Carcass wrote:
If \((x + 2)^2 = 9\) and \((y + 3)^2 = 25\), then what is the maximum value of \(\frac{x}{y}\)?
(A) 1/8
(B) 1/2
(C) 5/8
(D) 5/2
(E) 10/3
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITIONGRE - Math Book(x + 2)² = 9
We have (something)² = 9
So, EITHER something = 3 OR something = -3
In other words, EITHER x + 2= 3 OR x + 2 = -3
If x + 2= 3. then
x = 1If x + 2= -3. then
x = -5(y + 3)² = 25
We have (something)² = 25
So, EITHER something = 5 OR something = -5
In other words, EITHER y + 3= 5 OR y + 3 = -5
If y + 3= 5, then
y = 2If y + 3= -5, then
y = -8We want to MAXIMIZE the value of
x/
yIf the x-value and y-value have OPPOSITE signs,
x/
y will be NEGATIVE
If the x-value and y-value have the SAME sign,
x/
y will be POSITIVE
So, need the x-value and y-value have the SAME sign
Try #1
x/
y =
1/
2Try #2
x/
y =
(-5)/
(-8) = 5/8
Since 5/8 is greater than 1/2, the correct answer is C