Algebra Question. (Two-way) Concept.

Question

If -24 = 3(1+r) what is the value of 7(1+r)

I am practicing for the SAT on Khan Academy, and this was one of the questions that showed up. The answer I got was -56, which was the correct answer, but the method I had used was completely different from the workings Khan Academy had. I just want to know if my working is correct to.

Answer choices:

A. -56

B. -24

C. -8

D. -9

My workings:

First equation

-24 = 3(1+r)

-24 = 3 + 3r

-24 = 6r

-24/6r = 6r/6r

-4= r

Second equation:

7(1 + r)

7 + 7r

14r

Then I plugged in the answer choices as x = 14r.

-56=14r

-56/14r = 14r/14r

-4=r

Which was the same answer I got for the first equation, so I knew my answer was -56. Is my method another right way to get the answer?

Answer 1

-24 = 3(1+r)

The easiest way to find the value of 7(1 + r) is to start by dividing by 3:

-24/3 = 3(1 + r)/3

-8 = (1 + r)

Now multiply by 7:

7(-8) = 7(1 + r)

-56 = 7(1 + r)

and you're done.

In your approach, you added 3 + 3r to get 6r but that is not right. 3 and 3r are not like-terms and cannot be added. Instead, you should do the following:

-24 = 3(1+r)

The best thing to do next is just divide both sides by 3, but you can distribute the 3 like you did - it's just a bit more complicated. If you do, you'll get:

-24 = 3 + 3r

Subtract 3 from both sides:

-27 = 3r

Now divide by 3:

-27/3 = 3r/3

-9 = r (So r does not equal -4)

So:

7(1 + r) = 7 (1 + (-9)) = 7(1 - 9) = 7(-8) = -56

Answer 2

There's an error in your calculation:

YOU CAN'T ADD 3 + 3r AND GET 6r (they're not "like" terms). Likewise 7 + 7r can't be added. It should be:

-24 = 3(1 + r)

-24 = 3 + 3r

-27 = 3r

r = -9

So 7(1 + r) = 7(1 - 9) = 7(-8) = -56

But you could have used a ratio, as the 2nd equation is 7/3 times as large as the 1st one (the 1 + r remains

constant for both).

7/3 = x/-24 Cross-multiply

3x = -168

x = -56.

Answer 3