2x+3y= 22------ 1

3x+2y= 28------ 2

We will do elimination by equating coefficient:

In this we multiply either 1 or both the equations by suitable numbers to ensure that the coefficient of one variably is equal.

In this case we will multiply equation 1 by 3 and 2 by 2 to eliminate x. We get

6x+9y= 66------- 3

6x+4y= 56----- 4

Subtract equation 4 from 3 we get

5y = 10 that gives y = 2, plug this value in any equation 1 or 2.

Let us plug the value in equation 1 we get

2x + 3 x 2 = 22

2x = 22 – 6 = 16

X = 8

Thus the solution is x = 8 y = 2.

We can also solve the equations by elimination by actual substitution:

Let us take the second set of equations:

3x-2y= 28 ---- [1] from this equation we have 3x= 28+2y OR x = (28+2y)/3

Plug in this value for x in other equation i.e., 4x+5y= -1----[2] , we will get:

4 x (28+2y)/3 +5y = -1 Multiplying by 3 we get:

112 + 8y + 15y = -3

23y = -115 that gives y = -5 Plugging this value in any equation ( say equation 1 we get:

3x-2(-5)= 28 OR 3x +10 = 28 that gives x = 6

Thus the solution is x = 6, y = -5