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

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

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

Using the emimination method, let us add (1) and (2):

==> 3x = 9

==> x= 9/3 = 3

Now using the substituting method substotite in (1):

x+ y = 5

==> 3 + y = 5

==> y= 5-3 = 2

==> y= 2

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.

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)

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............(1)

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

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

so x=3

put x=3 in (1) y=2

Answer {x=3;y=2}

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