need guidance on an algebra problem
posted by Nikki .
20 x+3(y1)=11
2(xy)+8y = 28
multiply out
20 x+3y3=11  Is this correct?
2x+14y = 28
Using the addition method, would you multiply the first problem by 2 and the second by 20?
No. In the second multiply out, you should have gotten
2x 1y+8y=28 or
2x+7y=28
No, in the addition method
20x+3y=14
2x+14y=28
multiply the second equation by ten, then subtract one from the other.
Bob, I read the questions as
x+3(y1)=11 > x+3y = 14 and
2(xy)+8y = 28> 2x2y+8y=28 or x+3y=14
which are two identical equations.
I think the 20 was the question number.
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