Question

solve the pair of linear equation by elimination method x/a+y/b=a+b;x/a square+y/b square=2

Answer 1

Answer.

Given (x/a) + (y/b) = a + b
Multiply by (1/a) both sides, we get
(x/a2) + (y/ab) = 1 + (b/a) → (1)
Given other linear equation as (x/a2) + (y/b2) = 2  → (2)
Subtract (2) from (1), we get
(x/a2) + (y/ab) = 1 + (b/a)
(x/a2) + (y/b2) = 2 
----------------------------------
(y/ab) − (y/b2) = (b/a) − 1
⇒ y[(b − a)/ab2] = (b − a)/a
∴ y = b2
Put y = b2 in (x/a2) + (y/b2) = 2 
⇒ (x/a2) + (b2/b2) = 2 
⇒ (x/a2) + 1 = 2 
⇒ (x/a2)  = 1
∴ x = a2

Answer 2

Explanation:

Given:

$\frac{x}{a} + \frac{y}{b} = a + b \: \: \: ...eq1 \\ \\ \frac{x}{ {a}^{2} } + \frac{y}{ {b}^{2} } = 2 \: \: \: ...eq2$

To find: Value of x and y.

Solution:

Step 1: Multiply eq1 by 1/a and subtract both equations

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

Take LCM and solve

$\frac{y}{b} ( \frac{b - a}{ab} ) = \frac{b - a}{a} \\ \\ \frac{y}{b} ( \frac{ \cancel{ b - a}}{ \cancel ab} ) = \frac{ \cancel{b - a}}{\cancel a} \\ \\ \frac{y}{b} ( \frac{1}{b} ) = 1$

cross multiply,remain y in LHS

$\frac{y}{ {b}^{2} } = 1 \\ \\ y = {b}^{2}$

Step 2: Put the value of y in eq2

$\frac{x}{ {a}^{2} } + \frac{ {b}^{2} }{ {b}^{2} } = 2 \\ \\ \frac{x}{ {a}^{2} } + 1 = 2 \\ \\ \frac{x}{ {a}^{2} } = 2 - 1 \\ \\ \frac{x}{ {a}^{2} } = 1 \\ \\ x = {a}^{2}$

Final answer:

$\bold{\red{x = {a}^{2}}} \\ \\ \bold{\red{y = {b}^{2} }}$

Hope it helps you.

To learn more on brainly:

10/x+y+2/x-y=4;15x+y-5x-y=-2

https://brainly.in/question/10711299