Question

algebra 2 problem, much help needed

Answer 1

hmm Is it $65 and $70?

Answer 2

not sure it won't tell me the right answer :(

Answer 3

From the problem you are given the following:\[15x + 5y = 1325\]\[x+y=135\]
Using a system of equations:\[\left[\begin{matrix}15 & 5 \\ 1 & 1\end{matrix}\right]\left(\begin{matrix}1325 \\ 135\end{matrix}\right)\] Multiply the second equation by 5 and subtract it from the first.

Answer 4

or can i just subtract the y and move it to the other side therefore x=135-y and then substitute that in?

Answer 5

When using this method you are trying to eliminate one of the variables. By multiplying the one equation by 5 you get \[\left[\begin{matrix}15 & 5 \\ 5 & 5\end{matrix}\right]\left(\begin{matrix}1325 \\ 675\end{matrix}\right)\] which leaves you 10x=650, x=65, then substitute.

Answer 6

The answer to your question is yes, I worked it out as you suggested and came up with the same answers. :)

Answer 7

what was the answer? :)

Answer 8

I showed you my answer for x, what did you get for y? :)

Answer 9

but for my problem what would be x and what would be y?

Answer 10

To answer that you need to examine the question asked and compare it to how you set up the equation. In the case of my example, the mechanic that put in 15 hours is 'x' and the mechanic that put in 5 hours is 'y'.
The question tells you that:
(a mechanics hours * some rate) + (another mechanic hours * a second rate) equaled $1325
also that 'some rate + a second rate' equaled $135
So you see it is all in how you set your problem up.