Math CentralQuandaries & Queries


Question from kimberly, a student:

at+b=ar-c, solve for a


Let's do it first with specific numbers for t, b, r and c and then see if we can mimic the steps when these variables don't have specific values.

5a + 7 = 3a - 9, solve for a.

I would first add -3a to both sides to get

5a + 7 - 3a = 3a - 9 - 3a

and rearrange to get

5a - 3a + 7 = -9.

Next add -7 to each side and also notice that 5a and -3a have a common factor of a. Thus the equation becomes

(5 - 3)a = -9 -7.

Since I have specific numeric values I can perform he arithmetic to get

2a = -16.

Finally divide both sides by 2 to get the answer

a = -16/2 = -8.

Ok, now back to the original problem.

at + b = ar - c, solve for a.

The first step is to add -ar to both sides.

at + b - ar = ar - c - ar.

Again a rearrangement gives

at - ar + b = -c.

Notice here that at and -ar have a common factor of a.

Complete the solution mimicking the steps above and let me know what you get for a.


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