Video: Extraneous Solutions with Rationals

Find the solution set of the equation 2/(π‘₯ + 2) + 5/(π‘₯ βˆ’ 5) = 35/(π‘₯Β² βˆ’ 3π‘₯ βˆ’ 10).

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Video Transcript

Find the solution set of the equation two divided by π‘₯ plus two plus five divided by π‘₯ minus five is equal to 35 divided by π‘₯ squared minus three π‘₯ minus 10.

Before we try to solve the equation, it is worth considering whether there are any values of π‘₯ that give us undefined values. To work out which input values generate undefined output, we find the values of π‘₯ that give us zero denominators.

As the denominator of the right-hand side of the equation factorizes to π‘₯ minus five multiplied by π‘₯ plus two, our undefined values work here when π‘₯ minus five equals zero or when π‘₯ plus two equals zero. Therefore, π‘₯ equals five and π‘₯ equals negative two cannot be solutions to the equation as they give undefined values.

We will now solve the equation in the standard way in the knowledge that π‘₯ equals five and π‘₯ equals negative two cannot be solutions to the equation. The common denominator of the left-hand side of the equation is π‘₯ plus two multiplied by π‘₯ minus five. Therefore, this simplifies to two multiplied by π‘₯ minus five plus five multiplied by π‘₯ plus two divided by π‘₯ plus two multiplied by π‘₯ minus five.

Expanding the parentheses or brackets simplifies the numerator to two π‘₯ minus 10 plus five π‘₯ plus 10. Simplifying this further by grouping like terms gives us seven π‘₯ divided by π‘₯ plus two multiplied by π‘₯ minus five. Factorizing the denominator of the right-hand side of the equation gives us π‘₯ plus two multiplied by π‘₯ minus five.

As the denominators of the left-hand side and right-hand side of the equation are equal, this equation would normally be easy to solve. We will put the numerators equal to each other. Seven π‘₯ equals 35. Dividing both sides of the equation by seven gives us an answer of π‘₯ equals five. Therefore, we would assume that π‘₯ equals five is a valid solution to the equation.

However, we know that the equation is undefined at π‘₯ equals five. This means that there are no solutions to the equation or the solution set is empty. As the equation is undefined at π‘₯ equals five, there is no solution to the equation two divided by π‘₯ plus two plus five divided by π‘₯ minus five is equal to 35 divided by π‘₯ squared minus three π‘₯ minus 10.

We could check this by substituting π‘₯ equals five back into the initial equation. When we do this, the second and third fraction are undefined as the denominator is equal to zero.

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