Question

find X and Y by elimination method​

Answer 1

Basic Concept :-

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 reduce rhe equations to one variable.

• Step 3: Solve this new equation for this variable.

• Step 4: Substitute the value of this variable into either Equation 1 or Equation 2 above and solve for other variable.

Let's solve the problem now!!

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

Given equation is

$\rm :\longmapsto\:\dfrac{x}{3} + \dfrac{y}{4} = 11$

$\rm :\longmapsto\:\dfrac{4x + 3y}{12} = 11$

$\bf\implies \:4x + 3y = 132 - - - (1)$

Second given equation is

$\rm :\longmapsto\:\dfrac{5x}{6} - \dfrac{y}{3} = - 7$

$\rm :\longmapsto\:\dfrac{5x - 2y}{6} = - 7$

$\bf\implies \:5x - 2y = - 42 - - - (2)$

Now, multiply equation (1) by 2, we get

$\bf\implies \:8x + 6y = 264 - - - (3)$

Now, multiply equation (2) by 3, we get

$\bf\implies \:15x - 6y = - 126 - - - (4)$

On adding equation (3) and equation (4), we get

$\rm :\longmapsto\:23x = 138$

$\bf\implies \:x = 6 - - (5)$

Now substituting the value of x = 6, in equation (1), we get

$\rm :\longmapsto\:\:4 \times 6 + 3y = 132$

$\rm :\longmapsto\:\:24+ 3y = 132$

$\rm :\longmapsto\:\:3y = 132 - 24$

$\rm :\longmapsto\:\:3y = 108$

$\bf\implies \:y = 36$

Verification :-

Given first line is

$\rm :\longmapsto\:\dfrac{x}{3} + \dfrac{y}{4} = 11$

On substituting x = 6 and y = 36, we get

$\rm :\longmapsto\:\dfrac{6}{3} + \dfrac{36}{4} = 11$

$\rm :\longmapsto\:2 + 9 = 11$

$\bf\implies \:11 = 11$

Hence, Verified

Consider Second line

$\rm :\longmapsto\:\dfrac{5x}{6} - \dfrac{y}{3} = - 7$

On substituting x = 6 and y = 36, we get

$\rm :\longmapsto\:\dfrac{5 \times 6}{6} - \dfrac{36}{3} = - 7$

$\rm :\longmapsto\:5 - 12 = - 7$

$\bf\implies \: - 7 = - 7$

Hence, Verified

$\: \: \: \: \: \: \: \: \: \pink{\underbrace{\purple{\boxed{ \bf \:Solution \: is \: x = 6 \: \: and \: \: y = 36}}}}$