Solving Rational Equations
Rational equations have variables in denominators — solvable with care, but you must rule out values that break the fractions.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
Clear the denominators by multiplying through, then solve. Crucially, check each answer: any value that would make an original denominator zero is excluded.
How to solve one, step by step
Example: solve $\frac{1}{x} + \frac{1}{2} = \frac{3}{x}$.
- Multiply every term by $2x$: $2 + x = 6$.
- Solve: $x = 4$. Since $4 \neq 0$, it's valid.
The mistakes that cost points
- Never checking excluded values. An algebraic solution that makes a denominator zero is extraneous — throw it out.
- Cross-multiplying with more than two fractions. Cross-multiplication only works for a single fraction equal to a single fraction.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The equation $\dfrac{k}{x + 2} = \dfrac{3}{x - 1}$ has the solution $x = 4$. What is the value of $k$?
- A$k = 6$
- B$k = 3$
- C$k = \dfrac{1}{2}$
- D$k = 18$
Show solution
Answer: A, $k = 6$. Substitute $x = 4$: $\dfrac{k}{4 + 2} = \dfrac{3}{4 - 1}$. Simplify each denominator: $\dfrac{k}{6} = \dfrac{3}{3}$. Simplify the right side: $\dfrac{k}{6} = 1$. Multiply both sides by 6: $k = 6$. Check denominators at $x = 4$: $4 + 2 = 6 \neq 0$ and $4 - 1 = 3 \neq 0$. Solution is valid.
Medium
The equation $\dfrac{k}{x^2 - 4} = \dfrac{3}{x - 2}$ has the solution $x = 5$. What is the value of $k$?
- A$k = 7$
- B$k = 21$
- C$k = 3$
- D$k = 9$
Show solution
Answer: B, $k = 21$. Factor the denominator: $x^2 - 4 = (x-2)(x+2)$. Substitute $x = 5$: $\dfrac{k}{(5)^2 - 4} = \dfrac{3}{5 - 2}$. Simplify: $\dfrac{k}{21} = \dfrac{3}{3} = 1$. So $k = 21$. Check excluded values: $x^2 - 4 = 0$ when $x = \pm 2$; since $x = 5 \neq \pm 2$, and $x - 2 = 0$ when $x = 2 \neq 5$, the solution is valid.
Hard
The equation $\dfrac{2}{x - k} + \dfrac{3}{x + k} = \dfrac{k + 11}{x^2 - k^2}$ has the solution $x = 4$, where $k > 0$. What is the value of $k$?
- A$k = \dfrac{11}{2}$
- B$k = \dfrac{9}{2}$
- C$k = 9$
- D$k = 4$
Show solution
Answer: B, $k = \dfrac{9}{2}$. Factor the right side denominator: $x^2 - k^2 = (x - k)(x + k)$. Multiply every term by $(x - k)(x + k)$: $2(x + k) + 3(x - k) = k + 11$. Substitute $x = 4$: $2(4 + k) + 3(4 - k) = k + 11$. Expand: $8 + 2k + 12 - 3k = k + 11$. Simplify: $20 - k = k + 11$. Solve: $9 = 2k$, so $k = \dfrac{9}{2}$. Check excluded values: $x = k$ would be excluded; since $4 \neq \dfrac{9}{2}$, and $4 \neq -\dfrac{9}{2}$, the solution $x = 4$ is valid.
Find your exact gaps
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