SOLUTION: hey can someone plz help me solve this system of equations {{{2x^2-6x+4=0}}} 3y-12=-4x thank you

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Question 233407: hey can someone plz help me
solve this system of equations
2x%5E2-6x%2B4=0
3y-12=-4x
thank you
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
system%282x%5E2-6x%2B4=0%2C3y-12=-4x%29

The first equation contains only one unknown, x, so we
can solve it for x separately:

2x%5E2-6x%2B4=0

Divide every term by 2

%282x%5E2%29%2F2-%286x%29%2F2%2B4%2F2=0%2F2

x%5E2-3x%2B2=0

Factor the left side:

%28x-2%29%28x-1%29=0

Set each factor equal to 0:

x-2=0
x=2

x-1=0
x=1

Now we substitute in the other equation

3y-12=-4x%29

Substituting x=2

3y-12=-4x%29
3y-12=-4%282%29%29
3y-12=-8
3y=4%29
y=4%2F3

So one solution is (2,4%2F3)

Substituting x=1

3y-12=-4x%29
3y-12=-4%281%29%29
3y-12=-4
3y=8%29
y=8%2F3

So the other solution is (1,8%2F3)

Edwin