Basic Single-Variable Algebra - High School Math

If the original price of the T-shirt is , increasing the price by 50% means that the new price is 150% of , or .

If the price is then decreased by 50%, the new price is 50% of

or

The ratio of to is then:

The 's in the numerator and denominator cancel, leaving , or

.

Find the common multiple between the numerator and denominator.

divide numerator and denominator by 3:

divide numerator and denominator by 7:

divide numerator and denominator by 4:

Cannot be divided any more- lowest terms.

The mathematical expression given in the question is . Adding together like terms, , this can be simplified to . The expression can be factored as . For every positive integer , must be a multiple of 5. If , then , which is not an integer multiple of 2, 8, 10, or 15. Therefore, the correct answer is 5.

Multiply both sides by to get:

Distribute the :

Combine like terms:

Divide both sides by :

Divide both sides by :

. However, cannot be simplified any further because the terms have different exponents.

(Like terms are terms that have the same variables with the same exponents. Only like terms can be combined together.)

When multiplying rational expressions, we simply have to multiply the numerators together and the denominators together. (Warning: you only need to find a lowest common denominator when adding or subtracting, but not when multiplying or dividing rational expression.)

In order to simplify this, we will need to factor and . Because looks a little simpler, let's start with it first.

We can easily factor out a four from both terms.

.

Next, notice that fits the form of our difference of squares factoring formula. In general, we can factor as . In the polynomial we will let and . Thus, .

Now, we can see that.

We then factor . This also fits our difference of squares formula; however, this time and . In other words, . Applying the formula, we see that

. Now, let's take our factorization one step further and factor , which we already did above.

.

Be careful here. A common mistake that students make is to try to factor . There is no sum of squares factoring formula. In other words, in general, if we have , we can't factor it any further. (It is considered prime.)

We will then put all of these pieces of information in order to simplify our rational expression.

Lastly, we cancel the factors that appear in both the numerator and the denominator. We can cancel an and a term.

.

The answer is .

The domain of a function consists of all of the values of x for which f(x) is defined. When determining the domain of a funciton, the three most important things we want to consider are square roots, logarithms, and denominators of fractions. These tend to signal places where the function is not defined.

First, let's look at the term. Remember we can only find the square root of nonnegative values. Thus, everything under the square root symbol must be greater than or equal to zero. This tells us that, for this function, .

Second, we need to look at the natural logarithm. The natural logarithm can only be applied to positive numbers (which don't include zero). Thus, everything within the paranethesis of the natural logarithm must be greater than zero.

There are several ways to solve this inequality. One way is to factor the left side and examine the factors. We know that because of the difference of squares factorization formula.

.

This statement will only be true in two situations; either both factors must be positive, or both must be negative.

We can see that if , then the factor will be positive, but the factor will be negative. If we were to multiply a negative and a positive number, we would get a negative number. Thus, is not larger than zero when .

Let's consider the interval . In this case, both and would be positive. Thus, when .

Third, consider the interval . In this case, the first factor will be negative, and the second will be positive, so their product would be negative, and would not be greater than zero.

To summarize, only if .

We can see now that f(x) is only defined if and .

There is one more piece of information we need to consider--the denominator of f(x). Remember that a fraction is not defined if its denominator equals zero. Thus, if the denominator is equal to zero at a certain value of x, we can't include this value of x in the domain of f(x).

We can set the denominator equal to zero and solve to see if there are any values of x where the denominator would be zero.

Rewrite this as an exponential equation. In general, the equation can be rewritten as , provided that a is positive.

If we put into exponential form, we obtain

We can solve this for x.

So, let's put all of this information together. We know that f(x) is only defined if ALL of these conditions are met:

The only interval for which this is true is if x is greater than (and not equal to) zero but less than (and not equal to) 1. Thus, the domain of f(x) is .

The answer is .

To solve this problem, plug into and simplify. Then plug that expression into :

When multiplying an even number of negatives, you get a positive.

When multiplying an odd number of negative, you get a negative.

When multiplying an odd number of negatives, the answer is negative.

When multiplying an even number of negatives, the answer is positive.

Change the mixed numbers into improper fractions by multiplying the whole number by the denominator and adding the numerator:

Dividing by a fraction is the same as multiplying by the reciprocal, so the problem becomes .

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .

1st equation:

2nd equation:

Subtract the 2nd equation from the 1st equation to eliminate the "2y" from both equations and get an answer for x:

Plug the value of into either equation and solve for :

The break-even point occurs when the .

The equation to solve becomes

so the break-even point is .

So the equation to solve becomes , or must be sold to make a profit of .

The two equations in this system can be combined by addition or subtraction to solve for and . Isolate the variable to solve for it by multiplying the top equation by so that when the equations are combined the term disappears.

Divide both sides by to find as the value for .

Substituting for in both of the two equations in the system and solving for gives a value of for .

Rewrite as a compound statement and solve each part separately:

The solution set is