1. ## 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

2. 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?

3. 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.

4. 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

5. 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.

6. 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.