# High School 10th Grade Algebra II - Systems of Equations?

I'm learning how to solve systems of equations algebraically by using one of two methods. The first is the substitution method, and the second is the linear combinaton, or method of elimination. I need to solve this system of equation by using the substitution method where one of the two variables in the first...
show more
I'm learning how to solve systems of equations algebraically by using one of two methods. The first is the substitution method, and the second is the linear combinaton, or method of elimination. I need to solve this system of equation by using the substitution method where one of the two variables in the first equation needs to have a coefficient of 1.

4x+12y=4

5x-y=11

Everytime I do the problem, I can't get the answer provided in the back of the book. I would like a full walk-through of how to do the problem if possible! The answer in the book is (-2,1).

Thank you!

4x+12y=4

5x-y=11

Everytime I do the problem, I can't get the answer provided in the back of the book. I would like a full walk-through of how to do the problem if possible! The answer in the book is (-2,1).

Thank you!

Update:
Anirb37 - YES! I have been getting that answer. It's really confusing, and frustrating.

Update 2:
Jim - You remind me of a friend I have who is a junior but takes senior calculus. I wish I was that gifted in math! Thanks for the help I'm in the process of using what you gave me to figure my problem out.

Update 3:
Okay. So I tried what you gave me, Jim, and I had no luck. Here's what I did (maybe I'm doing something wrong I don't even know about):
4x + 12y = 4
4x = 4 -12y
(Divide all by 4)
x = 1 - 3y
Sub in (1-3y) for x in 2nd eq.
5(1 - 3y) - y = 11
5 - 15y -y = 11
5 - 16y = 11
-16y = 6
y =...
show more
Okay. So I tried what you gave me, Jim, and I had no luck. Here's what I did (maybe I'm doing something wrong I don't even know about):

4x + 12y = 4

4x = 4 -12y

(Divide all by 4)

x = 1 - 3y

Sub in (1-3y) for x in 2nd eq.

5(1 - 3y) - y = 11

5 - 15y -y = 11

5 - 16y = 11

-16y = 6

y = -6/16

Y is supposed to equal 1. Which makes no sense. I'm convinced that there is a mistake in my book. It's been known to happen. When I plug in the supposed correct answer (-2,1). It fits in for the 1st equation, but not for the second:

4(-2) + 12(1) = 4

-8 + 12 = 4

4 =4 (Fits)

5(-2) - 1 = 11

-10 - 1 = 11

-11 = 11 (Doesn't fit)

Unless I'm doing something wrong, this is baffling.

4x + 12y = 4

4x = 4 -12y

(Divide all by 4)

x = 1 - 3y

Sub in (1-3y) for x in 2nd eq.

5(1 - 3y) - y = 11

5 - 15y -y = 11

5 - 16y = 11

-16y = 6

y = -6/16

Y is supposed to equal 1. Which makes no sense. I'm convinced that there is a mistake in my book. It's been known to happen. When I plug in the supposed correct answer (-2,1). It fits in for the 1st equation, but not for the second:

4(-2) + 12(1) = 4

-8 + 12 = 4

4 =4 (Fits)

5(-2) - 1 = 11

-10 - 1 = 11

-11 = 11 (Doesn't fit)

Unless I'm doing something wrong, this is baffling.

Update 4:
Well, I just thought I'd clarify that I made one of the stupidest mistakes in algebra and forgot a negative sign. The second equation should be -11 not 11. Sorry for all the confusion. The answer (-2,1) works out.

Follow

2 answers
2

Are you sure you want to delete this answer?