9t^2-r^2=17

Test Preparation SAT Preparation
1

<p>9t^2-r^2=17
What is t+r?</p>

<p>Answer: 11?</p>

<p>How do you do it? Algebraically?</p>

2

<p>Factor the difference of squares:
(3t - r) (3t + r) = 17</p>

<p>t and r are positive integers, so because 17 is prime, the only way to get a product of 17 is if:
3t - r = 1
3t + r = 17</p>

<p>substitution/combination yields 6t = 18 => t = 3
Therefore, r = 8.</p>

3

<p>(3t-r)(3t+r) = 17
We are given that both t,r are integers, so that means 3t-r and 3t+r are integers.
Therefore, the only integers that multiply together to give 17 are:
(1,17)
(17,1)
Testing either case:

  1. setting 3t-r = 1 and 3t + r = 17
  2. setting 3t-r = 17 3t + r = 1
    Since r, s are both positive integers, it doesn’t make sense that 3t - r is greater than 3t + r, so eliminate case 2.
    We are left with a system:
    (1) 3t - r = 1
    (2) 3t + r = 17
    Manipulate equation 1:
    3t = r +1
    Substitute r+1 for 3t into equation 2
    r + 1 + r = 17
    2r + 1 = 17
    2r = 16
    r = 8
    Use r= 8 and substitute into the first equation
    3t - 8 = 1
    3t = 9
    t = 3</p>

<p>Thus,
r + s = 8 + 3 = 11</p>

<p>Hope that helps.</p>

4

<p>Gah I got to the (3t+r)(3t-r) = 17 step during the test, but I spent like 5 minutes thinking about how to get past that. Oh well wasn’t meant to be.</p>

5

<p>In that case, you should have just plugged in numbers, or tried some other strategy.</p>

<p>I got stuck there for a second too, then realized that t and r are positive integers. That was the key to the problem for me.</p>

6

<p>Was this problem on a 20 minute section or a 25 minute section?</p>