Question

3x-y=11
2x+5y=-4

systems of linear equations using substitution
find X and Y

Answer 1

same as before: subbing the first equation for y seems easy, so let's do that:
y=3x-11
now sub that into equation 2:
2x+5(3x-11)=-4
2x+15x-55=-4
17x=51
x=3
now plug that answer back into eqn. 1:
3(3)-y=11
9-y=11
y=2

Answer 2

you can use method of elimination to check your answer here
3x-y=11 mult by 5 gives 15x-5y=55
2x+5y=-4 2x+5y=-4 add to cancel out the y components
----------
17x =51
x=51/17=3 ......ans now sub this to eq
3x-y=11
3(3)-y=11
9-11 =y=-2 ...ans

Answer 3

oh yeah, y=-2... my bad

Answer 4

x=-3
y=-19

Answer 5

what?

Answer 6

3x-y=11
2x+5y=-4

If we multiply equation 1 by 5, then we can add the 2 equations together and cancel the y's. Let me show you.

5(3x-y=11) -> 15x - 5y = 55 (New equation 1)

15x-5y=55
2x+5y = -4

Now if we add these together, we have:
17x = 51
By dividing both sides by 17, we have x = 51/17, which gives us x =3.
Now if we subsitute 3 for x in either equation, we can find y. We will do it in both equations just to prove it.

3(3)-y=11 -> 9 - y = 11
-y = 2 -> y = -2

Now if we do the same for equation 2, we have:
2(3) + 5y = -4
6 + 5y = -4
5y = -10
Dividing both sides by 5, we have: y = -2

Hope this helps.

Please Like Tutor Sean on Facebook
( http://www.facebook.com/pages/Tutor-Sean/250732554967420).
It will help me help others, and I offer free help there with a community of other college tutors.

Also check out http://www.tutorsean.net for free videos on math and related subjects.

Answer 7

That's not the substitution method though, which the question asked for. There are, again, many ways to solve linear systems of equations, the most powerful of which can be learned through Linear Algebra. I would suggest practicing as many as possible.

Answer 8

Okay, can you refresh me on the substitution method??? Because I was taught that Gaussian Elimination is the same as substitution. I am a mathematics major. However, I admit that I could be wrong. Please help me understand.

Answer 9

Well, I am using the terminology given in high school algebra, which usually only teaches two methods: the "Substitution" as opposed to the "Addition" method. In the Addition method you solve it how you did above, which is clearly the same as Gaussian Elimination; you're multiplying and adding to get rid of variables. This is quite different from how you can see I solved the problem, in which I substituted expressions with variables into a different equation. I cannot see how Gaussian Elimination is not much more similar to the Addition method than the Substitution one, but since they all give the same answer I don't really see a strong contradiction in saying that Gaussian Elimination is the same as Cramer's Rule, or whatever method you choose. The difference is semantic I would say.

Answer 10

Thanks! When I was in school (homeschooled), we did not differentiate the two. Glad you were able to explain that. As a math and computer science tutor, I am always looking for new angles to approach a topic.

Answer 11

This is so hard I still cant understand it