Question 50182This question is from textbook Beginning Algebra
: I tried to use the solver on this website
(SOLVE linear system by SUBSTITUTION)
and I could not work this kind of problem. Do not know what I am doing wrong.
8x-4y=16
y=2x-4
can someone please help???
This question is from textbook Beginning Algebra
Answer by junior403(76)   (Show Source):
You can put this solution on YOUR website! This isn't too bad.
We can solve this system of equations using the elimination method.
But first we need to put each equation in standard form so that they look the same. You have...
8x - 4y = 16
y = 2x - 4
The top equation is already in standard form so we just need to get the 2x on the left side of the equation by subtracting it from both sides.
8x - 4y = 16
-2x + y = -4
Now we are ready to begin eliminating variables.
First we need to decide which coefficiant to work with in order to cancel out or "eliminate" the other.
For instance, we have 8x and -2x.
What do we need to do to the -2x in order to cancel out the 8x?
What if we multiply the second (bottom) equation by 4?
8x - 4y = 16
4(-2x + y = -4)
When we distribute, we get...
8x - 4y = 16
-8x + 4y = -16
Now we can add our system and eliminate the x (for now) and solve the equation for y. Like so...
8x - 4y = 16
-8x + 4y = -16
_______________
0 = 0
This is as far as you can go.
Since the outcome is 0 = 0, we can write the answer as...
{(x,y)\ y = 2x - 4}
This can be read as:
The variables x and y such that y = 2x -4.
I hope this helps.
Good Luck!
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