Question

Q1.solve the pair of linear equations. x/a+y/b=a+b and x/a^2+y/b^2=2 (a,b not equal to 0)​

Answer 1

Step-by-step explanation:

Given:

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

To find: Value of x and y.

Solution:

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

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

Take LCM and solve

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

cross multiply,remain y in LHS

$\begin{gathered} \frac{y}{ {b}^{2} } = 1 \\ \\ y = {b}^{2} \\ \\ \end{gathered}$

Step 2: Put the value of y in eq2

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

Final answer:

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

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