Question

5(x-2)+3y=1;3(x+1)+5(y-3)=1 solve this equation by elumination method class 10​

Answer 1

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

Given pair of equations are

5(x - 2) + 3y = 1 -----[1]

3(x + 1) + 5(y - 3) = 1 -----[2]

Now,

Equation (1) can be reduced further as

$\rm :\longmapsto\:5(x - 2) + 3y = 1$

$\rm :\longmapsto\:5x - 10+ 3y = 1$

$\rm :\longmapsto\:5x+ 3y = 1 + 10$

$\rm :\longmapsto\:5x+ 3y = 11 - - - - - (3)$

Now,

Equation (2) can be further reduced as

$\rm :\longmapsto\:3(x + 1) + 5(y - 3) = 1$

$\rm :\longmapsto\:3x +3 + 5y - 15 = 1$

$\rm :\longmapsto\:3x+ 5y - 12 = 1$

$\rm :\longmapsto\:3x+ 5y = 1 + 12$

$\rm :\longmapsto\:3x+ 5y = 13 - - - (4)$

Now, we have equations in simplified form as

$\rm :\longmapsto\:5x+ 3y = 11- - - (3)$

and

$\rm :\longmapsto\:3x+ 5y = 13 - - - (4)$

Now,

Multiply equation (3) by 3 and equation (4) by 5, we get

$\rm :\longmapsto\:15x+ 9y = 33- - - (5)$

and

$\rm :\longmapsto\:15x+ 25y = 65 - - - (6)$

On Subtracting equation (5) from equation (6), we get

$\rm :\longmapsto\:16y = 32$

$\bf\implies \:y = 2$

Now, Substitute y = 2 in equation (4), we get

$\rm :\longmapsto\:3x + 5 \times 2 = 13$

$\rm :\longmapsto\:3x + 10= 13$

$\rm :\longmapsto\:3x= 13 - 10$

$\rm :\longmapsto\:3x= 3$

$\bf\implies \:x = 1$

Hence, the solution set of given equation is

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{\underbrace{\purple{ \boxed{ \bf{x = 1}}} \: \: \: \pink{and} \: \: \: \purple{ \boxed{ \bf{y = 2}}}}}$

Verification :-

Consider equation (1)

$\rm :\longmapsto\:5(x-2)+3y=1$

Put the value of x = 1 and y = 2, we get

$\rm :\longmapsto\:5(1-2)+3(2)=1$

$\rm :\longmapsto\: - 5 + 6 = 1$

$\rm :\longmapsto\: 1= 1$

Hence, Verified

Consider Equation (2)

$\rm :\longmapsto\:3(x+1)+5(y-3)=1$

On substituting the value of x and y, we get

$\rm :\longmapsto\:3(1+1)+5(2-3)=1$

$\rm :\longmapsto\:3(2)+5( - 1)=1$

$\rm :\longmapsto\:6 - 5 = 1$

$\rm :\longmapsto\:1 = 1$

Hence, Verified