I need some help solving these problems through elimination.
1) 4x+3y=-1
2x+5y=3
2) 12x-5y=9
3x-8y=-18

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- amistre64

Is that the intersection of 4 lines? or 2 seperate questions?

- anonymous

2 seperate questions. If yiu can guide me through the first, maybe I can get the second?

- amistre64

elimination just means that we modify one equation thru multiplication in order to find a value that can cancel out a variable in the other equation.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

4x+3y=-1
2x+5y=3 <- we can multiply this by -2 to get: -4x
should we?

- anonymous

Ok..

- amistre64

4x+3y=-1
(2x+5y=3) (-2)
4x +3y =-1
-4x-10y =-6 <-- now we add them together
-------------

- anonymous

So would y be -1

- amistre64

4x +3y =-1
-4x-10y =-6
------------
0 -7y = -7 <-- now we solve for y
y = 1 is correct

- anonymous

not -1. 1?

- anonymous

ok. got it

- amistre64

Since y = 1, we can use that to find "x" in the equations...

- anonymous

Do I just plug the y into one of the equations? or both?

- amistre64

either one is fine; they both meet at a specific point if they are not parallel lines; and that point has y=1 in common; so use it in any of them to find the common x value

- anonymous

Ok, so if I plug it into the first equation I got x=1 is rhat right?

- anonymous

-1

- amistre64

-1 is good to me :) this is a generic setup so dont be fooled by the position of the solution in my graph.....

##### 1 Attachment

- anonymous

Oh... thanks, but thankfully my teacher isnt making us graph :)

- amistre64

You see where the lines cross they have x and y in common right?

- anonymous

Yes, I do see that.

- amistre64

the solution to a set of equations is "where the lines have X and Y in common". Or at a single point.

- anonymous

Thanks for your help. I want to see if I can get the answer to the second one on my own. If I need help will you be around for a few?

- amistre64

i should be... :)

- anonymous

Which one should I try to cancel out? Does it matter?

- amistre64

doesnt matter; pick one and run with it :)

- anonymous

YAy... I think I got it (3,3)
Can you help me with this one or tell me if I'm right. This one is using subsitution.
6x-y=5
y=11x I got x=-5
y=-55
I dont think its right?

- amistre64

6x - 11x = 5
-5x = 5
x = -1

- amistre64

since y = "11x" we substitute that value into the other equation:
6x -y = 5
6x - (11x) = 5 ...and solve

- anonymous

Thank you. I see where I went wrong.

Looking for something else?

Not the answer you are looking for? Search for more explanations.