b^2(a+b)}=\dfrac{v}{a^2(a+b)}=\dfrac{1}{a^2b^2(a+b)}\\\\\\\Rightarrow u=\dfrac{b^2(a+b)}{a^2b^2(a+b)}=\dfrac{1}{a^2},\\\\\\v=\dfrac{a^2(a+b)}{a^2b^2(a+b)}=\dfrac{1}{b^2}.$

Hence, we will get from here that

$x=\dfrac{1}{u}=a^2,~~~y=\dfrac{1}{y}=b^2.$

Thus, the required solution is (x, y) = (a², b²).