Question

what if you cant eliminate the y because there is the same symbol and instead of the opposite?

Answer 1

Please post the actual math problem you are working on.

Answer 2

x+3y=5
5x+2y=-1

Answer 3

so what i did was 2(x+3y=5) 2x+6y=10
then I did 3(5x+2y=-1) 15x+6y=-3

As you can see the 6's do not cancel each other out

Answer 4

U can multiply one of them by -1

Answer 5

How do I know what to multiply by going forward?

Answer 6

What do u mean by going forward

Answer 7

Multiply the first equation by 5

5x + 15y = 25
5x + 2y = -1

Subtract the second equation from the first:
13y = 26

Answer 8

Then solve for y. Afterwards find the value of x.

Answer 9

What I mean is after you help me this time how will I know the next time what I multiple by (going forward).

Answer 10

The truth is, depending on what system you have, you might choose to solve using either elimination or substitution.

Answer 11

You have to use the concept of LCM to figure out what to multiply each equation by if you want to use elimination method.

Answer 12

The problem is requesting that I use the elimination process

Answer 13

Say for instance if you have

2x + 3y = 5
3x + 2y = 7

Answer 14

If you want to use elimination to eliminate the x variables, you know that the least common multiply of 2 and 3 is 6. This means that to eliminate the x variable, you want the terms with x to be 6x

Answer 15

To do that, you have to
multiply both sides of the first equation by 3
multiply both sides of the second equation by 2

Answer 16

When you do that you will get:
6x + 9y = 15
6x + 4y = 14

Answer 17

Then subtract the second equation from the first:
5y = 1

Answer 18

so what i did was 2(x+3y=5) 2x+6y=10
then I did 3(5x+2y=-1) 15x+6y=-3

As you can see the 6's do not cancel each other out

as you know, you want opposite signs on the 6y.
so if you get this far, multiply one of the equations by -1 (both sides)
and you can then add the two equations.

Or, you can subtract one equation from the other.

Answer 19

Then solve for y:
y = 1/5

Answer 20

@phi, given the circumstances, I believe my approach would help the user.

Answer 21

Hero, I was just about to email before Phi jumped in, but the way Phi is going the problem is the same way I am doing the problem, so can you help me from that point that Phi stop @ or Phi can you keep going because when you say I need to multiply one by -1 I chose to multiple the last by -1 which cancels out the 6's and then leaves me with 2x=10 and -15x=3. So where do I go from here?

Answer 22

The concepts aren't that dissimilar from each other

Answer 23

You get to this point:

6x + 9y = 15
6x + 4y = 14

Then multiply either one of the equations by -1 then you get:

6x + 9y = 15
-6x - 4y = -14

Now you can add them together to get the same thing
5y = 1

and y = 1/5

Answer 24

o, did you get x=-1

Answer 25

But the basis of either method is the concept of LCM. Just try to understand it.

Answer 26

If you want to check your work, you can plug the values back in to BOTH equations.

Answer 27

That's how you check your solutions to make sure they are correct.

Answer 28

I was trying to show you an EXAMPLE of how to solve it before @Phi jumped in.

Answer 29

Thanks but I am still confused. I plugged -1 back into both equations and I got -3 =y for the second equation and y=2 for the first equation

Answer 30

You must get the same y

Answer 31

Otherwise you have done something wrong

Answer 32

So does this mean there is no solution or many solutions?

Answer 33

If you don't get the same x and same y for both equations, you have done something wrong.

Answer 34

If there was no solution, you wouldn't even be able to solve for x or y.

Answer 35

If there was infinite solutions, then you would end up with something like 0 = 0.

Answer 36

As long as you are able to solve for x and y, (which in this case, you will be) then for both equations, the values of x and y will be the same. That's what makes the two equations a system.

Answer 37

Because the same (x,y) point values satisfy both equations

Answer 38

At some point you got confused while solving or made a sign mistake. That's the only reason you got two different values.

Answer 39

okay Hero that make sense, however are you saying that the x and y will be the same answer

Answer 40

LCM is fine, but not required ... it only helps to avoid fractions

Answer 41

=So when trying to fine the LCM for he equations am I trying to find one for the yintercepts

Answer 42

@amstre64, LCM is required if you are using elimination method, which is what the user is using to try to solve this.

Answer 43

There's another method, called substitution of course.

Answer 44

x+3y=5
5x+2y=-1

eliminating the xs is simple enough,

the ys .. we could multiply the bottom by -3/2

\[ ~~~~~~~~~~ x+3y=5 \\
\frac{-3}{2}(5x)+\frac{-3}{2}(2y)=-1(\frac{-3}{2})\]


\[ ~~~~~~~~~~ x+3y=5 \\
-\frac{15}{2}x-3y=\frac{3}{2}\]

Answer 45

I should have just explained both methods. I feel @amistre64 would be more helpful for calculus and higher students.

Answer 46

LCM is suggested to avoid the fractions ... since fractions are consider to be scary for some reason is all

Answer 47

so what I did was 2(x+3y=5) then 3(5x+2y=-1) so after doing the multiplication I get 2x+6y=10
15x+6y=-3. So do I then multiple on of them by -1 so that the 6's can cancel out each other? If so when I did that I ended up with
2x=10
-15x=3, so do it become -13x=13, so x=-1 once you divide?

Answer 48

Or does x=5 for the 1st equation 2x=10 and then -1/5 for the 2nd equation. Do I flip the receprocal (misspelled)?

Answer 49

@Hero can you look at my last two reply's and help me finish this off? I would appreciate it.

Answer 50

I'm just wondering something...how did you end up with 2x = 10 again? I think you have the wrong idea of what to do as far as elimination is concerned.