Rational Equation Solver

Intro: Clear denominators carefully and verify domain restrictions.

Worked example

• Solve $\displaystyle \frac{2}{x-1}+\frac{3}{x+2}=1$



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• Goal: Solve the rational equation $\dfrac{2}{x-1}+\dfrac{3}{x+2}=1$.



• Step 1 — Domain restrictions: Denominators cannot be zero. So require $x-1\ne0$ and $x+2\ne0$. Excluded values: $\boxed{x\ne1,\;-2}$.



• Step 2 — Identify a common denominator (LCD): $(x-1)(x+2)$.



• Step 3 — Multiply both sides by the LCD to clear fractions:



• $$ (x-1)(x+2)\left[\frac{2}{x-1}+\frac{3}{x+2}\right] = (x-1)(x+2)\cdot1. $$



• Step 4 — Cancel factors term-by-term:



• • With $\dfrac{2}{x-1}$, the $(x-1)$ cancels leaving $2(x+2)$.



• • With $\dfrac{3}{x+2}$, the $(x+2)$ cancels leaving $3(x-1)$.



• • Right side is $(x-1)(x+2)$ unchanged.



• So we get: $$2(x+2) + 3(x-1) = (x-1)(x+2).$$



• Step 5 — Expand both sides:



• • Left: $2x+4 + 3x-3 = 5x+1.$



• • Right (FOIL): $(x-1)(x+2)=x^2+x-2.$



• Thus: $$5x+1 = x^2 + x - 2.$$



• Step 6 — Bring all terms to one side (standard quadratic form):



• $$0 = x^2 + x - 2 - (5x+1) = x^2 - 4x - 3.$$



• Step 7 — Solve the quadratic $x^2 - 4x - 3 = 0$ using the quadratic formula:



• Discriminant: $\Delta = (-4)^2 - 4(1)(-3) = 16 + 12 = 28.$



• $$x = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}.$$



• Step 8 — Check against domain restrictions ($x\ne1,-2$):



• • $2+\sqrt{7} \approx 4.6458$ (allowed).



• • $2-\sqrt{7} \approx -0.6458$ (allowed).



• Both are valid (neither equals $1$ nor $-2$).



• Step 9 — (Optional numeric verification)



• • For $x=2+\sqrt{7}$: compute LHS $= \dfrac{2}{(2+\sqrt{7})-1}+\dfrac{3}{(2+\sqrt{7})+2}$ which evaluates numerically to $1$ (to rounding).



• • For $x=2-\sqrt{7}$: similarly, LHS evaluates to $1$ (to rounding).



• Final Answer: $$\boxed{\{\,2+\sqrt{7},\;2-\sqrt{7}\,\}}.$$

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Rational equation

• Exclude x = 1 and x = -2.


• Multiply both sides by (x−1)(x+2) to get 2(x+2)+3(x−1)=(x−1)(x+2).


• Left: 5x+1; Right: x^2 + x − 2; so 5x+1 = x^2 + x − 2 → x^2 − 4x − 3 = 0.


• Use quadratic formula: x = 2 ± √7.


• Both solutions are allowed since neither equals 1 nor −2.

Related

• Partial Fractions