Question

solve x& y by cross multiplication method x+y=a+b. &. ax-by=a^2-b^2​

Answer 1

Answer:

x+y = a+b

x = a+b-y

x = a+b-y,y∈ℝ,a∈ℝ,b∈ℝ

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Answer 2

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

Given pair of lines are

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

and

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

Now,

Using Cross Multiplication method, we have

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

So,

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

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

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

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

Multiply by a + b, we get

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

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

Verification :-

Consider Equation (1)

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

On Substituting x = a and y = b, we get

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

Hence, Verified

Consider Equation (2)

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

On Substituting the values of x and y, we get

$\rm :\longmapsto\:a(a) - b(b) = {a}^{2} - {b}^{2}$

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

Hence, Verified