Week 16 – precalculus 11

Solving Rational Equations

To solve a rational equation we must cancel out the denominators. To do this we must multiply by each denominator, this will cancel them out but because we have to do the same for everything in the equation, we will have then multiply the numerators by the values that were not canceled out. Before we do anything we must find the non-permisible values, values that cannot be in the equation because they will make a denominator zero, which is not allowed. If any of the solutions for x is a non-permissible value, it is extraneous and is not a real solution. To explain this we must show an example. It should also be mentioned that if there are only two fractions in the equation, we can just cross multiply to get the answer. The goal of the equation is to isolate and find x. We may have to solve a quadratic or linear equation to get x.

We will start with a 2 simple examples and then do a more complex one.

We will start with \frac {3}{4}=\frac{3}{x}

We must first find the non-permissible values for x, which in this case would only be 0. We can then cross multiply


and we will divide by 3 to isolate x, giving us: 4=x, since this does not match any of our non-permissible values, it is a solution.

Now we will try \frac{2x+1}{x-4}=\frac{4x-3}{2x+1}

First we will find the non-permissible values, in this case, -1/2, 3/4, and 4

Now we could either cross mutliply or eliminate the denominators, because we have already done cross multiplying, we will cancel the denominators. In this case cross multiplying is much easier though. We will multiply both sides by (2x+1)(x-4). One of each will cancel out the denominator on each side and we will multiply the numerator with the remaining one, giving us:


We will then multiply

(4x^2+ 4x + 1)=(4x^2 - 19x + 12)

We will now isolate x onto one side, this will cancel out the quadratic part or the x squared, if it did not we would bring everything to one side and factor. We will be left with:

23x= 11

The final answer will be x = \frac {11}{23} This does not match any of the non-permissible values so it is a real solution.

Finally we will do a harder question using v for x:

\frac {1}{v} + \frac{1}{v-3} = \frac {v-2}{v-3}

We must first find the non-permissible values which are 0 and 3

Luckily, two of the denominators are the same, we will only need to multiply each set by (v-3) and (v) to eliminate the denominators.

This will leave us with

1(v-3) + 1(v) = v(v-2)

Then it will become

v - 3 + v = v^2 - 2v

We will want to bring everything to one side because it’s quadratic, becoming:

0=v^2 - 4v + 3

We can then factor this and get


v will then equal 3 and 1 and because 3 matched one of our non-permissible values, the only real solution is 1.

Solving a Rational Equation is not hard but one must think the question through and recall methods from previous units.

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