randRangeNonZero(-10, 10) randRangeNonZero(-10, 10) randFromArray([A, B, randRangeNonZero(-10, 10)]) A * B -A - B (function() { var coefficient1 = 0; var coefficient2 = COEFFICIENT; var constant1 = 0; var constant2 = CONSTANT; var simplify; if (rand(2) < 1) { coefficient1 = randRange(-COEFFICIENT, COEFFICIENT); coefficient2 = COEFFICIENT - coefficient1; simplify = coefficient1 !== 0 && coefficient2 !== 0; } else { constant1 = randRange(-CONSTANT, CONSTANT); constant2 = CONSTANT - constant1; simplify = constant1 !== 0 && constant2 !== 0; } return { coefficient1: coefficient1, coefficient2: coefficient2, constant1: constant1, constant2: constant2, simplify: simplify } })() new RationalExpression([[1, {x: 2}], [TERMS.coefficient1, 'x'], TERMS.constant1]) new RationalExpression([[-TERMS.coefficient2, 'x'], -TERMS.constant2]) TERM2.multiply(-1)

Solve for `x`.

`\dfrac{TERM1}{x - NUMERATOR} = \dfrac{TERM2}{x - NUMERATOR}`

A B
Fill in the blanks with the solution(s) of the equation. If there is only one solution, leave the second box empty.
`x =` a `x =` a
However, the original expression is undefined when `x = NUMERATOR`.
Therefore, the only solution is `x = A`.
Therefore, the only solution is `x = B`.
NUMERATOR === A || NUMERATOR === B ? "<code>x = " + NUMERATOR + "</code>" : "No extraneous solutions" shuffle([ "<code>x = " + NUMERATOR + "</code>", "<code>x = " + (-NUMERATOR) + "</code>", ])

What is the extraneous solution to this equation?

`\dfrac{TERM1}{x - NUMERATOR} = \dfrac{TERM2}{x - NUMERATOR}`

ANSWER

- POSSIBILITIES[0]
- POSSIBILITIES[1]
- No extraneous solutions

At `x = NUMERATOR`, the denominator of the original expression is `0`.

Since the expression is undefined at `x = NUMERATOR`, it is an extraneous solution.

The original expression is defined at `x = A` and `x = B`, so there are no extraneous solutions.

Multiply both sides by `x - NUMERATOR`:

`\qquad \dfrac{TERM1}{x - NUMERATOR} (x - NUMERATOR) = \dfrac{TERM2}{x - NUMERATOR} (x - NUMERATOR)`

`\qquad TERM1 = TERM2`

Subtract `TERM2` from both sides:

`\qquad TERM1 - (TERM2) = TERM2 - (TERM2)`

`\qquad TERM1 + TERM2NEG = 0`

`\qquad TERM1.add(TERM2NEG) = 0`

Factor the expression:

`\qquad (x - A)(x - B) = 0`

Therefore `x = A` or `x = B`