3 = 4 6 = = 18 4 = 9 2 = = = 25 = 8

Transcription

1 page 1 1. Perform the operations as indicated. (a) 6 8 (b) We apply order of operations We apply order of operations. The common denominator is First we perform the subtraction in the numerator Multiplication in the denominator: we simply multiply the numerator by the numerator, the denominator by the denominator, and then simplify Thus we have so far 6 8 To divide is to multiply by the reciprocal. Simplify each of the following. (a) a a 8a a a a ( ) a use rule (ab) n a n b n use rule (a n ) m a nm a () a 8 use rule a n a m a n+m 8a 11

2 page (b) a ab ab 16a 7 b 8 (c) ( x) y x y y x a ab ab a ( ) a b ab use rule (ab) n a n b n use rule (a n ) m a nm a ( 8) a b 6 ab use that multiplication is commutative ( 8) a a ab 6 b use rule a n a m a n+m 16a 7 b 8 ( x) y x y use rule (ab) n a n b n ( ) x y x y x y x y x y x y simplify by canceling out x y xy simplify by canceling out y y x (d) (x ) (x ) 7x We rst apply the distibutive law and then combine like terms (x ) (x ) distribute x 6 10x + combine like terms 7x (e) (x + ) (x ) x + 1x 9 We multiply two-term polynomials by two-term polynomials by the FOIL method. F stand for rst with rst, O for outer terms, I for inner terms, and L for last terms. (x + ) (x ) x F x O x + 1x 9 +1x I 9 L combine like terms (f) (a 1) a 10a + 1 to square something means to write it down twice and multiply. Then it is a FOIL problem. F stand for rst with rst, O for outer terms, I for inner terms, and L for last terms. (a 1) (a 1) (a 1) a F a O a 10a + 1 a I +1 L combine like terms

3 page (g) x + y x y 9x 10 16y x + y x y 9x 10 F 1x O 9x 10 16y +1x I 16y L combine like terms The expressions x + y and x y are called conjugates. Becuase of the same terms and alternating signs, O and I cacel out when "FOIL"-ing, giving us the di erence of two squares. (h) (x y) x + x y + x y + x y + xy + y x 6 y 6 We apply the law of distributivity (x y) x + x y + x y + x y + xy + y x x + x x y + x x y + x x y + x xy + x y y x y x y y x y y x y y xy y y x 6 + x y + x y + x y + x y + xy x y x y x y x y xy y 6 x 6 y 6. Solve each of the following equations. Make sure to check your solutions. (a) x this is a very simple equation, much like x + 1 7, only the numbers are fractions. But the principles and operations regarding equations are the same. x + subtract 8 from both sides; We check: 8 x 9 x 6 divide both sides by RHS 8 ( 6) LHS Thus our solution, 6 is correct. (b) (x ) (x + ) 1 1 We check: if x (x ) (x + ) 1 distribute 6x 9 x 1 combine like terms 1, then x 1 1 add 1 x 1 LHS (x ) (x + ) ( ( 1) ) ( ( 1) + ) ( ) ( + ) ( ) ( 1) RHS

4 page (c) w (w ) we rst apply the law of distributivity to simplify the right-hand side. w (w ) w w 10 subtract w from both sides w 10 add 10 to both sides w divide both sides by w We check: if w, then LHS w RHS (w ) Thus our solution, w is correct (d) 8 (x ) ( x) x 8 (x ) ( x) x multiply out parentheses 8x 1 + 6x x combine like terms 1x 9 x subtract x 1x 9 0 add 9 1x 9 divide by 1 x We check our solution by evaluating the left hand side and the right hand side of the original equation with x. RHS 8 ( ) ( ()) 8 0 ( 6) 8 0 ( 1) LHS 0 ( ) Since the left-hand side equals to the right-hand side, our solution x is correct. (e) 7(j ) + 8 (j + ) + j no solution 7(j ) + 8 (j + ) + j distribute on both sides 7j + 8 j j combine like terms 7j 7 7j + 10 subtract 7j 7 10 There is no value of j that could make the statement 7 10 true. Thus there is no solution of this equation. An equation of this type is called a contradiction.

5 page (f) (x ) ( 7x) (x + 1) (x 10) all numbers are solution In case of subtracting algebraic expressions, we either subtract the opposite, or distribute 1: We will use the second method here. (x ) ( 7x) (x + 1) (x 10) (x ) 1 ( 7x) (x + 1) 1 (x 10) distribute 6x x x + x + 10 combine like terms x + 1 x + 1 subtract x 1 1 Since x disappeared from the equation, we are left with an unconditional statement, this time unconditionally true. All values of x will make 1 1 be true, and thus all numbers are solution of this equation.. An equation of this type is called an identity. (g) x x x 1 x x x we express the left hand side as a fraction x x x express fractions with common denominator 1 (x 1) (8 x) 1 ( x) + multiply both sides by (x 1) + (8 x) 1 ( x) mulitply out parentheses (distribute) 9x + 16x 6 1x combine like terms 7x + 9 1x 6 add 1x to both sides x subtract 9 from both sides x 6 divide by x 1 We check our solution by evaluating the left hand side and the right hand side of the original equation with x 1. RHS ( 1) ( 1) LHS ( 1) ( ) Since the left-hand side equals to the right-hand side, our solution x 1 is correct. (h) x + x + 1 (x + 1) 1 all numbers are solution x + x + 1 (x + 1) 1 express fractions with common denominator (x ) (x + ) 1 (x + 1) multiply both sides by 1 (x ) + (x + ) 1 (x + 1) mulitply out parentheses (distribute) 9x 6 + x + 0 1x + 1 combine like terms 1x + 1 1x + 1

6 page 6 Since the left hand side and the right hand side are identical, every number will work if substituted. Thus this is an identity, all numbers are solution(s).. Intercepts. (a) Find the x-intercept of the line given by x y 10. ( ; 0) We substitute y 0 into the equation of the line and solve for x. If y 0, x? x 0 10 Thus the y intercept is ( ; 0). x 10 divide both sides by x ) we found the point ( ; 0) (b) What is the y-intercept of the line with equation x + y 0? (0; 1) The y intercept of a graph is the point where it intercects the y axis. We can nd this point by substituting x 0 into the equation of the line and solve for y. If x 0, y? (0) + y y 0 Thus the y intercept is (0; 1). y 0 divide both sides by y 1 ) we found the point (0; 1) x + 1. Graph the lines x y 10 and y in the same coordinate system. to nd the coordinates of the point where the lines intersect. (; ) Use your graph we nd points by picking a value for x; and then substitute that value into the equation of the line and solve for y. Finally, when we have a few points, we graph them and connect the points. If x 0, y? (0) y 10 0 y 10 y 10 divide both sides by y ) we found the point (0; ) If x, y? () y 10 0 y 10 subtract 0 from both sides y 10 divide both sides by y ) we found the point (; )

7 page 7 If x, y? ( ) y 10 0 y 10 add 0 to both sides y 0 divide both sides by y 6 ) we found the point ( ; 6) x + 1 We now graph the other line, y. We nd points by picking a value for x; and then substitute that value into the equation of the line and solve for y. Finally, once we have a few points, we graph and connect them. If x 1, y? y y y 1 y ) we found the point (1; ) If x, y? y + 1 y y y ) we found the point (; ) If x, y? y + 1 y 6 y y ) we found the point (; ) We are now ready to graph. Clearly, the lines intersect at the point (; ). y x -

8 page 8 6. Word Problems. (a) A rectangle has a width which is seven inches less than its length. The perimeter is 6 inches. Find the sides. 8 in and 1 in Let us call the shorter side (the width) by x. Then the longer side, the length must be x + 7. We obtain the equation by expressing the perimeter. x + (x + 7) 6 distribute x + x combine like terms x subtract 1 from both sides x divide both sides by x 8 We denoted the width by x, thus the width is 8 in long. The length is then x in long. (b) A couch went on a 1% sale. The sale price is $ 697. Find the original price. $ 80 A 1% sale means that we have to pay 8% of the original price. The question is: 8% of what number is 697? (is) 697 F (of) x We substitute these values into the formula (is) F (of) and solve for x Thus the original price is $ 80. (is) F (of) x divide bith sides by x x (c) The di erence between two numbers is 7, their sum is 7. Find these numbers. 1 and Let us call the smaller number x. Then the larger number is x+7, since the di erence between the two numbers is 7. The equation then expresses the sum of these numbers x smaller number + x + 7 larger number 7 solve for x x subtract 7 x 0 divide by x 1 Thus the smaller number, labeled x is 1. The larger number was labeled x + 7; so it must be Thus the numbers are 1 and. We check: the di erence between and 1 is (1) 7; and their sum is indeed Thus our solution is correct.

9 page 9 (d) A certain triangle s longest side is one centimeter less than six times the shortest side. The other side is ve times the shortest side. The perimeter is thirty- ve centimeters. Find the length of the longest side. 17 cm Let x denote the shortest side. Then the longest side is 6x 1; and the other side is x. We obtain the equation by expressing the perimeter of the triangle. Then we solve for x. x shortest side + 6x 1 longest side + x other side combine like terms 1x 1 add 1 to both sides 1x 6 divide both sides by 1 x Now we know that x. Since the longest side was denoted by 6x 1, it must be 6 () cm long. (e) Ann and Betty are roommates. The monthly rent is $ 90. The amount paid by Ann is $ 10 less than twice the amount paid by Betty. How much do they each pay for rent? $ 0 and $ 0 Let x denote the amount that is paid by Betty. Then Ann must pay x 10 per month. The equation expresses the monthly rent as the sum of the two payments x smaller amount + x 10 larger amount 90 solve for x x add 10 x 160 divide by x 0 Thus the amount paid by Betty is $ 0. The amount paid by Betty was labeled as x 10, so Betty must pay (0) Thus the payments are $ 0 and $ 0. We check: $ 0 ($ 0) $ 10 and $ 0 + $ 0 $ 90. Thus our solution is correct. (f) The population of a town has decreased from to What percent of a change does this represent? 8% decrease The decrease in the population is The question is then: 700is what percent of ? (is) 700 F x (of) We substitute these values into the formula (is) F (of) and solve for x. (is) F (of) 700 x divide bith sides by x x % Thus the change in population is 8%.

10 page 10 (g) A bank teller has 7 more ve-dollar bills than ten-dollar bills. The total value of the money is $1000: How much of each denomination of bill does he have? 1 ten-dollar bills and 98 ve-dollar bills Let us denote the number of ten-dollar bills by x: Then we have x+7 many ve-dollar bills. The equation expresses the value of the bills. 10x amount in 10-bills + (x + 7) amount in -bills 1000 distribute 10x + x combine like terms 1x subtract 1x 76 divide by 1 x 1 Thus we have 1 tens and ves: We check: and 1 (10) + 98 () 1000: Thus our solution; 1 ten-dollar bills and 98 ve-dollar bills; is correct. (h) One side of a rectangle is ft shorter than three times the other side. Find the sides if the perimeter is 6 ft. 9 ft and ft Let us denote the shorter side by x: Then the other side is x : The equation expresses the perimeter of the rectangle. (x) + (x ) 6 multiply out parentheses x + 6x 8 6 combine like terms 8x 8 6 add 8x 7 divide by 8 x 9 If the shorter side was denoted by x; we now know it is 9 ft. The longer side was denoted by x ; so it must be (9 ft) ft ft. Thus the sides of the rectangle are 9 ft and ft: We check: P (9 ft) + ( ft) 6 ft and ft (9 ft) ft: Thus our solution is correct. (i) Mary bought four less than three times the number of books that Jose did. Together they bought sixteen books. How many did Jose buy? Let us denote the number of books bought by Jose by x. Then Mary baught x many books. We obtain the equation by expressing the total number of books. Then we solve for x. Thus Jose bought books. x + x 16 combine like terms x 16 add to both sides x 0 divide both sides by x