Question

Solve using elimination method:
x/3 -y/4 = 11
5x/6 -y/3 = -7

Class 10 CBSE Board
Lesson:
Linear equations in two variables​

Answer 1

Answer:

I have taken out Y. From here on you can take out X as well.

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Answer 2

Solution :

$\bigstar$ By elimination method :

Given, two equation we can suppose;

$\bullet\:\sf{\dfrac{x}{3} + \dfrac{y}{4} =11...................(1)}\\\\\bullet\sf{\dfrac{5x}{6} -\dfrac{y}{3} =-7..................(2)}$

from equation (1),we get;

$\longrightarrow\sf{\dfrac{4x + 3y}{12} =11}\\\\\longrightarrow\sf{4x+3y = 132....................(3)}$

&

from equation (2),we get;

$\longrightarrow\sf{\dfrac{5x - 2y}{6} =-7}\\\\\longrightarrow\sf{5x-2y = -42....................(4)}$

Now;

We can multiply by 5 in equation (3),we get;

$\longrightarrow\sf{5(4x + 3y = 132)}\\\\\longrightarrow\sf{20x + 15y = 660...................(5)}$

We can multiply by 4 in equation (4),we get;

$\longrightarrow\sf{4(5x - 2y = -42)}\\\\\longrightarrow\sf{20x - 8y = -168.................(6)}$

∴Subtracting in equation (5) & (6), we get;

$\longrightarrow\sf{\cancel{20x-20x} + 15y - (-8y) =660 -(-168)}\\\\\longrightarrow\sf{15y + 8y = 660 + 168}\\\\\longrightarrow\sf{23y = 828}\\\\\longrightarrow\sf{y=\cancel{828/23}}\\\\\longrightarrow\bf{y=36}$

∴ Putting the value of y in equation (3),we get;

$\longrightarrow\sf{4x + 3(36) = 132}\\\\\longrightarrow\sf{4x + 108=132}\\\\\longrightarrow\sf{4x=132-108}\\\\\longrightarrow\sf{4x=24}\\\\\longrightarrow\sf{x=\cancel{24/4}}\\\\\longrightarrow\bf{x=6}$

Thus;

The value of x & y will be 6 and 36 .