MTH 100 College Algebra Essex County College Division of Mathematics Sample Review Questions 1 Created June 6, 2011

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1 MTH 00 College Algebra Essex County College Division of Mathematics Sample Review Questions Created June 6, 0 Math 00, Introductory College Mathematics, covers the mathematical content listed below. In order to place out of Math 00 or to prepare for the final examination in this course, one should be extremely comfortable with all of these items. NOTE: Calculators are permitted in this course, but should only be used when appropriate. The content listed below is covered in chapters through 8 and 0 in Algebra for College Students, Custom Version for Essex County College, Second Edition, by Bittinger & Ellenbogen (published by Addison Wesley, Inc.), the current textbook used in Math 00. Simplifying expressions involving negative integer and rational exponents. Simplifying, adding, subtracting, multiplying, and dividing rational expressions, radical expressions, and complex numbers. Multiplying and dividing polynomials. Factoring polynomials including greatest common factors (GCF), differences of squares, differences and sums of cubes, and quadratic trinomials. Evaluating expressions involving function notation. Finding the domains of functions. Solving equations, including ones containing rational or radical terms. Solving literal equations or formulas for a specified variable in terms of other variables. Solving inequalities and systems of equations. Solving quadratic equations by using the zero-product rule, the quadratic formula, and the square-root method in conjunction with completing the square. Solving application problems (including, but not limited to, total value, mixture, motion, work, area, and determine-the-number problems). Finding equations of lines through given points, with given slopes, perpendicular or parallel to given lines, and/or with given intercepts. Finding slopes and x and y-intercepts of lines. Finding vertices and lines of symmetry of parabolas and centers and radii of circles. Finding equations of circles with given centers and radii. Graphing lines, parabolas, and circles. Finding the distance between two points. This document was prepared by Ron Bannon using L A TEX ε and is a slight modification of Susan Gaulden s MTH 00 Content worksheet.

2 Review Problem Set At Essex County College you should be prepared to show all work clearly and in order, ending your work by boxing the answer. Furthermore, justify your answers algebraically whenever possible. These questions are for review only, and placement tests are not limited to these problems alone. Solutions and work are provided for each question. Please feel free to with questions or comments pertaining to this document.. Simplify. (a) ( x y z 0) ( x y z 0) ( ) y ( x = = x y ) = x4 9y 6 (b) 4a b 5 c 6a c 4a b 5 c 6a c = a5 b 5 c (c) ( m 4 ) n 6mn ( m 4 ) ( n = ) = ( m 5 n ) = 8m 5 n 6mn m 5 n. Simplify. (a) ( 5x y 4) ( 5x y 4) = 5 x y (b) ( 8a 6 b 9 c ) ( 8a 6 b 9 c ) = 4a 4 b 6 c These are sample problems and you should not limit your study to just these problems.

3 . Simplify (a) (b) x 5x + 6 x + x 8 6x 7 4x 5 + x 4x 5 x 5x + 6 (x ) (x ) x = + x 8 (x + 4) (x ) = x x + 4, x 6x 7 4x 5 + x 6x 7 + x 8x 0 (4x 5) = = = 4x 5 4x 5 4x 5 (4x 5) =, x 5 4 (c) x + + 5x x + x + x + x + + 5x x + x + x + = = (x + ) + 5x (x + ) (x + ) (x + ) (x + ) 5x (x + ) + (x + ) (x + ) (x + ) (x + ) (x + ) (x + ) = x x + 5x (x + ) (x + ) = 5x + 8x + 5 (x + ) (x + ) (d) x 4 x + 6x + x 7x + 0 x 4 x + 6x + x 7x + 0 = = (x + ) (x ) (x + ) (x + ) (x ) (x 5) (x + ) (x 5), x, x = x + 4 x 5, x, x

4 (e) x + 4x + x x 0 x x x 9x + 0 x + 4x + x x 0 x x x 9x + 0 = x + 4x + x x 0 x 9x + 0 x x (x + ) (x + ) (x 5) (x 4) = (x + ) (x 5) (x + ) (x 4) = x +, x, x 4, x 5 x + 4. Simplify. (a) 9x + x 4 9x 9x + x 4 9x = 9x + x 4 9x 9x 9x = + x 9x 4, x 0 ( + x) = (x + ) (x ), x 0 = x, x 0, x (b) x x + x x x x + x x = = = x x + (x ) (x + ) x (x ) (x + ) x (x + ) (x ) (x ) + 5 (x + ), x, x x + x +, x, x 4

5 5. Simplify. (a) 4 m 7 n 8 p 4 m 7 n 8 p = n mp 4 m (b) 80x y 5xy 5 Note, the restriction here is that x 0 and y 0. 80x y 5xy 5 = 6x y 4 = 4 x y (c) = (d) = = 5 (e) = 00 = 4 00 = 0 (f) 5 4a 4 bc 4 5 8a b c 5 4a 4 bc 4 5 8a b c = 5 a 7 b c 5 = ac 5 a b (g) 0xy 5 5x y The restriction here is that x > 0 and y > 0. 0xy 5 5x y = 0xy 5 4y 5x y = 4 x = y x 5

6 (h) = = = 7 8 = Rationalize the denominator 9 0x y. The restriction here is that x 0 and y > x y = 0x y = x 5y = 5y x 5y = 5y 5y 0 x y 7. Simplify. (a) = i + 5 4i = 6i + 0i = 6i (b) ( + i) + (5 6i) ( + 4i) ( + i) + (5 6i) ( + 4i) = 7 8i (c) 5i (7 4i) 5i (7 4i) = 5i 0i = 5i + 0 = 0 + 5i (d) (4 + i) ( i) (4 + i) ( i) = 8i + i i = 5i + = 4 5i (e) 4 + i i 4 + i i = 4 + i i + i + i 0 + i 0 + i = = = 0 9 4i + i 6

7 8. Simplify. ( (a) x x + 4 ) ( x + 5x ) ( x x + 4 ) ( x + 5x ) = x 4 + 5x x 6x 0x + 4x + x + 0x 8 = x 4 x + 4x 8 (b) x 5x x 4 x 5x x 4 = (x + ) (x 4) (x 4) = x +, x 4 (c) ( x + 7x 4 ) (x + ) Found by long division. x + 6 x + 9. Factor. (a) 5x y 5x y 4 + 0x y 5x y 5x y 4 + 0x y = 5x y ( x y + 4xy ) (b) m 4 6 m 4 6 = ( m 4 ) ( m + 4 ) = (m ) (m + ) ( m + 4 ) (c) 7x 7x = (x ) ( 9x + x + ) (d) 5x + x + 5x + x + = (5x + ) (x + ) 7

8 (e) x + 64 x + 64 = (x + 4) ( x 4x + 6 ) 0. Evaluate. (a) f ( ) given that f (x) = x x + 5. f ( ) = ( ) ( ) + 5 = = 4 (b) g () given that g (x) = x x + 0. g () = + 0 = = 6 = 4 (c) h (0) given that h (x) = 4x + x + 8. h (0) = = 8 =. Find the domain of each function. (a) f (x) = 5 x R, x 0 (b) g (x) = x + 5 x R, x (c) h (x) = x x + 7 R 8

9 . Solve. (a) x 8 = 5 (x + 4) x x 8 = 5 (x + 4) x x 8 = 5x + 0 x x 8 = 4x = 4x x 8 = x 4 = x (b) So, x = 4. 6 x + 4 x = 5 So, x =. 6 x + 4 x = 5 x 6 x + x 4 x = x = 5x 0 = 5x = x (c) x + 4 = x = x + 4 x (x + 4) (x ) = (x + 4) (x ) (x + 4) (x ) (x ) = (x + 4) x 9 = x + 8 x x = x = 7 So, x = 7. 9

10 (d) 50 x 6 x = 0 x So, x =. 50 x 6 x = 0 x 50 x (x ) x x (x ) 6 x = x (x ) 0 x x 50 (x ) 6 = (x ) 0 50x 6x + = 0x 60 4x + = 0x 60 4x 0x = 60 4x = 9 x = (e) x = x x = x x + 7 = x 5 ( x + 7 ) = (x 5) x + 7 = x 0x = x x = (x ) (x 9) It appears that x = and x = 9 are solutions, however, x = is extraneous (i.e. doesn t work), so x = 9. (f) x = x = ( x ) = ( ) x = 7 x = 5 So, x = 5. 0

11 . Solve. (a) A = P ( + rt) for r. A = P ( + rt) A = P + P rt A P = P rt A P P t = r So, r = A P P t (b) v = d + d t or r = A P t t. for d v = d + d t vt = d + d vt d = d So, d = vt d. (c) K = rt r t for t K = rt r t K (r t) = rt Kr Kt = rt Kr = rt + Kt Kr = t (r + K) Kr r + K = t So, y = Kr r + K.

12 4. Solve. (a) 4x 7 > x + 4x 7 > x + 4x x > 7 + x > 8 x > 6 So, x > 6, and in interval notation, (6, ). (b) x (x 4) x + 5 (x + ) x (x 4) x + 5 (x + ) x x + 8 x + 5x + 5 x + 8 7x x x 6x x So, x [ ), and in interval notation,,. 5. Solve. (a) { x + y = y = x + 7 Using the method of substitution. x + y = x + (x + 7) = x + 4x + 4 = 5x + 4 = 5x = 5 x = So, x = and since y = x + 7 = ( ) + 7 =, y =. (b) { x y = 7 x + 5y = 8 Using the method of elimination, by multiplying the first row by and the second row by, and then adding the two equations together. { { (x y) = (7) 6x + 4y = 4 9y = 8 y =. (x + 5y) = ( 8) 6x + 5y = 4

13 Using y = to back-substitute into any of these equations to find x. x y = 7 x + 4 = 7 x = x = So, finally, the solution, in ordered-pair notation, is (, ) 6. Solve. (a) x 9 = 0 Using the square-root rule. x 9 = 0 x = 9 x = ± 9 x = ± (b) x = x x = x x + x = 0 x (x + ) = 0 So, by the zero product rule, x = 0 or x =. (c) x x 0 = 0 x x 0 = 0 (x + 5) (x ) = 0 So, by the zero product rule, x = 5 or x =. (d) x 8x + 9 = 0 Since x 8x + 9 is not factorable over the rational numbers, you ll need to either complete the square or use the quadratic formula. Here I am using the technique of completing the square. x 8x + 9 = 0 x 8x = 9 x 8x + 6 = (x 4) = 7 x 4 = ± 7 x = 4 ± 7 So, x = or x = 4 7.

14 (e) x 6x = 5 Since x 6x 5 is not factorable over the rational numbers, you ll need to either complete the square or use the quadratic formula. Here I am using the quadratic formula. x 6x = 5 x 6x 5 = 0 x = 6 ± 6 4 () ( 5) 4 = 6 ± 76 4 = 6 ± 9 4 So, x = + 9 or x = Find the equation of the line through the point (0, 5) with slope /. Graph the line. Using the point slope form. y 5 = (x 0) Or, the slope intercept form. y = x + 5 Or, standard form. Here s the graph. x + y = Figure : Graph of x + y = 5. 4

15 8. Find the equation of the the line through the point (, 5) line with slope. Graph the line. Using the point slope form. y 5 = (x + ) Or, the slope intercept form. Or, standard form. Here s the graph. y = x + x + y = Figure : Graph of x + y =. 9. Find the equation of the line through the points (4, ) and (, ). Graph the line. First you ll need to find the slope. m = y y x x = 4 = = Using the point slope form with the point (4, ). y = (x 4) Using the point slope form with the point (, ). y + = (x ) Or, the slope intercept form. Or, standard form. y = x x y = 5

16 Here s the graph Figure : Graph of x y =. 0. Find the equation of the line through the point (0, ) that is perpendicular to the line x y =. Graph the line. The slope of the line whose equation is x y = can be found easily by solving for y. x y = y = x So the slope of the line perpendicular to this line is /. Using the point slope form. y = (x 0) Or, the slope intercept form. y = x + Or, standard form. Here s the graph. x + y = Figure 4: Graph of x + y = 6. 6

17 . Find the equation of the line through the point (, 4) that is parallel to the line x y = 5. Graph the line. The slope of the line whose equation is x y = 5 can be found easily by solving for y. x y = 5 y = x 5 So the slope of the line parallel to this line is /. Using the point slope form. y 4 = (x ) Or, the slope intercept form. y = x + 7 Or, standard form. Here s the graph. x + y = Figure 5: Graph of x + y = 7.. Find the equation of the line through the point (0, 4) and (, 0). Graph the line. First you ll need to find the slope. m = y y x x = = 4 = 4 Using the point slope form with the point (0, 4). y 4 = 4 (x 0) Using the point slope form with the point (, 0). y 0 = 4 (x + ) 7

18 Or, the slope intercept form. y = 4 x + 4 Or, standard form. Here s the graph. 4x + y = Figure 6: Graph of 4x + y =.. Find the slope of the line given by x 4y = 6. The slope of the line whose equation is x 4y = 6 can be found easily by solving for y. x 4y = 6 y = 4 x So the slope of this line is Find the slope of the line through the points (4, ) and (, 5). Using the formula. m = y y x x = 5 4 = 4 6 = 5. Find the x-intercept and the y-intercept of the line given by x 4y =. To find the x-intercept, set y = 0 and solve for x. x = x = 4 So the x-intercept is (4, 0). To find the y-intercept, set x = 0 and solve for y. 4y = y = So the y-intercept is (0, ). 8

19 6. Determine the vertex and the line of symmetry of the parabola given by y = (x ) 4. Graph the parabola. By inspection the vertex is (, 4) and the axis-of-symmetry is x = Figure 7: Graph of y = (x ) Determine the vertex and the line of symmetry of the parabola given by y = x 4x +. Graph the parabola. Rewrite into standard form. y = x 4x + y = ( x 4x + 4 ) 4 + y = ( x ) By inspection the vertex is (, ) and the axis-of-symmetry is x = Figure 8: Graph of y = x 4x +. 9

20 8. Determine the equation of the circle with center (, ) and radius. Graph the circle. (x ) + (y + ) = Figure 9: Graph of (x ) + (y + ) = Determine the center and radius of the circle given by (x + ) + (y ) = 4. Graph the circle. The center is (, ) and radius Figure 0: Graph of (x + ) + (y ) = 4. 0

21 0. Determine the center and radius of the circle given by x + y + 6x + 4y + = 0. Graph the circle. Rewrite into standard form. x + y + 6x + 4y + = 0 x + 6x + y + 4y = x + 6x y + 4y + 4 = (x + ) + (x + ) = The center is (, ) and radius Figure : Graph of x + y + 6x + 4y + = 0.. Determine the distance between the points (, 5) and (5, ). Using the distance formula. (5 ) + ( + 5) = = 5 = 5. Determine the distance between the points (0, 6) and (4, ). Using the distance formula. (4 0) + ( 6) = = 65

22 . The sum of two numbers is 8. The difference between four times the smaller number and seven is equal to the sum of two times the larger number and five. Find the numbers. Let s be the smaller number and l be the larger number. Resulting in this system of equations: s + l = 8 4s 7 = l + 5. Solving this system, s = 8 and l = 0. So the two numbers are 8 and The square of a negative number is fifteen more than twice the negative number. Find the number. Let x be the negative number. Resulting equation is: x = x + 5. Solving this equation gives two solutions, x = 5 or x =. So the number is. 5. The degree measures of the angles in a triangle are three consecutive integers. Find the measure of the angles. Let x, x + and x + be three consecutive integers. Resulting in this equation: x + x + + x + = 80. Solving this equation, x = 59. So the measures of the angles are 59, 60, and The perimeter of a rectangle is 50 meters. The width of the rectangle is five meters less than the length. Find the length and width of the rectangle. Let l be the measure of the length of the rectangle, the width of the rectangle, in terms of l, is l 5. Resulting in this equation: l + (l 5) = 50. Solving this equation, l = 50. So the length is 5 meters, and the width is 0 meters. 7. The width of a rectangle is five meters less than the length. The area of a rectangle is 76 square meters. Find the length and width of the rectangle. Let l be the measure of the length of the rectangle, the width of the rectangle, in terms of l, is l 5. Resulting in this equation: l (l 5) = 76. Solving this equation, l = 6 or l =, the later being extraneous because we cannot have a negative length. So the length is 6 meters, and the width is meters.

23 8. A steel rod 90 centimeters long is to be cut into two pieces, each to be bent to make an equilateral triangle. The length of a side of one triangle is to be twice the length of a side of the other. How should the rod be cut? Let x be the measure of the length, in centimeters, where the rod is cut, resulting in two pieces x and 90 x centimeters. Resulting in this equation: x = 90 x Solving this equation, x = 0. So the 90 centimeter rod should be cut at 0 centimeters. 9. A 6-foot ladder is leaning against a building. How high on the building will the ladder reach when the bottom of the ladder is 5 feet from the base of the building? Let x be the measure, in feet, how high on the building the ladder will reach when the bottom of the ladder is 5 feet from the base of the building. Resulting in this equation: 6 = 5 + x Solving this equation, x = ±, but again, due to the physical nature of the problem, we can only accept the positive value. So the 5.99 feet is how high the ladder will reach. 40. Cheap Cars rents cars for $8.00 a day and $0.0 for every mile driven. Save-On Cars rents cars for $0.00 a day and $0.08 per mile driven. You want to rent a car for one week. For how many miles will it cost you less to rent from Cheap Cars than from Save-On Cars? Let x be the number of miles driven. The resulting inequality is: x < x. Solving this inequality, x < 700. So, you have to drive less than 700 miles for Cheap Cars to be less than Save-On Cars. 4. An investment of $,500 is divided between two simple interest accounts. On one account, the annual simple interest rate is 5%, and on the second account the annual simple interest rate is 7.5%. How much should be invested in each account so that the total interest earned from the two accounts is $5? Let x be the amount of money invested at 5%, and 500 x be what remains to invest at 7.5%. The resulting equation is: 0.05x (500 x) = 5. Solving this equation, x = 900. So, of the $,500 you would need to invest $,900 at 5% and $,600 at 7.5% This is approximately 5 feet, 8 inches.

24 4. A motorboat leaves a harbor and travels at an average speed of 9 miles per hour (mph) toward a small island. Two hours later, a cabin cruiser leaves the same harbor and travels at an average speed of 8 mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cruiser be alongside the motorboat? Let x be the number of hours after the cabin cruiser leaves. equation is: 9 (x + ) = 8x. The resulting Solving this equation, x =. So, it will take hours. 4. A goldsmith combined an alloy that cost $4.0 per ounce with an alloy that cost $.80 per ounce. How many ounces of each alloy were used to make a mixture of 00 ounces costing $.50 per ounce? Let x be the number of ounces of the more expensive alloy, and 00 x be the number of ounces of the cheaper alloy. The resulting equation is: 4.0x +.8 (00 x) = Solving this equation, x = 56. So you will need to use 56 ounces of the expensive alloy and 44 ounces of the cheaper alloy. 44. A chemist needs to make liters of an 8% acid solution by mixing a 0% acid solution and a 5% acid solution. How many liters of each solution should the chemist use? Let x be the number of liters of the 0% acid solution, and x be the number of liters of the 5% acid solution. The resulting equation is: 0.0x ( x) = 0.08 Solving this equation, x =.. So you will need to use. liters of the 0% acid solution and 0.8 liters of the 5% acid solution. 45. One grocery clerk can stock a shelf in 0 minutes, whereas a second clerk requires 0 minutes to stock the same shelf. How long would it take to stock the shelf if the two clerks worked together? Let t be the number of minutes that the two clerks work together on the given task. The resulting equation is: 0 t + 0 t =. Solving this equation, t =. So, it will take minutes for the two clerks, working together, to stock the shelf. 4