How do you solve systems of equations/ multiplying division problems?

Question

Few things actually

1: How would one go about solving a system of equations such as 3a-2b=-6

a-b=-1

2:Also how do you multiply two division equations with one equation over another?

Answer 1

3a-2b=-6

a-b=-1

multiply second equation by 3

3a - 3b = -3

Subtract

3a-2b=-6

3a - 3b = -3

0 + b = -3

b = -3

substitute b to any of the two equations

a-b=-1

a -(-3) = -1

a + 3 =-1

a = -1 - 3

a =-4

a= -4 and b = -3

Answer 2

Ok the first one is a simple substitution.  Set the second equation equal to a and sub it into the first one.

Like this:

a = b -1

Subbing into Equation 1:

3(b-1) - 2b = -6

3b - 3 - 2b = -6

b - 3 =  -6

b = -3

Now put b back in to get a:

a -(-3) = -1

a = -4

For the second one you need to be more clear on what you're asking.

Answer 3

Find b:

3a - 2b = - 6

2b = 3a + 6

b = (3a + 6)/2

a - b = 1

b = a + 1

Find a:

3a + 6 = 2(a + 1)

3a + 6 = 2a + 2

a = - 4

Find b:

= - 4 + 1

= - 3

Answer: a = - 4, b = - 3

Proof:

3(- 4) - 2(- 3) = - 6

- 12 + 6 = - 6

- 6 = - 6

Proof:

- 4 - (- 3) = - 1

- 4 + 3 = - 1

- 1 = - 1

Answer 4

1.

If 3a - 2b = -6 and a - b = -1

a = -1 + b

3(-1 + b) - 2b = -6

-3 + 3b - 2b = -6

b = -3

a = -1 + (-3)

a = -4

2.

don't know what you're on about...