If (x + 2)^2 = 9 and (y + 3)^2 = 25, then what is the maximum value of

Let’s determine values for x and y.

(x + 2)^2 = 9

√(x + 2)^2 = √9

|x + 2| = 3

We need to determine the value of x when (x + 2) is positive and when (x + 2) is negative.

x + 2 = 3

x = 1

OR

-(x + 2) = 3

-x - 2 = 3

-x = 5

x = -5

So x = 1 or x = -5.

In a similar fashion, we determine the value of y.

(y + 3)^2 = 25

√(y + 3)^2 =√25

|y + 3| = 5

We need to determine the value of y when (y + 3) is positive and when (y + 3) is negative.

y + 3 = 5

y = 2

OR

-(y + 3) = 5

-y - 3 = 5

-y = 8

y = -8

y = 2 or y = -8

We can maximize the value of x/y if x and y are both positive or if x and y are both negative. If x and y are both positive, then x/y = 1/2. If x and y are both negative, then x/y = -5/-8 = 5/8. Since 5/8 > 1/2, the maximum value of x/y is 5/8.

Answer: C

You must first solve for x and y. This can be done in a few different ways.

\((x+2)^2=9\)
\(x^2 + 4x + 4 = 9\)
\(x^2 + 4x -5 = 0\)
\((x+5)(x-1) = 0\)
\(x = -5, 1\)

Alternatively, you can take the square root of each side

\(\sqrt{(x+2)^2}=\sqrt{9}\)
\((x+2)=3\)
\(-(x+2)=3\)
\(x=1, -5\)

For y:
\((y + 3)^2 = 25\)
\(\sqrt{(y+3)^2}=\sqrt{25}\)
\(y+3=5\)
\(-(y+3)=5\)
\(y= 2, -8\)

Keep positive by using both negative or both positive.

Options are \(\frac{1}{2}\) or \(\frac{-5}{-8}= \frac{5}{8}\).

\(\frac{5}{8}\) is larger. Option C.

(x + 2)^2 = 9
(x + 2) = +/-3
x = 1,-4

(y + 3)^2 = 25
(y + 3) = +/-5
y = 2,-8

x/y = max. (1/4, -5/-8) = 5/8
we didn't considered negative values -4/2 and 1/-8 as they are already smaller than positive values.

What is the minimum value of y/x if I say the data given in the question above are the same?

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