SOLUTION: Solve this system of linear equations using substitution: x - 7y = -11; and 5x + 2y = -18

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Question 1190990: Solve this system of linear equations using substitution: x - 7y = -11; and 5x + 2y = -18
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Let's solve the first equation for x
x - 7y = -11
x = -11 + 7y

We'll then plug this into the other equation.
Wherever you see an x, replace it with (-11+7y)
The term "substitute" effectively means "replace".
So,
5x + 2y = -18
5(-11+7y) + 2y = -18
-55+35y + 2y = -18
-55+37y = -18
37y = -18+55
37y = 37
y = 37/37
y = 1

We can now find x based on that
x = -11+7y
x = -11+7(1)
x = -11+7
x = -4

Solution: x = -4 and y = 1

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Check:

Plug x = -4 and y = 1 into the first original equation
x-7y = -11
-4-7(1) = -11
-4-7 = -11
-11 = -11

Repeat for the other original equation as well
5x + 2y = -18
5(-4) + 2(1) = -18
-20 + 2 = -18
-18 = -18
For each equation above, we get a true statement (i.e. the same number on both sides).
The solution is fully confirmed.


Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve this system of linear equations using substitution:
x - 7y = -11
5x + 2y = -18
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     x - 7y = -11      (1)
    5x + 2y = -18      (2)


From equation (1),  express  x = 7y - 11  and substitute it into equation (2). You will get

    5*(7y - 11) + 2y = -18.


Thus you have an equation for one single unknown y.   Simplify it and find y


    35y - 55 + 2y = -18

    35y + 2y = 55 - 18

        37y  =    37.

          y  =    37%2F37 = 1.


Now substitute this value of y into equation (1) to get


    x - 7*1 = -11,

    x       = -11 + 7 = -4.


Answer.  The solution is  x= -4;  y= 1.


    Check the solution on your own by substituting the found values into the original equations.

Solved.

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On the Substitution method,  see the lesson
    - Solution of the linear system of two equations in two unknowns by the Substitution method
in this site.