Algebra (solving linear equations by substitution)..... help?????

Question

ok so my equations are
2c - d = - 2 and 4c + d = 20.
(I am supposed to solve the system of linear equations using the substitution method)
I keep trying to do it and I keep winding up with answers like
(- 12, 17) but I know thats not right because (I'm in internet school) when I put it in for my homework it says it is incorrect and I won't get the points for my homework unless they are all correct. Please help me. Please don't just give me the answer, I need to know how you got the answer because I have to show my steps....your help is much appreciated....

It says its correct, Thank You so much!!!

Thank you MathBoi, your answer really helped...

Answer 1

This is really quite easy to do. Just express one variable in terms of another. Then substitute that variable in the other equation to determine the other variable value. Once you have done that, you can plug the second variable's value into the first equation to find the first variable's value.

2c - d = -2
-d = -2 -2c
d = 2 + 2c because we have multiplied both sides of the equation by -1

Now put the value for d in terms of c into the second equation:

4c + (2 + 2c) = 20
4c + 2c + 2 = 20
6c + 2 = 20
6c = 20 - 2
6c = 18
c = 3

Now we can substitute our calculated value for c, which is 3, into the first equation and solve for the actual value of d:

2(3) - d = -2
6 - d = -2
-d = -2 - 6
-1(-d) = -1(-2 - 6)
d = 2 + 6
d = 8

Now plug c = 3 and d = 8 into either one of those equations. If it makes it true, then we have the right numbers.

4(3) + 8 = 20
12 + 8 = 20
20 = 20 This checks out.

2(3) - 8 = -2
6 - 8 = -2
-2 = -2 This checks out also. So we must have the right numbers.

Answer 2

The answer you are looking for is (3, 8). I run an educational site and have a free learning module covering the substitution method at http://www.college-cram.com. You can find the learning module for a step by step solution to your problem at http://www.college-cram.com/library/mysearch.htm?string=substitution+method&search=Search

College-cram.com is a free resource for students covering freshman and sophomore classes in math, business, science, and foreign languages

Answer 3

Here's what I did.

4c+d=20
d=20-4c --->

2c-d=-2
2c-(20-4c)=-2
2c-20+4c=-2
6c-20=-2
6c=18
c=3--->

d=20-4(3)
d=20-12
d=8

My answers are: d=8 and c=3

Does this work?

Answer 4

Multiply #one million by ability of two >>4x - 6y = 12 Subtract #2 from this >> x = 4 replace this into #one million >> 8 - 3y = 6 upload 3y to the two factors & deduct 8 from the two factors >> 8 - 6 = 3y i.e.3y = 2 Divide the two factors by ability of three >> y = 2/3

Answer 5

The guy above me did it correclty, thats how I solved it.