Question

What is the solution to the system of equations?

Use the Linear combination method.

5x + 2y = -4
-3x + 2y = 12

A (2,-7)
B (0,-2)
C (-2,3)
D (4,0)

Answer 1

5x+2y + (-6x-2y) = 12 + (-14)
5x - 6x + 2y-2y = 12-14
-x = -2
x = 2

Substitute the value of x into either of the original equation to find y.

5(2) + 2y = 12
10 + 2y = 12
2y = 10
y = 1

The solution is (x,y) = (2, 1)

Answer 2

that one is suppose to be 3

Answer 3

Let me see if I can do this in a fashion that will actually show you how to do it if you don't know how:

\[5x + 2y = -4\]\[
-3x + 2y = 12\]

We want to combine those two equations in such a way that we get a simpler equation (with fewer variables) that we can solve directly.

Notice how both equations have \(2y\) in them. If we multiply one of the equations by \(-1\) (multiplying each term!) we will have an equivalent equation, but all of the signs will be changed from negative to positive or positive to negative. Let's say I do that to second equation:

\[(-1)*(-3x)+(-1)(2y) = (-1)(12)\]\[3x-2y=-12\]

Now let's line that up underneath our first equation (the one we didn't alter):

\[5x+2y=-4\]\[3x-2y=-12\]Let's add those two together, just like a big addition problem:

\[5x+2y=-4\]\[3x-2y=-12\]--------------\[5x+3x + 2y+(-2y) = -4 + (-12)\]\[8x+0y= -16\]\[8x=-16\]
That is easily solved for the value of \(x\).

Now we take the value of \(x\) and we plug it into either of the original equations and solve for the value of \(y\). The combination of \((x,y)\) is our solution.

However, before we go tell anyone that we have the solution, we must plug our values of \(x,y\) into all of the original equations and make sure they work. It is entirely possibly to get values of \(x,y\) which satisfy one equation, but not the other. There's a technical term for this kind of solution. It's called "wrong" :-)

Answer 4

So, the trick to linear combination is looking at your equations and multiplying by the right numbers so that one of the variables in both equations has coefficients that will combine to 0 when you combine the equations, removing that variable and giving you a simpler equation to solve.

Answer 5

@Derrickcurry

even after your correction your answer is STILL wrong

your option does not even appear in the list - and it certainly does not satisfy either equation