Question

(a-b)x+ (a+b)y =a2-2ab-b2and (a+b) (x+y)=a2+b2

Answer 1

Given Question :- Solve for x and y :

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

snd

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

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

Given pair of linear equations is

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

and

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

can be rewritten as

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

On Subtracting equation (1) from equation (2), we get

$\rm \: (a + b)x - (a - b)x={a}^{2}+{b}^{2} - {a}^{2} + 2ab + {b}^{2}$

$\rm \: (a + b - a + b)x= 2{b}^{2} + 2ab$

$\rm \: 2bx= 2{b}^{2} + 2ab$

$\rm \: 2bx= 2b(b + a)$

$\bf\implies \:\boxed{ \bf{ \:x \: = \: a + b \: \: }} - - - (3)$

On substituting the value of x in equation (2), we get

$\rm \: (a + b)(a + b) + (a + b)y = {a}^{2} + {b}^{2}$

$\rm \: {(a + b)}^{2} + (a + b)y = {a}^{2} + {b}^{2}$

$\rm \: {a}^{2} + {b}^{2} + 2ab + (a + b)y = {a}^{2} + {b}^{2}$

$\rm \: 2ab + (a + b)y = 0$

$\rm \: (a + b)y = - 2ab$

$\bf\implies \:\boxed{ \bf{ \: \: y = \: - \: \dfrac{2ab}{a + b} \: \: }}$

Hence,

$\begin{gathered}\begin{gathered}\bf\: Solution \: is \: \begin{cases} &\bf{x \: = \: a + b} \\ \\ &\bf{y \: = \: - \: \dfrac{2ab}{a + b} } \end{cases}\end{gathered}\end{gathered}$

Answer 2

$\begin{gathered}\blue{\boxed{\underline{\boxed{\bold {x = \dfrac{(a^2 + b + ab)}{(a+b)}}}}}}\\\text{And,}\\\blue{\boxed{\underline{\boxed{\bold{y = -\dfrac{2ab}{(a+b)}}}}}}\\\end{gathered}$

Step-by-step explanation:

$\begin{gathered}\text{Given pair of linear equations in two variables:}\\\sf(a-b)x + (a+b)y = a^2 - 2ab - b^2\\\sf(a+b)(x+y) = a^2 + b^2 \\\text{Simplifying the second equation, we get}\\\sf(a+b)x + (a+b)y = a^2 + b^2\\\text{A standard pair of linear equations in two variables is}\\\text{of the form:}\\ \sf{A_{1}x + B_{1}y = C_{1}}\\\text{and}\\ \sf{A_{2}x + B_{2}y = C_{2}}\\\text{On comparing the given pair of linear equations}\\\\\text{in two variables with a standard pair of linear}\\\end{gathered}$

$\begin{gathered}\text{Given pair of linear equations in two variables:}\\\sf(a-b)x + (a+b)y = a^2 - 2ab - b^2\\\sf(a+b)(x+y) = a^2 + b^2 \\\text{Simplifying the second equation, we get}\\\sf(a+b)x + (a+b)y = a^2 + b^2\\\text{A standard pair of linear equations in two variables is}\\\text{of the form:}\\ \sf{A_{1}x + B_{1}y = C_{1}}\\\text{and}\\ \sf{A_{2}x + B_{2}y = C_{2}}\\\text{On comparing the given pair of linear equations}\\\text{in two variables with a standard pair of linear}\end{gathered} \\ \begin{gathered}\\\text{equations in two variables, we get}\\\sf\begin{array}{| c | c | c | c | c| c |}{1-6} A_1 & \sf (a-b) & \sf B_1 & \sf (a+b) & \sf C_1 & \sf (a^2-2ab-b^2) \\ \sf{1-6} A_2 & \sf (a+b) & \sf B_2 & \sf (a+b) & \sf C_2 & \sf (a^2 + b^2) \\ \sf{1-6}\end{array}\\\text{Using the cross multiplication formula for solving}\\\text{a pair of linear equations in two variables, we get} \\ \sf\dfrac{x}{B_1 C_2 - B_2 C_1} = \dfrac{y}{C_1 A_2 - C_2 A_1} = -\dfrac{1}{A_1 B_2 - A_2 B_1}\\\end{gathered} \\ \begin{gathered} \sf\implies \dfrac{x}{(a+b)(a^2 + b^2) - (a+b)(a^2 - 2ab - b^2)}\\\sf= \dfrac{y}{(a^2 - 2ab - b^2)(a+b) - (a^2 + b^2)(a-b)}\\\sf= - \dfrac{1}{(a-b)(a+b) - (a+b)(a+b)}\\ \sf{\implies \dfrac{x}{a^3 + ab^2 + a^2b + b^3 - (a^3 -2a^2b - ab^2 + a^2b - 2ab^2 -b^3)}}\\\sf= \dfrac{y}{a^3 -2a^2b - ab^2 + a^2b - 2ab^2 - b^3 - (a^3 + ab^2 -a^2b - b^3)}\\ \sf= - \dfrac{1}{a^2 - b^2 - (a^2+ b^2+ 2ab)}\\ \sf\implies \dfrac{x}{a^3 + b^3 + a^2b + ab^2 - (a^3 - b^3 -a^2b - 3ab^2)}\end{gathered} \\ \begin{gathered}= \dfrac{y}{a^3 - b^3 - a^2b - 3ab^2 - a^3 - ab^2 + a^2b + b^3} \\ \sf= - \dfrac{1}{a^2 - b^2 -a^2 - b^2 - 2ab}\\ \sf\implies \dfrac{x}{a^3 + b^3 + a^2b + ab^2 -a^3 + b^3 +a^2b + 3ab^2}\\ \sf= \dfrac{y}{-4ab^2} = - \dfrac{1}{-2b^2 - 2ab}\\\sf\implies \dfrac{x}{2b^3 + 2a^2b +4ab^2} = -\dfrac{y}{4ab^2} = \dfrac{-1}{-2b(a+b)}\\ \sf\implies \dfrac{x}{2b(b + a^2 + ab)} = -\dfrac{y}{4ab^2} = \dfrac{1}{2b(a+b)}\end{gathered} \\\begin{gathered}\text {On comparing \dfrac{x}{2b(b + a^2 + ab)} with \dfrac{1}{2b(a+b)} , we get}\\ \sf{x = \dfrac{2b(a^2 + b + ab)}{2b(a+b)}}\\ \sf\implies \boxed{\underline{\boxed{\bold{x = \dfrac{(a^2 + b + ab)}{(a+b)}}}}}\\ \sf{\text{And on comparing -\dfrac{y}{4ab^2} with \dfrac{1}{2b(a+b)} , we get}}\\\sf{y = \dfrac{-4ab^2}{2b(a+b)}}\\\sf\implies \boxed{\underline{\boxed{\bold{y = -\dfrac{2ab}{(a+b)}}}}}\end{gathered}$