Question

4x +y/3 = 8/3; x/2 +3/4y = -5/2 Solve the pair of linear equation by substitution method .find x and y

Answer 1

Hi ,

4x + y/3 = 8/3

multiply each term with 3 , we get

12x + y = 8

=> y = 8 - 12x ---( 1 )

x/2 + 3y/4 = -5/2----( 2 )

Substitute y = 8 - 12x in equation ( 2 ),

x/2 + ( 3/4 )[ 8 - 12x ] = -5/2

=> x/2 + 3[ 8/4 - 12x/4 ] = -5/2

=> x/2 + 3( 2 - 3x ) = -5/2

multiply each term with 2 , we get

=> x + 6( 2 - 3x ) = -5

=> x + 12 - 18x = -5

=> -17x = -5 - 12

=> -17x = -17

=> x = ( -17 )/( -17 )

x = 1

put x = 1 , in equation ( 1 ) , we get

y = 8 - 12 × 1

y = 8 - 12

y = -4

Therefore ,

x = 1 , y = -4

I hope this helps you.

; )

Answer 2

Answer:

$\mathbb{\green{answer = \: x = 1 \: and \: y = - 4}}$

Step-by-step explanation:

=》 4x+y/3=8/3

Multiply by 3

so, we get,

=12x+y=8

= y=8-12x –––– 1

And

= x/2+3y/4= -5/2 -------2

Now put the value of y in eq-1

=》 x/2+(3/4)(8-12x)=-5/2

=》x/2+3(8/4-12x/4)=-5/2

=》x/2+3(2-3x)=-5/2

multiply each term by 2

=》x+6(2-3x)=-5

=》x+12-18x=-5

=》-17x=-17

=》x=1

Putting x=1 in equation-----1

Y=8-12×1

Y=-4

X=1 and Y=-4