Question

Solve the system of equations : ax + by =1, bx + ay = 2ab/(a² + b²).

Answer 1

Answer:

put the value of y in eq 1 , we get the value of x also

Answer 2

When the equations are solved, the value of $\bold{x=\frac{a}{a^{2}+b^{2}}}$ and $\bold{y=\frac{b}{a^{2}+b^{2}}}$.

Given:

$\bold{a x+b y-1=0}$

$\bold{b x+a y-\frac{2 a b}{a^{2}+b^{2}}=0}$

Solution:

Here, the equation are solved by applying cross multiplication method.

The formula used to solve the equation is:

$\bold{\frac{\mathrm{x}}{\mathrm{b}_{1} \mathrm{c}_{2}-\mathrm{b}_{2} \mathrm{c}_{1}}=\frac{\mathrm{y}}{\mathrm{c}_{1} \mathrm{a}_{2}-\mathrm{c}_{2} \mathrm{a}_{1}}=\frac{1}{\mathrm{a}_{1} \mathrm{b}_{2}-\mathrm{a}_{2} \mathrm{b}_{1}}}$

The values of the variables are:

$\bold{a_{1}=a}$

$\bold{b_{1}=b}$

$\bold{c_{1}=-1}$

$\bold{a_{2}=b}$

$\bold{c_{2}=-\frac{2 a b}{a^{2}+b^{2}}}$

On substituting the values in the formula, we get,

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

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

On equating, x and y separately, we get,

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

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

Now, the values of x and y is:

$\bold{x=\frac{a}{a^{2}+b^{2}}}$

$\bold{y=\frac{b}{a^{2}+b^{2}}}$