Question

x+y = a+b
ax - by = a² - b²
solve using cross multiplication method​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:x + y = a + b$

and

$\rm :\longmapsto\:ax - by = {a}^{2} - {b}^{2}$

By using Cross Multiplication method,

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf 1 & \sf a + b & \sf 1 & \sf 1\\ \\ \sf - b & \sf {a}^{2} - {b}^{2} & \sf a & \sf - b\\ \end{array}} \\ \end{gathered}$

Now,

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

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

$\rm :\longmapsto\: \sf \:\dfrac{x }{{a}^{2}+ba} =\dfrac{y}{ab + {b}^{2}} = \dfrac{1}{b + a}$

$\rm :\longmapsto\:\sf \: \dfrac{x}{a(a + b)} = \dfrac{y}{b(a + b)} = \dfrac{1}{a + b}$

On multiply each term by a + b, we get

$\rm :\longmapsto\:\sf \: \dfrac{x}{a} = \dfrac{y}{b} = 1$

$\bf\implies \:x = a \: \: \: \: and \: \: \: \: \: y = b$