4x + 13y = 40 4x + 3y = -40

Question

e.mccormick · Answer

System of equations?

e.mccormick · Answer

If so, did they say how you needed to solve it?  Elimination?  Substitution?  Or just any way that works?

anonymous · Answer

Use addition or substitution to find the value of y for this set of equations.

e.mccormick · Answer

OK, do you know how those methods work?

anonymous · Answer

y would equal 8 n x would be -6

anonymous · Answer

kinda like idk which one to do first

e.mccormick · Answer

Well, substitution is to replace one thing with another.  Here, you have 4x in both equations, and you want to solve for y.  So, pick one of those equations and solve it to the point where 4x is on one side and everything else is on the other.

anonymous · Answer

you multiply the second equation by a negative then add it to the first

e.mccormick · Answer

What skinny23 just described is the addition or elimination method.

To continue subsitution, once you knew what 4x in one equation was equal to, you could replace it in the other equation.  Then solve for y.

Both methods get the same answer.

anonymous · Answer

alright thank you guys verry mucho!

e.mccormick · Answer

Substitution:
because they both have 4x, I can take a shorter form of substitution.  Normally you have to solve for x or y.  This time I will do 4x.

$4x + 13y = 40 $
$4x + 3y = -40$

If I take the second equation:
$4x + 3y = -40\implies 4x = -40 - 3y$

Now plug that into the first, replacing the 4x:
$(-40 - 3y) + 13y = 40 $
$13y- 3y = 40+ 40 $
$10y = 80 $
$y = 8 $


Or, the addition/elimination method:
$4x + 13y = 40 $
$4x + 3y = -40$

$-1$ multiplied through the second equation:
$4x + 13y = 40 $
$-4x - 3y = +40$

Now add the two together:
$0x + 10y = 80$

And that ends the same.

e.mccormick · Answer

There you go.  A summary of it all.