Question

RS aggarwal maths.
Class - 10
Ex- 3B
question num 36
pls solve.
Answer is x=1, y=1​

Answer 1

Answer:

take a=3x+y and b=3x-y

Step-by-step explanation:

by taking l.c.m and by using the above method

x=y=1

hope this helps

Answer 2

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

Given pair of linear equation are

$\rm :\longmapsto\: \dfrac{1}{3x + y} + \dfrac{1}{3x - y} = \dfrac{3}{4} - - - (1)$

and

$\rm :\longmapsto\: \dfrac{1}{2(3x + y)} - \dfrac{1}{2(3x - y)} = - \: \dfrac{1}{8} - - - (2)$

Let we assume that,

$\red{\rm :\longmapsto\:\dfrac{1}{3x + y} = u - - - (3)}$

$\red{\rm :\longmapsto\:\dfrac{1}{3x - y} = v - - - (4)}$

So,

Equation (1) and (2) reduces to,

$\rm :\longmapsto\:u + v = \dfrac{3}{4}$

$\rm :\longmapsto\:4u + 4v = 3 - - - - (5)$

and

$\rm :\longmapsto\:\dfrac{u}{2} - \dfrac{v}{2} = - \dfrac{1}{8}$

$\rm :\longmapsto\:4u - 4v = - 1 - - - (6)$

Now, on adding equation (5) and (6), we get

$\rm :\longmapsto\:8u = 2$

$\bf\implies \:u = \dfrac{1}{4} - - - (7)$

On substituting the value of u in equation (5), we get

$\rm :\longmapsto\:1 + 4v = 3$

$\rm :\longmapsto\:4v = 3 - 1$

$\rm :\longmapsto\:4v = 2$

$\bf\implies \:v = \dfrac{1}{2} - - - (8)$

Now, on Substituting the values of u and v in equation (3) and (4), we get

$\rm :\longmapsto\:3x + y = 4 - - - (9)$

and

$\rm :\longmapsto\:3x - y = 2 - - - (10)$

On adding equation (9) and (10), we get

$\rm :\longmapsto\:6x = 6$

$\bf\implies \:x = 1 - - - (11)$

On substituting the value of x in equation (9), we get

$\rm :\longmapsto\: 3 + y = 4$

$\bf\implies \:y = 1$

$\purple{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Hence-\begin{cases} &\sf{x = 1} \\ &\sf{y = 1} \end{cases}\end{gathered}\end{gathered}}$

Basic Concept Used :-

The Elimination Method

Step 1: Multiply each equation by a suitable number so that the two equations have the same leading coefficient.

Step 2: Subtract the second equation from the first to eliminate one variable

Step 3: Solve this new equation for other variable.

Step 4: Substitute the value of variable thus evaluated into either Equation 1 or Equation 2 and get the value other variable.