SAT Nonlinear Equations & Systems Question 174 | Free SAT Prep

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Question 174·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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The equation

\left(\frac{k+3}{k-3}\right)^2=\frac{a}{b}

relates the positive real numbers $a$ , $b$ , and $k$ , with $k>3$ . Which equation correctly expresses $k$ in terms of $a$ and $b$ ?

For equations where a rational expression is squared and you must solve for a variable in the fraction, first take the square root of both sides and introduce a simple symbol like $r$ for the square root to clean up the algebra. Always remember that taking a square root produces both a positive and negative option; then use any given conditions (such as $k>3$ making a ratio positive) to discard the invalid sign. From there, treat the problem as a linear equation in $k$ : cross-multiply, collect like terms, factor out the variable, and carefully manage signs when simplifying the final fraction to match one of the answer choices.

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Undo the square

The equation has $\left(\dfrac{k+3}{k-3}\right)^2$ . What operation will remove the square and make the equation simpler?

Introduce a simpler symbol

Let $r = \sqrt{\dfrac{a}{b}}$ to avoid carrying the square root around. Rewrite the equation in terms of $r$ and $k$ and then solve for $k$ .

Use the condition on k

When you take a square root, you get both a positive and a negative possibility. How does the condition $k>3$ help you decide whether $\dfrac{k+3}{k-3}$ equals $r$ or $-r$ ?

Isolate k carefully

After you set $\dfrac{k+3}{k-3} = r$ , cross-multiply and collect all the $k$ terms on one side. Factor out $k$ and solve, watching for sign changes when you move terms across the equals sign.

Desmos Guide

Choose valid test values for a and b

For a quick verification, choose positive values that are consistent with $k>3$ . Since $\frac{k+3}{k-3}>1$ when $k>3$ , use values with $\frac{a}{b}>1$ , such as $a=4$ and $b=1$ .

Graph the original relationship

In Desmos, enter y=((x+3)/(x-3))^2 and y=4. Find the intersection whose $x$ -coordinate is greater than $3$ ; this coordinate is the valid value of $k$ for the selected values of $a$ and $b$ .

Check the answer choices

Using $\sqrt{\frac{a}{b}}=\sqrt{4}=2$ , enter these expressions in Desmos:

A: 3*(1+sqrt(4))/(sqrt(4)-1)
B: 3*(1-sqrt(4))/(sqrt(4)+1)
C: 3*(sqrt(4)-1)/(sqrt(4)+1)
D: 3*(1+sqrt(4))/(1-sqrt(4))

Compare each value with the valid intersection coordinate from the graph. The matching expression verifies the formula.

Step-by-step Explanation

Undo the square with a square root

The equation is

\left(\frac{k+3}{k-3}\right)^2 = \frac{a}{b}.

Because $a$ and $b$ are positive, $\frac{a}{b}>0$ , so we can take the square root of both sides:

\frac{k+3}{k-3} = \sqrt{\frac{a}{b}} \quad \text{or} \quad \frac{k+3}{k-3} = -\sqrt{\frac{a}{b}}.

To make the algebra cleaner, let

r = \sqrt{\frac{a}{b}}.

Then the two possible equations become $\frac{k+3}{k-3} = r$ or $\frac{k+3}{k-3} = -r$ . We will decide which one is valid using the condition on $k$ .

Use the condition $k>3$ to choose the correct sign

Since $k>3$ , both $k+3$ and $k-3$ are positive, so their ratio $\frac{k+3}{k-3}$ is also positive.

But $r = \sqrt{\frac{a}{b}}$ is positive, and $-r$ is negative. Therefore, the only equation that can be true is

\frac{k+3}{k-3} = r.

We can now ignore the negative-root case, because it would give a value of $k$ that does not satisfy $k>3$ .

Solve the linear fractional equation for k

Starting from

\frac{k+3}{k-3} = r,

cross-multiply:

k+3 = r(k-3).

Distribute $r$ on the right:

k+3 = rk - 3r.

Move all $k$ terms to one side and constants to the other:

k - rk = -3r - 3.

Factor $k$ on the left and $-3$ on the right:

k(1 - r) = -3(r + 1).

Now solve for $k$ by dividing both sides by $(1-r)$ :

k = \frac{-3(r+1)}{1-r}.

Simplify the expression and substitute back for r

The expression

k = \frac{-3(r+1)}{1-r}

can be simplified by multiplying the numerator and denominator by $-1$ (which does not change the value):

k = \frac{3(r+1)}{r-1}.

Now substitute back $r = \sqrt{\frac{a}{b}}$ :

k = \frac{3\left(1 + \sqrt{\frac{a}{b}}\right)}{\sqrt{\frac{a}{b}} - 1}.

This matches choice A, so the correct equation is

k = \frac{3\bigl(1 + \sqrt{a/b}\bigr)}{\sqrt{a/b} - 1}.

Strategy

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation