What is the solution (x, y) to the system of equations below? 3x+4y=-23 2y-x=-19

A) (-5,-2)
B) (3,-8)
C) (4,-6)
D) (9,-6)

We are going to perform The Elimination Method to find the values of x and y.

Here we have two equations.

3x + 4y = -23 ( Take this as equation 1)

2y - x = 19 Here we can write it as -x + 2y = 19 ( Take this as equation 2)

Before subtracting, we need to check whether either of the coefficients of the variables in both the equations are same in both the both the equations.

Here we find that both the variable in both the equations do not have same coefficients.

So all we have to do is, multiply the coefficient of any variable of the second equation to the first and the same process to the second one also. ( You need to check whether anything is working)

So, from the second equation, I choose 2 ( coefficient of y)

Therefore, 2( 3x + 4y= -23)

It will be 6x + 8y = -46

Now from the first equation i chose 4 ( coefficients of y in both the equations to be cross multiplied.

Therefore, 4( -x + 2y =19)

It will be -4x + 8y = 76

Now we got equal coefficients of y

Now, lets begin our Elimination method. ( subtracting the two equations )

6x + 8y = -46

-( - 4x + 8y = 76)

Therefore we will get 10x = 30

By normal division, You will get x = 3

Now all we have to do is to substitute the value of x (3) to any of the equation. ( One before the multiplication process)

So I choose equation 2=

2y -x = 19

we know that x = 3

So 2y + 3 =19

2y = 19 - 3

2y = 16

Now by Transposing method,

y = 16/2= 8

Now we got the values for x and y

Therefore ( x,y )

= (-3, 8)

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