You may clear the denominators in an equation. You may not in an inequality.
Solve for x
2x − 1 = 3x
Cross-multiply: 2x = 3(x − 1) = 3x − 3, so −x = −3 and x = 3. Check it against the restrictions x ≠ 1 and x ≠ 0 — 3 breaks neither, so it stands. B is the trap: x = 1 makes the original left side undefined, so even if it fell out of the algebra it could never be an answer. Substitute to confirm: 22 = 1 and 33 = 1 ✓
They come from the ORIGINAL denominators, and they survive every later step.
f(x) = x + 5x² − 4
a rational function
x² − 4 = (x + 2)(x − 2)
factor the bottom
x ≠ 2, x ≠ −2
the domain is everything else
A rational function is a fraction of two polynomials. A rational equation sets two of them equal. A rational inequality asks which x make one bigger or smaller than the other — and that last one behaves differently, as step 4 shows.
x3 + 12 = 56
LCD is 6
6(x3) + 6(12) = 6(56)
every term, not just some
2x + 3 = 5
the fractions are gone
x = 1
no restriction to check here
Multiplying only the terms that have fractions is the commonest wreck in this topic. The 56 on the right is a term too, and skipping it changes the equation into a different one that happens to look tidier.
So the check is part of the method, not an optional tidy-up.
1x − 4 = xx − 4, x ≠ 4
restriction first
1 = x
multiply both sides by (x − 4)
x = 1 ✓
1 ≠ 4, so it survives
Now change the left side to 4x − 4. The same step gives x = 4 — which the restriction forbids. That is an extraneous root: created by the multiplying, not by the original equation, and the honest answer is "no solution".
Multiplying an inequality by a quantity whose sign you do not know is guessing whether to flip it. Both answers cannot be right. The way round it is to get everything on one side, make a single fraction, and read the signs off a chart.
When x sits on top of one fraction and under another, clearing the bottoms gives an x². Solve it as a quadratic, then check BOTH roots against the restrictions.
Solve 2x = xx + 4.
x ≠ 0, x ≠ −4
restrictions first
2(x + 4) = x · x
cross-multiply
x² − 2x − 8 = 0
everything on one side
(x − 4)(x + 2) = 0
factor
x = 4 or x = −2
neither is 0 or −4, so both stand
Check both: 24 = 12 and 48 = 12 ✓; 2−2 = −1 and −22 = −1 ✓. A quadratic can give two answers, one answer and one extraneous root, or none.
One root thrown out: x²x − 2 = 4x − 2
x ≠ 2
restriction
x² = 4
multiply by (x − 2)
x = 2 or x = −2
both square to 4
x = −2 only
2 is forbidden
Check: (−2)²−4 = −1 and 4−4 = −1 ✓. The algebra gave two roots; the restriction kept one.
A fraction answer: 52x = 3
x ≠ 0
restriction
5 = 6x
multiply by 2x
x = 56
not a whole number, and that is fine
Check: 2x = 53, and 5 ÷ 53 = 3 ✓.
A number on the other side of an inequality: xx − 1 ≥ 2
xx − 1 − 2 ≥ 0
move the 2 across; do not multiply
x − 2(x − 1)x − 1 = 2 − xx − 1 ≥ 0
one fraction
boundaries x = 2 (top zero), x = 1 (bottom zero)
test x = 0, 1.5, 3: −, +, −
1 < x ≤ 2
2 included, 1 never
Check x = 1.5: 1.50.5 = 3, which is at least 2 ✓. Multiplying by (x − 1) at the start would have needed the sign of x − 1, which we do not know.
x − 2x + 1 ≥ 0
already one fraction
zeros at x = 2, undefined at x = −1
the two boundary points
x < −1: −− = +
test x = −2
−1 < x < 2: −+ = −
test x = 0
x > 2: ++ = +
test x = 3
Note the two different brackets. At x = 2 the fraction is 0, and the sign is "≥", so 2 is included. At x = −1 the fraction is undefined, so −1 is excluded no matter what the sign says. A zero on the top can be included; a zero on the bottom never can.
Periodical exams ask for the excluded values, a rational equation to solve and check, or a short inequality; entrance tests favour an equation whose algebra produces a forbidden value, so the answer is "no solution" or only one root. Write the restrictions first, clear the bottoms, and solve. The 10-second check is to substitute each answer into the ORIGINAL equation — if a bottom becomes zero, cross that answer out.
Solve for x
x + 2x − 3 = 2
Multiply both sides by (x − 3), noting x ≠ 3: x + 2 = 2(x − 3) = 2x − 6. Then −x = −8, so x = 8. Check: 105 = 2 ✓ C is the excluded value and can never be a solution. A comes from forgetting to distribute the 2, giving x + 2 = 2x − 3.
Hint: write x ≠ 3 down before you start. The 2 on the right has to be multiplied by the whole bracket (x − 3), both terms.
