Solve this system of equations: 5x+4y=-8 and 2x-y=-11

Question

anonymous · Answer

method or answer?

anonymous · Answer

Method and answer

anonymous · Answer

or maybe the answer, then show work

anonymous · Answer

http://www.wolframalpha.com/input/?i=5x%2B4y%3D-8+and+2x-y%3D-11

anonymous · Answer

multiply second equation by 4, add to first equation

amistre64 · Answer

y = (za-tc)/(ad-bc)

anonymous · Answer

$$5x+4y=-8$$
$$8x-4y=-44$$
$$13x=-52$$
$$x=-4$$

amistre64 · Answer

5  4 =-8
2 -1 =-11

y = (5(-11)-2(-8))/(5(-1)-2(4))

anonymous · Answer

then replace x by -4 in either equation, solve for y

amistre64 · Answer

i get y = 3 if i did it right :)

anonymous · Answer

5x+4y=-8 
2x-y=-11

get the bottom to have some common system like below:

5x+4y=-8
8x-4y=-44

the 4's cancel out then add going down:

13x = -52

divide both sides by 13 to get x.

x = -4

now plug in -4 in to the first equation

5(-4)+4y=-8
-20+4y=-8
+20        +20

4y=12

divide both sides by 4.

you get:

y = 3

So your answer is:

x = -4, y = 3

amistre64 · Answer

4   5 =-8
-1  2 =-11

x = (4(-11)+1(-8))/(4(2)-5(-1)) = -4

yay... it works