X = 3 and Y = 2. Here is how we find this.

What we have to do is use the first equation to get a value for y. This is quite easy because all we must do is say

x + y = 5 so y = 5 -x

Now we plug that value into the second equation.

2x - (5-x) = 4

This becomes

2x - 5 + x = 4 (because you are subtracting negative x, you are in effect adding x)

3x - 5 = 4

3x = 9

x = 3

If x = 3, then 3 + y = 5 and y = 2

x+ y = 5

2x - y = 4

First multiply everything in the first equation by 2 .

By multiplying everything in the first equation by 2 , you should get :

2x + 2y = 10

2x - y = 4 now subtract the two equation .

By subtracting the two equations , you should get :

3y = 6 divide both sides by 3 .

By dividing both sides by 3 , you should get :

y = 2 which is your answer for " y "

Now plug " y " into one of the equation

x + 2 = 5 subtract both sides by 2 .

By subtracting both sides by 2 , you should get

x = 3 which is your answer for " x "

So " x " equals to 3 and " y " equals to 2

x+ y = 5............(1)

2x-y = 4.............(2)

(1)+(2) => 3x=9

so x=3

put x=3 in (1) y=2

Answer {x=3;y=2}

Because the equations of the system are linear, we'll use the matrix formed by the coefficients of the variables to calculate the determinant of the system:

1 1

det A =

2 -1

det A = -1-2=-3

det A = -3 different from zero.

Now, we'll calculate x:

x = det x/ det A

5 3

det x =

4 -3

det x = -15 - 12 = -27

x = -27/-3

**x = 9**

Now, we'll calculate y:

1 5

det y =

2 4

det y = 4 - 10

det y = -6

y = det y/ det A

y = -6/-3

**y = 2**

**The solution of the system is: {(9 , 2)}**

x+y = 5

2x-y = 4.

To solve for x and y:

From the 2nd equation, y = 2x-4. Put this in 1st equation. x+y =5.

x+ 2x- 4 =5.

3x=5+4 =3

x = 9/3 =3.

Put x= 3 and from 1st equation, x+y =5, we get: 3+y =5. So y = 5-3 =2.

So (x,y) = (3,2)