Math help, 9th grade level?

Question

6r-3=9r-6

and

a+2 = -2+a

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5

Answer 1

ok first question 6r-3=9r-6

take the three and add it to both sides which makes it 6r=9r-3

take the -3 and add it to both sides to make it 6r+3=9r

divide 6r and 9r by six to get r+3=1.5

subtract the 3 from both sides to get r=-1.5 which is the answer

Second question  a+2=-2+9

add 2 to both sides then you get a=9 which is the answer hope this helps

Answer 2

The person who posted before me is wrong. The first equation the first step is to isolate r to one side. So you start by doing this

1. Subtract 6r from both sides. So you end up with -3+3r - 6

2. Add 6 to both sides because you are still trying to isolate r. So that then becomes 3=3r

3. Divide each side by 3 and you get r=1

4. Final step is to check the answer. Do that by putting in 1 for r and seeing if the equation is true.

5. 6(1)-3=9 (1)-6

6-3=9-6

3=3 So that is the correct answer.

For the second problem here is how to start. I am assuming you mean this (a+2)/5=-2+a

1. Multiply both sides by 5. So it becomes a+2=5(-2+a)

2. Distribute the 5. a+2=-10 + 5a

3. Yet again try to get a by itself by subtracting an a from both sides.

That will get you 2=-10 +4a

4. Add ten to both sides and you get 12= 4a

5. Divide both sides by 4 and you get a=3

6. Plug 3 back into the original equation. (3+2)/5=-2 + 3

1=1 so that is the answer

Answer 3

first question

6r-3=9r-6

add three to both sides

6r=9r-3

Subtract 9r from both sides

-3r=-3

divide both sides by -3

r=1

Second Question

a+2 = -2+a

add two to both sides

a+4=a

subtract a from both sides

4=0

the answer is no solution