MAT 115 Course Prep
This collection of problems is meant to provide practice for a range of skills needed for MAT 115. Read and attempt each problem first; if you aren’t sure how to start a problem, explore the resources on the right to refresh your memory and try again. A dropdown button is found beneath each problem for you to compare both your logic and final answers. Keep track of the skills you aren’t comfortable with, and reacquaint yourself with them so you’re fully prepared for the topics you’ll grapple with soon. Get help in the Math Lab. Return to the Course Prep page.
Problem 1: Using the functions defined below, simplify each expression. If possible, find exact values.
Part a. 
Click here to show solution. 
Video: Worked Example: Evaluating Functions from Equation 
Part b. 
Click here to show solution. 

Part c. 
Click here to show solution. 

Part d. 
Click here to show solution.Since no value of is given, an exact value is not possible as an answer. 

Part e. 
Click here to show solution. 

Part f. 
Click here to show solution.since the top and bottom share a common factor of which cancels (assuming ). 

Part g. 
Click here to show solution.. 

Part h. 
Click here to show solution.by the distributive property. 
Problem 2: Solve the following equations for x, and verify your solution.
Part a. 
Click here to show solution.Divide both sides by 0.2: . To verify, plug this value of back into the original question and make sure that the statement is true. In this case, the choice of gives us . Since this is true, we know our solution is correct. To give you an idea of what it would look like if we somehow got the incorrect answer, pretend we solved the original equation and got . Then, in the verification step, we would have . This is obviously false, and so is not a valid solution to the equation. 
Video: Solving Linear Equations I 
Part b. 
Click here to show solution.Divide both sides by . This has the effect of multiplying each side by : , which reduces to show that . We verify by plugging this value back into the original question: . This is true, and so we’ve verified our answer. 

Part c. 
Click here to show solution.Begin by subtracting 8 from each side, yielding . Division by 6 gives . Verify the solution by plugging this value of back into the original equation: . This is true, and so we’ve verified our answer. 
Video: Solving Linear Equations II 
Part d. 
Click here to show solution.On the lefthand side, a factor of is common to each term in the numerator, and cancels with the found in the denominator assuming : . Simplification gives , and adding 6 to both sides gives . Verify the solution by substitution: . This is true, and so we’ve verified our answer. 

Part e. 
Click here to show solution.To isolate the , raise both sides to the power. This gives . We plug in this value of to verify the answer: . This is true, and so we’ve verified our answer. 
Video: Solving Equations with Rational Exponents 
Part f. 
Click here to show solution.In a similar fashion as the previous problem, raise both sides to the power, giving . Using a calculator, we find that this is approximately 1.447. To verify our answer, we can compute: . This shows that x = 1.447 is a correct approximation of the exact solution, which is x = . 

Part g. 
Click here to show solution.In earlier questions, was the only variable present, and “solving the equation” ended with a numerical solution. In cases like these, we must recall that “solving for ” is synonymous with isolating on one side of the equation. In this equation, divide both sides by : due to the cancellation on the lefthand side of the equation. We can still verify this solution via substitution for in the original equation: where the final step is possible due to the cancellation of ‘s on the left hand side. This statement is true, and so we’ve verified our answer. 

Part h. 
Click here to show solution.In a similar fashion as the previous problem, we simply aim to isolate on one side of the equation. Divide by on both sides: . Verify via substitution: , where the final step is possible due to the cancellation of ‘s on the right hand side. This statement is true, and so we’ve verified our answer. 
Problem 3: Expand the following expressions.
Part a. 
Click here to show solution.This can be rewritten as , which can be FOILed (recall this tells you the order in which to multiply terms – First, Outer, Inner, Last). This gives . 
Video: Multiplying Binomials 
Part b. 
Click here to show solution.Following the same logic as the last problem, the multiplication gives . 
Problem 4: Find the slope of the line passing through the following points.
Part a.  (0,5) and (4,19)
Click here to show solution.Slope (commonly denoted by the symbol ) can be found via the familiar phrase “rise over run”. More accurately, we are concerned with the change in rise over the change in run . In this case, this comes out to . 
Worked Example: Slope from Two Points 
Part b.  (2,11) and (3,3)
Click here to show solution.Following the same logic as the previous problem, we have . 

Part c.  and
Click here to show solution.This represents the general case: . 
Problem 5: Find the equation of the line passing through the following points.
Part a.  (0,5) and (4,19)
Click here to show solution.There are two lanes of thought here, corresponding to the two different generic forms of a line: slopeintercept form and pointslope form . In slopeintercept form, the slope and yintercept of the line need to be known (unsurprisingly). We calculated the slope between these points above: . Separately, when , we know that since is given as a point on this line. Therefore, the yintercept, commonly denoted with the symbol , is 5. The equation of the line is therefore . In pointslope form, the slope and any one point on the line must be known (unsurprisingly). Choose , for example: then which reduces to . Adding 5 to both sides gives the same answer as before: . Pointslope form is generally more helpful, as needing any point on the line is a less strict requirement than knowing the yintercept. The slope is needed in either case. 
Video: PointSlope and SlopeIntercept Equations 
Part b.  (2,11) and (3,3)
Click here to show solution.We don’t know the yintercept here, so use pointslope form with either of the points provided. For the purpose of this example, we’ll use ; recall from the earlier question that the slope between these points is 14. The equation of the line is . Adding 11 to both sides gives . (Note that from this manipulation, we can actually determine the yintercept to be 39, as the final simplified answer is in slopeintercept form with . This isn’t obvious from the information given initially.) 

Part c.  and
Click here to show solution.Use pointslope form again. For slope, recall that , and note that either point can be plugged into the pointslope formula. Both and are correct. 
Problem 6: Your annual salary is $40,000 for the year 2018. Use the information below to calculate your salary in 2019, 2020, and 2021.
Part a.  Assume your salary increases by $1500 per year.
Click here to show solution.Take the salary for each year and add 1500 to get the salary for the next year.


Part b.  Assume your salary increases by 3% per year.
Click here to show solution.Let’s start with the calculation for 2019. An increase of 3% over the salary in 2018 means that is earned in 2019. Factor 40,000 out of the left hand side to find . This logic can be applied recursively for the other future years:

Problem 7: Use the following graph to answer the questions below.
Problem 8: Rewrite each expression in the form: x^{a} (this is called baseexponent form).
Part a. 
Click here to show solution.We can move a factor from the denominator to the numerator as long as we change the sign of the exponent: If it seems like the 1 in the numerator disappeared, remember that multiplying by one won’t change the value of . Dividing by 1 also won’t change the value of , so we can just write: If we want to show all the steps, we could write: 
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Part b. 
Click here to show solution.We can rewrite radical expressions using fractional exponents. For a square root, we use the fraction 1/2: If this seems strange, think of it this way: 2^{0} = 1 and 2^{1} = 2 so we would expect that would produce a number between 1 and 2, and is between 1 and 2. 

Part c. 
Click here to show solution.We can distribute an exponent over division, which means the 4 becomes the exponent of the 1 in the numerator and the x^3 in the denominator: Now 1 multiplied by itself 4 times is still 1, but in the denominator we have to combine the exponents. Recall that successive exponents can be combined by multiplication: Finally, we can rewrite this by moving the x into the numerator, and changing its exponent from positive to negative: We could also have written out 4 copies of the original fraction: This could then be expanded as: 

Part d. 
Click here to show solution.Recall that the 3 is called the index of the root, and indicates that this is not a square root but a cube root. Following our earlier rule, we can rewrite this root using a 1/3 exponent: Exponent rules allow us to multiply successive exponents, so we calculate 2 times 1/3 to get 2/3 as the exponent: 
Problem 9: Expand the following expressions. Remember to use FOIL.
F.O.I.L. stands for First, Outer, Inner, Last.
When two binomials are being multiplied together, we can use FOIL to remember all the products that appear in the expansion. For example: is the product that results from multiplying the First terms together. is the product of the Outer terms. is the product of the Inner terms. is the product of the Last terms. 
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Part a. 
Click here to show solution.Following FOIL will give us the following: We can take each product and simplify it as: Now we can combine 6x and 4x to make 10x: 

Part b. 
Click here to show solution.Following FOIL will give us the following: We can take each product and simplify it as: Now we can combine 3x and 8x to make 5x: 

Part c. 
Click here to show solution.Following FOIL will give us the following: We can take each product and simplify it as: Notice how x times 7 gave us a positive result, because the negatives cancel each other. Now we can combine 7x and 2x to make 9x: 

Part d. 
Click here to show solution.Following FOIL will give us the following: We can take each product and simplify it as: Notice how the first term is positive, because it is the product of two negatives. This is also true of the third term, 1x. Also, instead of writing 1x, we usually just write x, and the coefficient of 1 is implied Now we can combine 5x and x to make 4x: 
Problem 10: Rewrite each expression in factored form.
Part a. 
Click here to show solution.List the factors of the constant term, 12: Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is 8, so we pick 2 and 6. 
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Part b. 
Click here to show solution.List the factors of the constant term, 8: Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is 9, so we pick 1 and 8. 

Part c. 
Click here to show solution.List the factors of the constant term, 15: Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is 2, so we pick 3 and 5, because adding them together gives us 3 + 5 = 2. 

Part d. 
Click here to show solution.List the factors of the constant term, 42: Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is 1, so we pick 6 and 7, because when we add them together we get 6 + 7 = 1. 

Part e. 
Click here to show solution.List the factors of the constant term, 12: Notice that we include pairs where both numbers are negative. We want to choose the pair that adds up to the coefficient of x, which is 7, so we need to pick two negative numbers. We pick 3 and 4 because adding them gives us 3 + 4 = 7 

Part f. 
Click here to show solution.List the factors of the constant term, 40: Notice that we include each pair twice, once with the smaller number negative and once with the larger number negative. We want to choose the pair that adds up to the coefficient of x, which is 3, so we pick 5 and 8, because when we add them together we get 5 + 8 = 3. 
Problem 11: Solve the linear inequality.
Part a. 
Click here to show solution.Notice that there are terms containing x on both sides of the inequality. First, we can subtract 4x from both sides so that we can combine these like terms: Now we can move all the constants to the other side, which means we must add 13 to both sides: Finally, we can divide both sides by 3 to isolate the variable: This says 6 is less than x, which is another way of saying that x is greater than 6. Thus, we have: 
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Problem 12: Express the following using interval notation.
Part a. 
Click here to show solution.The statement literally says that 7 is less than x, which is less than 13. In other words, x is greater than 7 and x is less than 13. This indicates a single interval, from 7 to 13. Because the original statement uses and not we know that 7 and 13 are NOT included in the interval. We use “soft brackets” (parentheses) to indicate that the interval includes all numbers between 7 and 13, but does NOT include the endpoints: 
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Part b. 
Click here to show solution.The statement literally says that 16 is less than or equal to x, which is less than or equal to 10. In other words, x not greater than 10, and is not less than 16. This indicates a single interval with endpoints at 16 and 10. Because the original statement uses and not we know that 16 and 10 ARE included in the interval. We use “hard brackets” [square braces] to indicate that the interval includes all numbers between 7 and 13 and ALSO includes the endpoints: 

Part c. 
Click here to show solution.Compared to the previous two problems, we simply need to use a hard bracket to start our interval, because the 8 is NOT included, and a soft bracket to end the interval, because the 4 IS included. 

Part d. 
Click here to show solution.The statement says x is greater than 56. This excludes the possibility that x = 56, so we will use a soft bracket. All the numbers greater than 56 together form a sort of interval with a left endpoint at 56, but no right endpoint. Since the interval starts at 56 and never ends, we can think of the right endpoint as being “at” infinity. Infinity is not a number, so we do not want to include it, which means we must use a soft bracket here as well. Using infinity as our right endpoint means that all numbers between 56 and infinity are included, even if both endpoints are not. 

Part e. 
Click here to show solution.The statement says that x is less than or equal to 42. This includes the possibility that x = 42, so we must use a hard bracket. Since all negative numbers less than 42 are included in this interval, we need to start our interval at negative infinity (but not include it). 42 is actually where our interval ends. In other words, negative infinity is the left endpoint of this interval, and 42 is the right endpoint. 
Problem 13: Solve using the quadratic formula to get exact answers (expressed using radicals).
Part a. 
Click here to show solution.Recall the quadratic formula: In the quadratic equation given above, we can see that a = 1, b = 8, and c = 5. Notice that, even thought there is no visible coefficient in front of the x^{2} term, we always assume a coefficient of 1, which is why a = 1. Also, the minus sign in front of the 8 must be included in the b value, which is why we have b = 8. Notice how we placed parentheses around every place where we plugged in a value. Now we want to simplify (8) = 8 and (8)^{2} = 64 and 4(1)(5) = 20 and 2(1) = 2. Since 64 – 20 = 44, and 44 = 4 times 11, we can write: 4 is a square number, so we can pull it out from under the square root as a 2: Now, we can divide by 2. Recall that each term in the numerator must be divided by 2: Finally, recall that the symbol means that we actually have two separate solutions for x: 
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