Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #112 : Solving Rational Expressions

If , what is ?

Possible Answers:

Correct answer:

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.

Example Question #17 : Solving Rational Expressions

If , find .

Possible Answers:

Correct answer:

Explanation:

When we first look at this problem, we might be panicking because just plugging  into the function makes us divide by , which we don't like.  For now, let's forget about the denominator and focus on the numerator.  If we look closely, we can see that we can factor the numerator:

Giving us:

As we can see, the  will cancel from the numerator and the denominator, clearing up our "divide by 0" problem:

Now we can solve:

Example Question #21 : Solving Rational Expressions

If  and , what is ?

Possible Answers:

Correct answer:

Explanation:

To begin, let's write the whole problem out plainly:

From here, in order to add the two fractions, we need to get the denominators to be the same.  To do this, we multiply the numerator and denominator of each fraction by the denominator in the OTHER fraction:

Now we can add the fractions together:

Now we can expand using FOIL:

Finally, we collect like terms in the numerator for a final answer of:

Example Question #681 : Intermediate Single Variable Algebra

If , what is ?

Possible Answers:

Correct answer:

Explanation:

First we're going to give the function a quick glance to see if we can simplify by factoring.  We can't, so in order to solve, we're going to replace all the 's with the number we want to solve for, :

Example Question #1823 : Algebra Ii

Solve for x: 

Possible Answers:

Correct answer:

Explanation:

Step 1: We need to square both sides of the equation. Squaring a square root will just leave me with whatever is inside the root.





Step 2: Move all terms on the left side to the right side.



Step 3: Factor the right side.  We want two numbers that multiply to  and add to 



Rewrite the trinomial as two binomals:



Step 4: Set each parentheses equal to  and solve for :




The solutions to this equation are -4 and 5.

Example Question #1824 : Algebra Ii

Solve for .

Possible Answers:

Correct answer:

Explanation:

To solve for the variable , isolate the variable on one side of the equation with all other constants on the other side. To accomplish this perform the opposite operation to manipulate the equation.

First cross multiply.

Next, divide by four on both sides.

 

Example Question #1825 : Algebra Ii

Solve:  

Possible Answers:

Correct answer:

Explanation:

The terms of the numerator cannot be added or subtracted without a common denominator.

Convert the first term with a denominator of .

Expand the first term and combine the fractions as one fraction.  Be sure to enclose the numerator of the second term in parentheses.

Simplify this fraction.

The answer is:  

Example Question #1826 : Algebra Ii

Solve for .

Possible Answers:

Correct answer:

Explanation:

To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

 

Cross multiply.

 

Divide  on both sides.

Example Question #1827 : Algebra Ii

Solve for .

Possible Answers:

Correct answer:

Explanation:

To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

 

Cross multiply.

 

Divide  on both sides.

Example Question #1828 : Algebra Ii

Solve for .

Possible Answers:

Correct answer:

Explanation:

To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

 

Cross multiply.

 

Divide  on both sides.

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