# solve by substitution

• Sep 30th 2006, 04:40 PM
shenton
solve by substitution
I tried many times to solve the following 2 problems but cannot get the answer. It looks simple enough yet so tough.

Solve each system of equation by substitution.

Problem 1.

2c - d + 2 = 0
3c + 2d + 10 = 0

Problem 2.

a + 4b = 3
5b = -2a + 3

• Sep 30th 2006, 04:59 PM
Quick
Quote:

Originally Posted by shenton
I tried many times to solve the following 2 problems but cannot get the answer. It looks simple enough yet so tough.

Solve each system of equation by substitution.

Problem 1.

2c - d + 2 = 0
3c + 2d + 10 = 0

You have equation: 2c - d + 2 = 0

add d to both sides: 2c + 2 = d

You have equation: 3c + 2d + 10 = 0

substitute: 3c + 2(2c + 2) + 10 = 0

multiply: 3c + 4c + 4 + 10 = 0

add: 7c + 14 = 0

subtract 14 from both sides: 7c = -14

divide both sides by 7: c = -2

go back to equation: 2c - d + 2 = 0

substitute: 2(-2) - d + 2 = 0

multiply: -4 - d + 2 = 0

add: -2 - d = 0

add 2 to both sides: -d = 2

multiply by -1: d = -2

can you do the other one now?
• Sep 30th 2006, 06:00 PM
shenton
Thanks Quick for coming to the rescue. This one is not as hard as it should be.

I was able to solve the second problem with the encouragement from your workings.

Thanks. :)
• Sep 30th 2006, 06:00 PM
OReilly
Quote:

Originally Posted by shenton
I tried many times to solve the following 2 problems but cannot get the answer. It looks simple enough yet so tough.

Solve each system of equation by substitution.

Problem 1.

2c - d + 2 = 0
3c + 2d + 10 = 0

Problem 2.

a + 4b = 3
5b = -2a + 3

Solution of problem 2:

From first equation we get a = 3-4b

We substitute a in second equation:
5b=-2(3-4b)+3
5b=-6+8b+3
5b=-3+8b
3=3b
b=1

We can now substitute b in first equation:
a=3-4(1)
a=-1

So, solution is a=-1 and b=1
• Sep 30th 2006, 06:03 PM
Quick
Quote:

Originally Posted by shenton
Thanks Quick for coming to the rescue. This one is not as hard as it should be.

I was able to solve the second problem with the encouragement from your workings.

Thanks. :)

:) It's always nice to see someone learn from this site rather than just get all the answers without being able to do them themselves.
• Sep 30th 2006, 06:06 PM
shenton
Thanks OReilly for your help and providing the workings. This is very useful in understanding the how the final answer is derived.

I think I will try some equation solving by "elimination". Hopefully, that won't cause me any headache.

Thanks. :)