Question

Graphing Systems of Equations Question

Answer 1

how do you solve this problem?

Answer 2

thank youc:

Answer 3

@maaryliin, one quick way to figure it out is to plug each point (-1,-5) and (4,0) in to each equation given in the answer choices. The result that is true for both equations is the the right choice.

Answer 4

but i dont know how to do it ive never done algebra 2 before:/ i need like step by step:/

Answer 5

There's a bit of a learning curve involved, but the process is not difficult. I'll go over it with out.

Answer 6

Again, recall that the solution points of graphs of both equations (whatever they are) are (-1,-5) and (4,0)

Answer 7

We will plug each point into the given systems of equations to test if they are true or false. We're interested in the system that yields a true result for a particular system.

Answer 8

Let's try answer choice A:

\[y = \frac{x + 4}{x + 2}\]\[y = x + 4\]

Answer 9

We will first try plugging in the point (4,0)

Answer 10

Remember, if we get a false result at any time, then we can automatically stop because system would not be true. Both points have to work for both equations in order for the system to be correct.

Answer 11

So here goes...plugging in:

Answer 12

\[0 = \frac{4 + 4}{4 + 2}\]
\[0 = 4 + 4\]

Answer 13

so the answer cant be negative number

Answer 14

Now we can stop there since \(0 \ne 8\)

Answer 15

We tried plugging in (4,0) and it did not work for system A

Answer 16

Notice that system B has one of the same equations from system A that didn't work. So we know System B does not work either.

Answer 17

This means that either system C or D is correct. So we can test those out

Answer 18

It makes sense to test point (4,0) first because we know that the fraction must equal zero in order to be true. And the linear expression also must equal zero.

Answer 19

But if we find a system that is true for one point, we have to also test the other point (-1,-5) as well

Answer 20

So here goes. For System C we have:

\[y = \frac{x - 4}{x + 2}\]
\[y = x + 4\]

Answer 21

Now we will test point (4,0) by plugging the values in to both equations:

Answer 22

\[0 = \frac{4 - 4}{4 + 2}\]
\[0 = 4 + 4\]

Answer 23

For the fractional equation we get
0 = 0 True

But for the linear equation we get
0 = 8 False

Answer 24

So now we're down to the very last choice. Again, we suspect that it will work, but still it must work for both BOTH points in order to be true.

Answer 25

For system D we have

\[y = \frac{x - 4}{x + 2}\]
\[y = x - 4\]

Answer 26

Testing point (4,0) we get

\[0 = \frac{4 - 4}{4 + 2}\]
\[0 = 4 - 4\]

Answer 27

Both equations yield a result of 0 = 0

Answer 28

So far, D works. Now we just have to test the point (-1,-5) to be doubly sure that the system works

Answer 29

Can you try it out please?

Answer 30

so it is the same but with -1.-5?

Answer 31

Yes, test it out with (-1, -5)

Answer 32

It's x = -1 and y = -5

Answer 33

but how do i set it up :/?