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
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
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)