Posted by **Cassie** on Friday, February 27, 2009 at 6:01pm.

Solve the system of equations using substitution.

x + 4y = -5

3x - 8y = 45

When I tried to solve it, I got x=-17 and y=-3. When I plugged these into the first equation, it didn't work out at all. Could someone please show me the steps so I can see where I went wrong? Thank you!

- Algebra -
**Reiny**, Friday, February 27, 2009 at 6:24pm
I doubled the first and added it to the second to get

5x = 35,

so x = 7, subbing that back I got y = -3

those values verify.

- Algebra -
**Cassie**, Friday, February 27, 2009 at 6:28pm
MC-- 7th grade algebra.

Thanks for your help, Reiny!

- Algebra -
**cc**, Friday, February 27, 2009 at 11:20pm
since you need to solve the system of equations by substitution, solve the first equation for x by subtracting 4y from both sides. You will get x = -5 - 4y.

Next substitute this expression into the x of the 2nd equation to get

3(-5 - 4y)- 8y =45

Now use the distributive property to get

-15 -12y - 8y =45

simplify to get

-15 -20y = 45 (when you combine to negatives you get a negative answer)

add 15 to both sides and you get

-20y = 60

divide both sides by -20 to get y = -3.

Now replace this answer into the first equation or the x = -5 - 4y equation to get x= -5 - 4(-3) = -5 + 12 = 7

answer (7,-3)

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