Question

Solve the system of equations.
3x - 4y = -10 and 5x + 8y = -2

Answer 1

(1, 2)


(2, 1)


(1, -2)


(-2, 1)

Answer 2

Why don't you multiply the first equation by 2, then add them together. That should eliminate y.

Answer 3

@chalieberzak
Wherever you have whpalmer, just chill and feel the awesomeness.

Answer 4

haha okay, ill try

Answer 5

@primeralph makes a compelling PR man, doesn't he? ;-)

Answer 6

I dont get the multiply the first equation part....

Answer 7

Okay, take your equation:

\[3x - 4y = -10\]
Now multiply everything in it by 2

\[2*3x - 2*4y = 2*-10\]
\[6x - 8y = -20\]

Notice how you now have \(-8y\) in this new equation, and in the other equation (the one not multiplied by 2) you have \(+8y\)? When you add the two equations together, the \(y\) terms go up in a puff of smoke, leaving you an equation just in terms of \(x\), which is easily solved. Then, you take your newly found value of \(x\) and plug it into any of the equations and solve for \(y\).

Answer 8

wwoowww, lot of info

Answer 9

We can do that multiplication because we are applying it to everything in the equation, so the relationship remains the same. Just like if you and your brother both have $20, you both have the same amount of money, and you'll still have an equal (though different from where you started) amount if I give each of you another $10. Another way to think of it is stacking weights on a teeter-totter — up until the point where it breaks, so long as we stick the same amount of weight the same distance from the middle, we can stack extra weight on both sides without causing it to tilt to one side or the other.

Answer 10

but i think im understanding

Answer 11

if you think about the equation, say we have

x + 2y = 5

say we picked x = 1 and y = 2 as a point that satisfies that equation. Now let's multiply the whole equation by 2:

2x + 4y = 10

we plug in our point (1,2):
2(1) + 4(2) = 10
2 + 8 = 10

still works!

Doesn't change the relationship of x and y.

Now, if we didn't multiply everything by 2 (say we forgot to multiply the right hand side), it no longer works:

2x + 4y = 5
2(1) + 4(2) = 5
2 + 8 = 5
uh, oops.

So, do the same thing to everything, and everyone's still happy.

Answer 12

How are we doing with the solving?

Answer 13

lol my brain is still in process mode.... but the first equation you said to multiply the first equation by 2 so thats

6x - 8y = -20

then add them together does that mean both equations or what?

Answer 14

I figured that might be where you were stuck.

Write the two equations out on two lines, with the like terms lined up over each other:
\[6x - 8y = -20\]\[5x+8y=~-2\]---------------

now we add straight down the columns, giving us

\[11x + 0y = -22\]or\[11x=-22\]You can solve that for \(x\), right?

Answer 15

This is why we multiplied the first equation by 2—we wanted to get -8 and +8 on the \(y\) terms so they would cancel out. We could also have done something similar to make the \(x\) terms cancel out, but there we didn't have one being the multiple of the other, so we'd have to multiply each equations by a different number.

Answer 16

okay

Answer 17

\[11x = -22\]Divide both sides by 11
\[\frac{11x}{11} = \frac{-22}{11}\]\[x=-2\]

Answer 18

Now we plug \(x=-2\) back into one of the earlier equations to find the value of \(y\):
\[5x+8y=-2\]\[5(-2)+8y=-2\]\[-10+8y=-2\]\[-10+8y+10=-2+10\]\[8y=8\]\[\frac{8y}{8} = \frac{8}{8}\]\[y=1\]So our final answer is (-2,1).

Answer 19

That is the method of evaluation. :P