Question 946·Hard·Math

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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 $x$ ?

A rational equation has a variable in a denominator. Here, the two fractions are reciprocals, so the ratio $y/(x-2)$ is the quantity to solve for. You can use Desmos at one convenient allowed value of $x$ to distinguish the answer pairs, then use the ratio to carry the result to every allowed $x$ . Keep both solutions; finding one does not determine the whole answer.

Hints

Hint 1
Since $x-2$ and $y$ are both nonzero, you can call $y/(x-2)$ a new quantity, $t$ . The other fraction is then $1/t$ . What does the given equation say about this ratio?
Hint 2
Choose $x=3$ , which is allowed and makes $x-2=1$ . Now $y$ equals the ratio $t$ . The resulting equation has two solutions, so look for both rather than stopping at the first Desmos result.
Hint 3
Each answer gives two values because of its $\pm$ sign. At $x=3$ , compare both values of a promising answer with the two values you found for $y$ .

Step-by-step

Approach 1: Test a fixed ratio with Desmos

Step 1
Turn the reciprocal fractions into one ratio
Let $t=\dfrac{y}{x-2}$ . The given conditions make $t$ nonzero, and $\dfrac{x-2}{y}=\dfrac1t$ , so the equation becomes $\frac1t+t=5$ . There is no $x$ in this equation. Once you find its possible ratios $t$ , each one gives $y=t(x-2)$ for every allowed $x$ .
Step 2
Find the greater ratio at a convenient value
Choose $x=3$ , so $x-2=1$ and $y=t$ . Type the original equation with $3$ in place of $x$ , $y_1$ in place of $y$ , and $\sim$ in place of $=$ . The restriction asks Desmos for the solution above $1$ . Under PARAMETERS, it shows $y_1\approx4.7913$ , the greater possible ratio.
Step 3
Find the other ratio
Add the same regression using $z_1$ for a separate solution and restrict it between $0$ and $1$ . Desmos shows $z_1\approx0.2087$ under PARAMETERS. Don't discard it: the equation asks for $y$ without a condition that selects only one branch.
Step 4
Match both ratios and restore x
At $x=3$ , each answer's factor $x-2$ becomes $1$ . Type the displayed expressions to compare their values. The pair $\dfrac{5+\sqrt{21}}2\approx4.7913$ and $\dfrac{5-\sqrt{21}}2\approx0.2087$ matches both Desmos results; the other answers' greater values do not. Since $y=t(x-2)$ , restore that factor: $y=\dfrac{(x-2)(5\pm\sqrt{21})}{2}$ . Choice A.

Approach 2: Derive the exact formula directly

Step 1
Clear the original denominators
For an exact check that works with $x$ still unknown, multiply both sides of the given equation by $y(x-2)$ . Neither factor is zero, so this preserves the equation: $(x-2)^2+y^2=5y(x-2)$ .
Step 2
Put the equation in quadratic form
Subtract $5y(x-2)$ from both sides: $y^2-5(x-2)y+(x-2)^2=0$ . This is a quadratic in $y$ : it has a $y^2$ term, while $x-2$ stays fixed as you solve for $y$ .
Step 3
Find the number under the square root
In the quadratic formula, $a=1$ , $b=-5(x-2)$ , and $c=(x-2)^2$ . The discriminant, $b^2-4ac$ , is the part under its square root. Factor out $(x-2)^2$ : $\begin{aligned}D&=[-5(x-2)]^2-4(1)(x-2)^2\\&=(25-4)(x-2)^2\\&=21(x-2)^2.\end{aligned}$ The $21$ , not $17$ , comes from subtracting $4$ from $25$ .
Step 4
Keep both square-root branches
Use $y=\dfrac{-b\pm\sqrt D}{2a}$ : $y=\frac{5(x-2)\pm\sqrt{21(x-2)^2}}{2}$ . Because a square root is nonnegative, $\sqrt{(x-2)^2}=|x-2|$ , the distance of $x-2$ from zero. So $y=\frac{5(x-2)\pm|x-2|\sqrt{21}}{2}$ .
Step 5
Write the pair in the requested form
If $x-2$ is negative, replacing $|x-2|$ with $x-2$ swaps the plus and minus answers; it does not change the pair. So the two possible values of $y$ are written together as $y=\dfrac{(x-2)(5\pm\sqrt{21})}{2}$ . Choice A.

Question 946·Hard·Math

Strategy

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Step-by-step

Lessons that teach this

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Strategy

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Step-by-step

Lessons that teach this