Question

solve for x and y (a-b)x+(a+b)y=a2-2ab-b2 (a+b)(x+y)=a2+b2.

I want the solution as quickly as possible

Answer 1

Hey user

Here is your answer :-

$(a - b)x + (a + b)y = {a}^{2} - 2ab - {b}^{2} \: \: \: \: \: \: \: ...(1) \\ \\ (a + b)(x + y) = {a}^{2} + {b}^{2} \\ \\ (a + b)x + (a + b )y = {a}^{2} + {b}^{2} \: \: \: \: \: \: \: ...(2) \\ \\ subtracting \: \: equations \: (1) \: from \: equation\: (2) we \: get \: \\ \\ (a + b)x - (a - b)x = 2ab + 2{b}^{2} \\ \\ ax + bx - ax + bx = 2b(a + b) \\ \\ 2bx = 2b(a + b) \\ \\ x = a + b \\ \\ putting \: the \: value \: of \: x \: in \: equation \: (2) \: we \: get \: \\ \\ (a + b )( a+ b) + ( a+ b)y = {a}^{2} + {b}^{2} \\ \\ {a}^{2} + 2ab + {b}^{2} + ( a+ b)y = {a}^{2} + {b}^{2} \\ \\ ( a+ b)y = - 2ab \\ \\ y = \frac{ - 2ab}{a + b}$

thank you.

Answer 2

Answer:

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

Step-by-step explanation:

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

$\\\text{equations in two variables, we get}\\\\\begin{array}{| c | c | c | c | c| c |}\cline{1-6} A_1 & (a-b) & B_1 & (a+b) & C_1 & (a^2-2ab-b^2) \\\cline{1-6} A_2 & (a+b) & B_2 & (a+b) & C_2 & (a^2 + b^2) \\\cline{1-6}\end{array}\\\\\\\text{Using the cross multiplication formula for solving}\\\\\text{a pair of linear equations in two variables, we get}\\\\\dfrac{x}{B_1 C_2 - B_2 C_1} = \dfrac{y}{C_1 A_2 - C_2 A_1} = -\dfrac{1}{A_1 B_2 - A_2 B_1}$

$\implies \dfrac{x}{(a+b)(a^2 + b^2) - (a+b)(a^2 - 2ab - b^2)}\\\\\\= \dfrac{y}{(a^2 - 2ab - b^2)(a+b) - (a^2 + b^2)(a-b)}\\\\\\= - \dfrac{1}{(a-b)(a+b) - (a+b)(a+b)}\\\\\\\\\implies \dfrac{x}{a^3 + ab^2 + a^2b + b^3 - (a^3 -2a^2b - ab^2 + a^2b - 2ab^2 -b^3)}\\\\\\= \dfrac{y}{a^3 -2a^2b - ab^2 + a^2b - 2ab^2 - b^3 - (a^3 + ab^2 -a^2b - b^3)}\\\\\\= - \dfrac{1}{a^2 - b^2 - (a^2+ b^2+ 2ab)}\\\\\\\\\implies \dfrac{x}{a^3 + b^3 + a^2b + ab^2 - (a^3 - b^3 -a^2b - 3ab^2)}$

$= \dfrac{y}{a^3 - b^3 - a^2b - 3ab^2 - a^3 - ab^2 + a^2b + b^3}\\\\\\= - \dfrac{1}{a^2 - b^2 -a^2 - b^2 - 2ab}\\\\\\\\\implies \dfrac{x}{a^3 + b^3 + a^2b + ab^2 -a^3 + b^3 +a^2b + 3ab^2}\\\\\\= \dfrac{y}{-4ab^2} = - \dfrac{1}{-2b^2 - 2ab}\\\\\\\\\implies \dfrac{x}{2b^3 + 2a^2b +4ab^2} = -\dfrac{y}{4ab^2} = \dfrac{-1}{-2b(a+b)}\\\\\\\implies \dfrac{x}{2b(b + a^2 + ab)} = -\dfrac{y}{4ab^2} = \dfrac{1}{2b(a+b)}$

$\text{On comparing \dfrac{x}{2b(b + a^2 + ab)} with \dfrac{1}{2b(a+b)} , we get}\\\\\\x = \dfrac{2b(a^2 + b + ab)}{2b(a+b)}\\\\\\\implies \boxed{\underline{\boxed{\bold{x = \dfrac{(a^2 + b + ab)}{(a+b)}}}}}\\\\\\\\\text{And on comparing -\dfrac{y}{4ab^2} with \dfrac{1}{2b(a+b)} , we get}\\\\\\y = \dfrac{-4ab^2}{2b(a+b)}\\\\\\\implies \boxed{\underline{\boxed{\bold{y = -\dfrac{2ab}{(a+b)}}}}}$