# Solve a and b if 2a=b and a-3b= 12

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2a = b

a -3b =12

To solve, set one of the equations equal to zero. The first equation is actually already set to 0.

b = 2a

Now you can plug this equation into the other one, by substituting the b in the equation with this.

a -3b = 12

a – 3(2a) = 12

Now all you have to do is distribute the 3 (multiply the number outside the parenthesis, 3 in this case, by everything inside).

a – 6a = 12

Combine like terms. Since we have 1 positive a and 6 negative ones, we subtract the two numbers without the signs

6-1 = 5

And take the sign of the larger number. Since 6 is large and 6 is negative, we are left with

-5a = 12

Now solve for a, which in this case means dividing both sides by -5

a = -12/5

Simplify

a = -2.4

Now that you have a, it’s easy to find b

b = 2a

b = 2(-2.4)

b= -4.8

Now you check your work by plugging your new a and b into either equation to see if it’s true.

-4.8 = 2(-2.4)

-4.8 = -4.8

And just for laughs let’s try the other equation

a -3b =12

-2.4 – 3(-4.8) = 12

Distribute the 3 again. Remember that when two negative numbers are multiplied they become positive.

-2.4 +14.4 = 12

12 = 12

Given the systems:

2a = b..........(1)

a -3b = 12 ..........(2)

We need to solve for a and b.

We will use the substitution method to solve.

We will substitue (1) into (2)

==> a - 3b = 12

==> a - 3 ( 2a) = 12

==> a - 6a = 12

==> -5a = 12

==> a = -12/5

==> b = 2a = 2*-12/5 = -24/5

**Then the answer is the pair ( -12/5, -24/5)**