solve 5x+2y=-1 and 3x-4y=11 using elimination

Question

I just need help quickly solving this problem, thank you in advance

Answer 1

Hi Gregory D.,

First we'll eliminate the y variable and solve for x by multiplying 2*(5x + 2y = -1) and then add it to equation 3x - 4y = 11.

2*(5x + 2y = -1) ----->>>10x + 4y = -2

10x + 4y = -2

+(3x - 4y = 11)

13x + 0 = 9

x = 9/13

Plug this into either original equation to find y.

5*(9/13) + 2y = -1

45/13 + 2y = -1

2y = -58/13

y = -29/13

I hope this helps, Joe.

Answer 2

These used to confuse me sooooo badly!

What we are going to do is add the two equations together. As of now, we'd end up with 8x - 2y = 10. Thaaaat doesn't help.

What we want is for one of the variables to be opposites so that they add up to 0 and cancel out, like 7x + (-7x), for example.

The way we do that is by multiplying one or both equations by whatever constant(s) needed to end up with opposites. Multiplying an entire equation by a constant does not change the graph (what we are looking for is the point where our graphs intersect, the point that makes BOTH equations true).

Well, if I look at 2y and -4y, all I have to do is multiply the top equation by 2 and I end up with 4y and -4y. Boom.

10x + 4y = = -4

3x - 4y = 11

Adding those together, we get 13x = 7.

Let me stop here and say perhaps you didn't copy the problem correctly, but either way, you get the idea, hopefully and can continue.

Just for fun, let's eliminate the x's. We have 5x and 3x. Their LCM is 15, so we could multiply the top equation by 3 and the bottom by -5, ooooor the top by -3 and the bottom by 5. Let's do the first.

15x + 6y = -3

-15x + 20y = -55

26y = -58

Continue

Again, please go back and check your typing of the problem. Either way, apply the elimination method. Hopefully I've helped!

Answer 3

Given:

5x + 2y = -1. Multiply by 2

3x - 4y = 11

——————

10x + 4y = -2

3x - 4y = 11. (Eliminate y)

——————

13x = 9

x = 9/13. Solve for y. 5(9/13) + 2y = -1. Y = -29/13

Solution (9/13,-29/13). The only obstacle in this problem is fractions.

Answer 4

First notice that 5 and 3 are relatively prime (meaning the greatest common divisor is 1), while 2 and 4 are not relatively prime. We have that 4 = 2*2, and so we need to multiply both sides of the first equation by 2. We have that 2*(5x+2y) = 10x+4y and 2*(-1) = -2. This gives us the equivalent equation, 10x+4y = -2.

Now we will assume both equations to be true. Then we have that (10x+4y) + (3x-4y) = (-2) + (11) must also be a true equation. By the associative and distributive properties, we have (10+3)x + (4-4)y = (-2+11). By doing this addition, we have 13x+0y = 9, or 13x = 9, or x = 9/13. We have found x.

Now take 5x+2y=-1, or any of the true equations we have, and replace x with 9/13.

We have then: 5(9/13) + 2y = -1, or (5*9)/13 + 2y = -1, or 45/13 + 2y = -1, or 2y = -1 -45/13, or 2y = -13/13 -45/13, or 2y = -58/13, or y = -58/(13*2), or y = -(58/2)/13, or y = -29/13. We have found y.

The prompt just wants you to report the two x and y values that work for the equations. Any time you have to solve two equations that are linear combinations of two variables, you will either have a unique solution, infinite solutions (e.g., x=y and 2x=2y), or no solutions (e.g., x + y =5 and x + y = 6). In this case, we have a unique solution.