### All Algebra II Resources

## Example Questions

### Example Question #1 : Solving Rational Expressions

Solve for :

**Possible Answers:**

**Correct answer:**

To solve this rational equation, start by cross multiplying:

Then, distribute the right side:

Finally, subtract from both sides and bring the over to the left side:

Dividing by gives the answer:

### Example Question #1 : Solving Rational Expressions

Solve for :

**Possible Answers:**

**Correct answer:**

The first step is to multiply everything by a common denominator. One way to do this is to multiply the entire equation by all three denominators:

Then, to solve for , use the quadratic formula:

### Example Question #1 : Solving Rational Expressions

Solve for , given the equation below.

**Possible Answers:**

No solutions

**Correct answer:**

Begin by cross-multiplying.

Distribute the on the left side and expand the polynomial on the right.

Combine like terms and rearrange to set the equation equal to zero.

Now we can isolate and solve for by adding to both sides.

### Example Question #1 : How To Find Out When An Equation Has No Solution

Solve the rational equation:

**Possible Answers:**

no solution

or

**Correct answer:**

no solution

With rational equations we must first note the domain, which is all real numbers except** **and** **. That is, these are the values of that will cause the equation to be undefined. Since the least common denominator of , , and is , we can mulitply each term by the LCD to cancel out the denominators and reduce the equation to . Combining like terms, we end up with . Dividing both sides of the equation by the constant, we obtain an answer of . However, this solution is NOT in the domain. Thus, there is NO SOLUTION because is an extraneous answer.

### Example Question #1 : How To Find The Solution Of A Rational Equation With A Binomial Denominator

Simplify:

**Possible Answers:**

**Correct answer:**

Factor out from the numerator which gives us

Hence we get the following

which is equal to

### Example Question #1 : How To Subtract Polynomials

Solve:

**Possible Answers:**

**Correct answer:**

First we convert each of the denominators into an LCD which gives us the following:

Now we add or subtract the numerators which gives us:

Simplifying the above equation gives us the answer which is:

### Example Question #3 : How To Find The Solution Of A Rational Equation With A Binomial Denominator

Solve for .

**Possible Answers:**

,

,

,

,

**Correct answer:**

,

The two fractions on the left side of the equation need a common denominator. We can easily do find one by multiplying both the top and bottom of each fraction by the denominator of the other.

becomes .

becomes .

Now add the two fractions:

To solve, multiply both sides of the equation by , yielding

.

Multiply both sides by 3:

Move all terms to the same side:

This looks like a complicated equation to factor, but luckily, the only factors of 37 are 37 and 1, so we are left with

.

Our solutions are therefore

and

.

### Example Question #4 : How To Find The Solution To A Rational Equation With Lcd

Solve for :

**Possible Answers:**

**Correct answer:**

Multiply both sides by :

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is . These integers are :

Set each factor equal to 0 and solve separately:

or

### Example Question #4 : Solving Rational Expressions

Simplify the following expression:

**Possible Answers:**

This expression is already simplified.

**Correct answer:**

The first step of problems like this is to try to factor the quadratic and see if it shares a factor with the linear polynomial in the denominator. And as it turns out,

So our rational function is equal to

which is as simplified as it can get.

### Example Question #1 : Solving Rational Expressions

Evaluate the following expression:

**Possible Answers:**

You cannot divide fractions.

**Correct answer:**

When dividing fractions, we multiply by the reciprocal of the second fraction. Therefore, the problem becomes:

Our final unfactored expression is therefore .

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