SOLUTION: I am studying for my 8th grade algebra final and i dont understand the way that on this linear system 4x+5y = 7 6x-2y = -18 how do i do this

Question 2753: I am studying for my 8th grade algebra final and i dont understand the way that on this linear system 4x+5y = 7
6x-2y = -18
how do i do this x = (7 -5y)/4
Substituting this expression into the second equation:
6(7+5y)/4-2y = -18

and from after that how do i simplify this 6(7+5y)/4-2y = -18
into this (-38)y = -114
I listen in class but i was absent on some very important days and this doesnt make sence
I thank anyone who could help.

Answer by WannabeCAgirl83(35) (Show Source): You can put this solution on YOUR website!
You�re right. It doesn�t make sense that way because there are errors... ;-)

Your first aim is to bring x all alone on one side of the equation in order to find out what it equals, so that you can substitute x by its equivalent and thereby �get rid� of x as an unknown variable. Here�s how to do it:

4x + 5y = 7 | - 5y on both sides
4x = 7 - 5y | / 4 on both sides
x = 7/4 - 5/4y

As you found out that x is equal to 7/4 - 5/4y, you can substitute it in the next equation:

6x - 2y = -18
6(7/4 - 5/4y) - 2y = -18

Thereby you �got rid� of x and are now only dealing with one unknown variable (y) instead of two (x and y).

Your second aim is now to �get rid� of y the same way you �got rid� of x: You have to bring y all alone on one side of the equation to find out what it equals. Here�s how:

6(7/4 - 5/4y) - 2y = -18
10 � - 7 �y - 2y = - 18
10 � - 9 � y = - 18 | - 10 �
- 9 �y = - 28 � | / -9 �
y = 3

As you found out that y is equal to 3, you can substitute it in one of your equations in order to get a value for x:

4x + 5y = 7
4x +5(3) = 7
4x + 15 = 7 | - 15
4x = -8 | / 4
x = -2

Now you �got rid� of both of your unknown variables as you found out that:

x = -2 and y = 3

Let�s check if the values for x and y are correct by putting them in the original equations:

4x + 5y = 7
4(-2) + 5(3) = 7 → true

6x - 2y = - 18
6(-2) - 2(3) = - 18 → true

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