If a = 4 + b, then

3a = 12 + 3b

Now subtract that from the second equation to eliminate a:

3a = 12 + 2b

- 3a = 12 + 3b

-------------------

0 = 0 + 5b Therefore, b=0

So the solution is a = 4 and b = 0

if the equations are a= 4+b and 3a= 12 +2b what is the value for x?

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I don’t know what ‘x’ is in reference to the equations. However, you have two equations and two unknowns, so the values for a and b can be determined. To do so, eliminate one of the variables to get the value for the other and then use the value found in the given equations to compute the first variable.

If a = 4 + b, then

3a = 12 + 3b

Now subtract that from the second equation to eliminate a:

3a = 12 + 2b

- 3a = 12 + 3b

-------------------

0 = 0 + 5b Therefore, b=0

If a = 4 + b, then

3a = 12 + 3b

Now subtract that from the second equation to eliminate a:

3a = 12 + 2b

- 3a = 12 + 3b

-------------------

0 = 0 + 5b Therefore, b=0

Then a = 4 + b becomes a = 4 + 0 = 4

So the solution is a = 4 and b = 0

So the solution is a = 4 and b = 0

Caleb M. | Mathematics TutorMathematics Tutor

The problem does not contain any x. Therefore, it does not matter what x is because it does not change the necessary values of a and b in those equations. Meaning, x is anything you want it to be. Really, the question is not well posed. I'm guessing the author of the question meant to say solve for a or b. Let's do that

a=4+b

3a=12+2b

Let's multiply the first equation by 3; that's allowed.

3a=12+3b

3a=12+2b

But then we have 3a's in both equations. So let's subtract the equations

3a=12+3b

- 3a=12+2b

- 3a=12+2b

0 = 0 +b

So that b=0. Then we can find 'a' using a=4+b. Since b=0, we have a=4+b=4+0=4. Therefore, a=4 and b=0.

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