Question

solve the following equations of elimination by substitution method iv)5x+3y=11 ,3x+5y=13​

Answer 1

Answer:

x = 1 and y = 2

Step-by-step explanation:

Let 5x + 3y = 11 - - - (1)

And 3x + 5y = 13 - - - (2)

Multiply (1) by 5 and (2) by 3

=> 25x + 15y = 55 - - - (3)

=> 9x + 15y = 39 - - - (4)

On subtracting (4) from (3)

=> 16x = 16

=> x = 1

Put This value of x in (1)

=> 3y + 5(1) = 11

=> 3y = 6

=> y = 2

Hence x = 1 and y = 2

Mark me

Answer 2

5x=11-3y
So x=11-3y/5
3(11-3y/5)+5y=13
(33-9y+25y)/5=13
33+16y=13
16y=20
Y=20/16=5/4
5x+3(5/4)=11
5x+15/4=11
5x=11-15/4
5x=44-15/4
5x=29/4
X=29/20 and y=5/4