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