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Question 15·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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If $x \ne 2$ and $y \ne 0$ satisfy $\dfrac{x-2}{y} + \dfrac{y}{x-2} = 5$ , what is $y$ in terms of $x$ ?

When a rational equation has the same linear expression in both denominators and you are asked for one variable in terms of the other, reduce it to a single-variable quadratic. Either (1) define a substitution like $t = \dfrac{y}{x-2}$ so the equation becomes something like $t + \dfrac{1}{t} = \text{constant}$ , then clear denominators and use the quadratic formula, or (2) clear denominators directly and treat the result as a quadratic in $y$ . This keeps the algebra organized and helps you avoid sign and discriminant mistakes while quickly matching to the answer choices.

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Use the repeated expression

Notice that $x-2$ appears in both denominators. Try defining a new variable for the ratio involving $y$ and $x-2$ so that the equation only has one variable.

Eliminate the fractions

If you let $t = \dfrac{y}{x-2}$ (or $t = \dfrac{x-2}{y}$ ), rewrite both terms in the equation using $t$ and then multiply to clear the denominator.

Recognize the quadratic

After substituting and clearing denominators, you should get a quadratic equation in $t$ . Use the quadratic formula and then convert back from $t$ to $y$ .

Desmos Guide

Set up the original relationship check

In Desmos, type the expression ((x-2)/y) + (y/(x-2)) and note that valid $(x,y)$ pairs must make this equal to 5. You will verify each answer choice by substitution rather than solving symbolically.

Create functions for each answer choice

For each option, define functions of $x$ using the + sign first. For example, for choice A define yA(x) = (x-2)*(5+sqrt(21))/2; for choice B define yB(x) = (x-2)*(5+sqrt(17))/2; similarly define yC(x) and yD(x) from their formulas (using the + in place of ±).

Check which choice satisfies the equation

For each candidate function, define a new function that plugs it into the original expression, e.g., fA(x) = (x-2)/yA(x) + yA(x)/(x-2). Then create a table of values for $x$ (such as $x=0,1,3,4$ , avoiding $x=2$ ) and look at $fA(x)$ , $fB(x)$ , etc. The correct answer choice will be the one whose corresponding function consistently gives $5$ for all tested $x$ -values.

Step-by-step Explanation

Turn the equation into a quadratic

Start with

\frac{x-2}{y} + \frac{y}{x-2} = 5,

where $x\ne 2$ and $y\ne 0$ .

Let

t = \frac{y}{x-2}.

Then

\frac{x-2}{y} = \frac{1}{t},

so the equation becomes

t + \frac{1}{t} = 5.

Multiply both sides by $t$ (allowed because $t\ne 0$ ):

t^2 + 1 = 5t,

which rearranges to the quadratic

t^2 - 5t + 1 = 0.

Solve the quadratic for the ratio

Use the quadratic formula on

t^2 - 5t + 1 = 0,

where $a=1$ , $b=-5$ , and $c=1$ .

The quadratic formula gives

t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Substitute $a$ , $b$ , and $c$ :

\begin{aligned} t &= \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(1)}}{2(1)} \\ &= \frac{5 \pm \sqrt{25 - 4}}{2} \\ &= \frac{5 \pm \sqrt{21}}{2}. \end{aligned}

\frac{y}{x-2} = t = \frac{5 \pm \sqrt{21}}{2}.

Express y in terms of x and match the choice

From

\frac{y}{x-2} = \frac{5 \pm \sqrt{21}}{2},

solve for $y$ by multiplying both sides by $(x-2)$ :

y = (x-2)\cdot \frac{5 \pm \sqrt{21}}{2} = \frac{(x-2)(5 \pm \sqrt{21})}{2}.

This matches choice A) $y = \dfrac{(x-2)(5\pm\sqrt{21})}{2}$ .

Question 15·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation