Question 221·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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In the equation

\frac{1}{x-2} + \frac{1}{y-2} = \frac{1}{3},

$x$ and $y$ are real numbers with $x \ne 2$ and $y \ne 2$ . Which equation correctly expresses $y$ in terms of $x$ ?

To express one variable in terms of another in a rational equation, look for the variable in a denominator. On a multiple-choice question, a quick route is to choose an allowed input, use Desmos to find the required output, and test every proposed formula. One matching input doesn't prove a formula, but one failed input rules it out. Watch for sign changes in a numerator or denominator.

Hints

Hint 1
Choose an allowed input that makes the given equation easier to read. At $x=3$ , the denominator $x-2$ becomes $1$ . Could a correct formula use that same input?
Hint 2
An intersection is a point two graphs share. Graph the given equation and $x=3$ , then click where they meet. What $y$ -value must every correct formula give at $x=3$ ?
Hint 3
Use substitution: put $3$ in for $x$ in each proposed formula. Any formula with a different output fails, even if it looks almost identical to the one that works.

Step-by-step

Approach 1: Test an allowed input with Desmos
Step 1
Find the output required at $x=3$
Type the given equation and $x=3$ into Desmos. This input is allowed because $3\ne2$ , and it makes $x-2=1$ . Click their intersection, where the graphs meet. Desmos shows $(3,0.5)$ , so a correct formula must give $y=0.5$ when $x=3$ .
Step 2
Compare every proposed formula
Type $x_1=3$ , then enter the four right-hand sides in the order shown, replacing $x$ with $x_1$ . Desmos prints $-0.5$ , $-15.5$ , $0.5$ , and $-0.5$ . Only $\frac{5x-16}{x-5}$ gives the required $0.5$ . The other three fail this substitution test, so the equation is $y=\frac{5x-16}{x-5}$ . Choice C.

Approach 2: Rearrange the equation for every allowed $x$
Step 1
Isolate the fraction containing $y$
To get the fraction containing $y$ by itself, subtract $\frac{1}{x-2}$ from both sides:
$\frac{1}{y-2}=\frac{1}{3}-\frac{1}{x-2}.$
Step 2
Combine the fractions on the right
Use the common denominator $3(x-2)$ , so both fractions have the same bottom:
$\frac{1}{y-2}=\frac{x-2}{3(x-2)}-\frac{3}{3(x-2)}.$
Subtract the tops:
$\frac{1}{y-2}=\frac{x-2-3}{3(x-2)}.$
Simplify the top:
$\frac{1}{y-2}=\frac{x-5}{3(x-2)}.$
Step 3
Undo the reciprocal
The reciprocal flips a fraction's top and bottom. Take reciprocals to remove the fraction around $y$ :
$y-2=\frac{3(x-2)}{x-5}.$
Keep $x\ne5$ : at $x=5$ , the previous line would say $\frac{1}{y-2}=0$ , which no real $y$ can satisfy.
Step 4
Get $y$ by itself
Undo the subtraction of $2$ by adding $2$ to both sides:
$y=2+\frac{3(x-2)}{x-5}.$
Step 5
Write the result as one fraction
Give $2$ the same denominator as the other fraction:
$y=\frac{2(x-5)+3(x-2)}{x-5}.$
Distribute in the top:
$y=\frac{2x-10+3x-6}{x-5}.$
Combine like terms:
$y=\frac{5x-16}{x-5}.$
This expresses $y$ in terms of $x$ , with $x\ne2,5$ . Choice C.

Question 221·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this