We have to find the solution of :

x + 2y = 4 ...(1)

2x - 3y = 5 ...(2)

2*(1) - (2)

=> 2x + 4y - 2x + 3y = 8 - 5

=> 7y = 3

=> y = 3/7

substitute in (1)

x = 4 - 3*2/7

=> 4 - 6/7

=> 22/7

**The solution of the system of equations is x = 22/7 and y = 3/7**

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

x = 4 - 2y

2(4 - 2y) - 3y = 5

8 - 4y - 3y = 5

8 - 7y = 5

-7y = 5 - 8

-7y = -3

y = -3/-7

y =3.14

x + 2(3/7) = 4

2(3/7) =0.86

x + .86 = 4

subtract .86

x = 4 - .86

y = 3.14

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

you can do this many ways one of those will be by subtracting 2y from x + 2y = 4

x = 4 - 2y

now plug in this number as x in the other problem:

2(4 - 2y) - 3y = 5

distribute the 2:

8 - 4y - 3y = 5

combine like terms

8 - 7y = 5

subtract the 8:

-7y = 5 - 8

-7y = -3

divide by -7

y = -3/-7

the negatives cancel each other out:

**y = 3/7 or .43**

to find x just plug in 3/7 as y in any of the equations above:

x + 2(3/7) = 4

2(3/7) = 0.857142857 or 0.86

simplify:

x + .86 = 4

subtract .86

x = 4 - .86

**x = 3.14**

Given the equations.

x+2y = 4..(1)

x-3y = 5..(2), we are required to solve for x and y.

We compute (1) - (2):

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

x+2y-x+3y = 4-5 = -1

x's cancels and we get

2y+3y = -1

5y = -1

y= -1/5.

Put y = -1/5 in (2): x -3y = 5

x - 3(-1/5) = 5

x+3/5 = 5

x = 5-3/5 = 22/5.

**x = 22/5 , y = -1/5**.