# Math Help - 4 number 2 variable problems?

1. ## 4 number 2 variable problems?

Hello, again. I have done all the 3 number 2 variable problems. However I am having immense problems with the 4 number 2 variable problems.

8-3c=2c-7

I thought

8-3c=2c-7
8-3c-2c=2c-7-2c
8-1c=7

And I'm lost.

Am I doing it wrong?
Any tips for future problems like this?

2. Originally Posted by hoiguy1208
Hello, again. I have done all the 3 number 2 variable problems. However I am having immense problems with the 4 number 2 variable problems.

8-3c=2c-7

I thought

8-3c=2c-7
8-3c-2c=2c-7-2c
8-1c=7

And I'm lost.

Am I doing it wrong?
Any tips for future problems like this?
$8-3c = 2c-7$

Add 3c and 7 on both the sides,

$8-3c+3c+7 = 2c-7+3c+7$

$15=3c$

$c = 5$

3. ## My attempt on the next problem

8-3c=2c-2

8-3c+3c+2=2c-2+3c+2
10-1c

Stuck again ;(

Do I have my formula or technique wrong?

4. Originally Posted by hoiguy1208
8-3c=2c-2

8-3c+3c+2=2c-2+3c+2
10-1c

Stuck again ;(

Do I have my formula or technique wrong?
your way is correct, there are some algebra problems:

$8-3c+3c+2=2c-2+3c+2$

$10 = 5c$ Note that : $2c+3c=5c$ , not 1c

$c = 2$

5. ## Do I get the concept now?

I think I am getting closer to getting this, hope I am not too much of bother.

3-4b=10b+10

3-4b+4b+10=10b+10+4b+10

13=14+10

Where exactly did I go wrong?

6. Originally Posted by hoiguy1208
I think I am getting closer to getting this, hope I am not too much of bother.

3-4b=10b+10

3-4b+4b+10=10b+10+4b+10

13=14+10

Where exactly did I go wrong?
Your second line messed you up. You are trying to get rid of the 10 on the right hand side. So you subtract 10, not add it.

$3-4b=10b+10$

$3-4b+4b-10=10b+10+4b-10$

$-7 = 14b$

$b = \frac{-7}{14} = \frac{-1}{2}$

7. ## Another attempt

7d-13=3d+7

7d-13+7d-7=3d+7+7d-7

-20=10d

Devide both by 2

D = -2?

On a side note: Thank you Harish for your continued help.

8. Originally Posted by hoiguy1208
7d-13=3d+7

7d-13+7d-7=3d+7+7d-7

-20=10d

Devide both by 2

D = -2?

On a side note: Thank you Harish for your continued help.
$7d - 13 = 3d + 7$

Okay. You are getting a little confused here. Remeber that you have to bring the the like terms together. That is bring "d"s on one side and the "numbers" on the other side.

Basically, you need to get rid of that 13 on the left hand side and 3d on the right hand side. Lets do it step by step.

To get rid of 13 on the left side, add 13 to both sides of the equation (since the equation has -13 on the LHS, you are adding 13 on both sides. Had it been only 13, you would be subtracting)

$7d-13 +13 = 3d +7+ 13$

$7d = 3d + 20$

now you need to get rid of the 3d on the right side. so subtract "3d" from both sides:

$7d - 3d = 3d + 20 -3d$

$4d = 20$

$d = 5$