# Algebra II : Solving Rational Expressions

## Example Questions

### Example Question #5 : How To Find The Solution To A Rational Equation With Lcd

Solve for :

Explanation:

Since the two fractions already have a common denominator, you can add the fractions by adding up the two numerators and keeping the common denominator:

Next you will algebraically solve for  by isolating it on one side of the equation. The first step is to multiply each side by :

Cancel out the  on the left and distribute out on the right. Then solve for  by subtracting  to the left and subtracting 10 to the right. Finally divide each side by negative 2:

### Example Question #6 : How To Find The Solution To A Rational Equation With Lcd

Solve the rational equation:

no solution;  is extraneous

Explanation:

With rational equations we must first note the domain, which is all real numbers except . (If , then the term  will be undefined.) Next, the least common denominator is , so we multiply every term by the LCD in order to cancel out the denominators. The resulting equation is . Subtract  on both sides of the equation to collect all variables on one side: . Lastly, divide by the constant to isolate the variable, and the answer is . Be sure to double check that the solution is in the domain of our equation, which it is.

### Example Question #11 : How To Find A Solution Set

Solve for :

There is no solution.

Explanation:

Subtract 1 from both sides, then multiply all sides by :

A quadratic equation is yielded. We can factor the expression, then set each individual factor to 0.

Both of these solutions can be confirmed by substitution.

### Example Question #671 : Intermediate Single Variable Algebra

Solve for :

Explanation:

In order to solve for , we need to consider this equation as two proportions being set equal:

Now, we can cross-multiply.

Distribute the  on the right side:

Subtract  from both sides to start combining like terms:

Now, this is just a two-step equation. Start by subtracting  from both sides:

Divide by .

### Example Question #672 : Intermediate Single Variable Algebra

Solve for :

Imaginary solution

and

Explanation:

First, cross-multiply:

Once we simplify we are left with the following two quadratic equations:

In order to solve a quadratic, we need to have it equal to zero and then we can use the quadratic formula. What we need to do now is combine like terms. We can subtract all of the terms on the left from the like terms on the right:

This gives us:

Now we can use the quadratic formula:

Where the quadratic equation follows the pattern:

Therefore, we can use the terms in our quadratic equation and rewrite the equation as follows:

Since we can re-write as and , our answer becomes

### Example Question #673 : Intermediate Single Variable Algebra

Solve for :

Imaginary solution

Explanation:

Consider this problem as 2 proportions set equal:

Now, we can cross-multiply.

Using FOIL (method of simplifying binomials by multiplying following the pattern of first terms, outer terms, inner terms, and last terms), we can multiply the binomials on the right giving us:

In order to solve this equation, we need to have one side of the equation equal to .

Add  to both sides.

Now, we can solve by using the quadratic formula, or by factoring. If we factor, we get:

Solving for each equation leaves us with the following answers:

and

### Example Question #674 : Intermediate Single Variable Algebra

Solve for :

Imaginary solution

Explanation:

In order to solve, we need to first cross-multiply.

Using FOIL (method of simplifying binomials by multiplying following the pattern of first terms, outer terms, inner terms, and last terms), we can multiply the binomials on the left giving us:

In order to continue solving this quadratic, we need to subtract  from both sides so that the quadratic is equal to .

Now, we can solve using the quadratic formula:

Where the quadratic equation follows the pattern:

Therefore, we can use the terms in our quadratic equation and rewrite the equation as follows:

Since we can re-write as  and , our answer becomes:

### Example Question #675 : Intermediate Single Variable Algebra

Solving Rational Expressions

Solve the below equation for

Explanation:

When solving two rational expressions that are set equal to each other Cross Multiply.

In this case we will multiply 4 by the the (3x-2), and also multiply 5 by the (2x+7)

Distribute on both sides to get:

Subtract 10x from both sides to get:

Add 8 to both sides to get:

Divide both sides by 2 to get the final solution:

Finally, double check for extraneaous solutions. Anytime you might get a zero in the bottom of a fraction, this is considered extraneaous because it is a mathematical impossibility to divide by zero.

In this case there are no x values in the denominators of the original problem, so there are no extraneaous solutions and the only solution will be:

### Example Question #11 : Solving Rational Expressions

If , what is ?

Explanation:

We start by taking the original function, and replacing all the 's with .  We end up with:

Then we can solve like normal:

If we noticed it, we also could have factored the numerator into:

The  terms would have canceled leaving:

Solving that would have been much easier:

Either way, we would still get the exact same answer.

If , find .