Math 99 Review for Exam 3

Transcription

1 age 1 1. Simlify each of the following eressions. (a) ab a b + 1 b 1 a 1 b + 1 Solution: We will factor both numerator and denominator and then cancel. The numerator can be factored by grouing ab {z a b {z + 1 a (b 1) 1 (b 1) (a 1) (b 1) The denominator factors by the di erence of squares theorem. Thus the fraction can be simli ed as b 1 (b + 1) (b 1) ab a b + 1 b 1 (a 1) (b 1) (b + 1) (b 1) a 1 b + 1 (b) Solution: we will factor whatever we can and then cancel ( 6) ( + 6) ( + 6) ( 6) (c) 16q 16 + q ( + ) ( ) q + q ( + ) + Solution: ( + ) ( + ) and ( + ) ( + ) are easy. With the other eressions, we start with the greatest common factor. 16q 16 + q + q {z 16 16q grouing {z ( + q) 16 ( + q) factor out + q 16 ( + q) di erence of squares ( + ) ( ) ( + q) + + q + q + q + + q {z {z We have all the ieces now: grouing ( ( + q) + ( + q)) factor out + q ( + ) ( + q) 16q 16 + q q + q ( + ) ( ) ( + q) ( + ) ( + ) ( + ) ( + ) ( + ) ( + q) ( + ) ( ) ( + ) 1 +

2 age (d) + Solution: This is a subtraction of fractions. The denominators are the same, the only di culty is that we are subtracting eressions instead of numbers. The second air of arentheses is essential. + () ( + ) ( ) (e) Solution: () + ( 1 + ) (f) (g) Solution: Solution: We will rst work out 1 and then multily that by () Rationalize the denominator in each of the following eressions. (a) (b) Solution: To rationalize the denominator, we will multily both the numerator and denominator by Solution: To rationalize the denominator, we will multily both the numerator and denominator by the conjugate of the denominator, which is The denominator is 1 since

3 age (c) Solution: To rationalize the denominator, we will multily both the numerator and denominator by the conjugate of the denominator, which is Find the eact value of + 6 if. Solution: We work out rst Now we substitute into + 6: Factor 1 + by comleting the square. ( + 8) ( + 8) ( ) Solution: We rearrange the terms rst and then factor out the leading coe cient Half of the linear coe cient is 1, thus we work out smuggle in to comlete the square. + 1 Thus we need to smuggle in rst to see what we need to {z !

4 age We bring the last two numbers to the common denominator ! 61 16! 1 (16) 1 (16)! Since , we factor via the di erence of squares theorem ! ( + 8) ( + 8) ( ) We may distribute into the second factor. Then we get + 1 ( + 8) ( + 8) ( ) We FOIL to check:. Factor comletely each of the following: ( + 8) ( ) (a) a mn 1abm 6abmn + 10a m am (n + m) (a b) Solution: a mn 1abm 6abmn + 10a m the GCF is am am (an 1bm 6bn + 10am) rearrange am an {z 6bn +10am {z 1bm am (n (a b) + m (a b)) am (n + m) (a b)

5 age (b) a b a + b (a + b ) ( + 1) ( 1) Solution: a b a + b the GCF is a b a + b rearrange a {z a +b {z b a 1 + b 1 a + b 1 We are not done yet since ( 1) ( 1 ) further factors via the di erence of squares theorem. Thus the answer is a + b 1 a + b 1 a + b ( + 1) ( 1) (c) 16a + 16b a b (9 + ) ( + ) ( ) (a + b) Solution: 16a + 16b a b the GCF is 81a {z + 81b a {z b 81 (a + b) (a + b) 81 (a + b) We are not done yet, since 81 9 ( ) further factors via the di erence of squares theorem. 81 (a + b) 9 (a + b) (a + b) One factor still further factors: 9 ( + ) ( ) (a + b) 9 + (a + b) 9 + ( + ) ( ) (a + b) (d) ( ) ( ) Solution: we will factor by grouing. First we conduct the "q game". q 8 1st coe cient times rd coe cient + q 6 nd coe cient We start by eressing 8 as a roduct of two numbers. there are only two airs, 1 with 8 and with. Since the roduct q is ositive, and q have to have the same sign. Since the sum + q is negative, they both have to be negative. We only need to consider 1 with 8 and with. Clearly with work as and q. We use these numbers to eress the second term as the sum of two terms, and then factor by grouing {z {z + 8 ( ) ( ) ( ) ( )

6 age 6 We check by multilication: ( ) ( ) Thus our result is correct. (e) a a (a ) (a + 1) Solution: we will factor by grouing. First we conduct the "q game". q 6 1st coe cient times rd coe cient + q nd coe cient We start by eressing 6 as a roduct of two numbers. there are only two airs, 1 with 6 and with. Since the roduct q is negative, one number must be ositive, the other one must be ositive.. Since the sum + q is negative, the negative sign has to be in front of the larger number. We only need to consider 1 with 6 and with. Clearly 1 with 6 work as and q. We use these these numbers to eress the second term as the sum of two terms, and then factor by grouing. We check by multilication: a a a {z + a 6a {z a (a + 1) (a + 1) (a ) (a + 1) (a ) (a + 1) a + a 6a a a Thus our result is correct. (f) b b (b ) (b + 1) Solution: we will factor by grouing. First we conduct the "q game". q 0 1st coe cient times rd coe cient + q 1 nd coe cient We start by eressing 0 as a roduct of two numbers. the ossible airs are, 1 with 0, with 10, and with. Since the roduct q is negative, one number must be ositive, the other one must be ositive.. Because the sum + q is negative, the negative sign has to be in front of the larger number. We only need to consider 1 with 0, with 10, and with. Clearly with work as and q. We use these these numbers to eress the second term as the sum of two terms, and then factor by grouing. We check by multilication: Thus our result is correct. b b b {z + b b {z b (b + 1) (b + 1) (b ) (b + 1) (b ) (b + 1) b + b b b b

7 age 7 6. Solve each of the following equations. Make sure to check your solution(s). (a) ; 0; and Solution: We reduce one side to zero, then factor, and then aly the zero roerty factor out GCF divide both sides by We will factor by comleting the square. Half of the linear coe cient is ; and thus we will work with ( ) 10 + We smuggle in {z ( ) ( ) 0 0 ( + 0) ( 0) 0 ( + ) ( ) 0 (b) Alying the zero roerty we obtain ; 0; and as the solutions identity, all real numbers are solution. Solution: ( 1) ( 1) eress everything as a fraction bring everything to the common denominator add fractions on right hand side multily out arentheses combine like terms multily by Because the left hand side is now identical to the right hand side, this equation is an identity, and all real numbers are solution.

8 age 8 (c) j j + 0 and Solution: j j + subtract j j or subtract 0 or 6 divide by 0 or Thus the solution are 0 and. LHS j We check. If 0; then (0)j + jj + + RHS and if, then LHS j ( )j + j 6j + j j + + RHS Thus both numbers are solutions. (d) ( 7) ( + 1) Solution: 1 ( 7) ( + 1) ( 7) ( + 1) bring fractions to common denominator ( 7) ( + 1) 1 1 multily both sides by 1 10 ( 7) 1 ( + 1) multilty out arentheses subtract subtract 1 8 divide by 1 We check: LHS ( 1 7) ( 8) RHS ( 1 + 1) ( 0) Thus our solution, 1 is correct. (e) 7 + ( + ) ( 1) ( + 1)

9 age 9 Solution: 7 + ( + ) ( 1) ( + 1) multily the olynomials on both sides combine like terms We check our result: subtract subtract + 1 subtract 1 LHS 7 ( ) + (( ) + ) ( ( ) 1) ( 1) ( 9) RHS ( ( ) + 1) ( 1 + 1) ( 11) 11 Thus the solution, is correct. (f) 8a + a 7; Solution: since this equation is of a higher degree than 1, our only method is to reduce one side to zero, factor, and then aly the secial zero roerty. 8a + a subtract, rearrange a + 8a 0 the GCF is a + a 1 0 We will factor a + a 1 by grouing. First we conduct the "q game". q 1 1st coe cient times rd coe cient + q nd coe cient We start by eressing 1 as a roduct of two numbers. the only ossible airs are, 1 with 1 and with 7: Since the roduct q is negative, one number must be ositive, the other one must be ositive.. Because the sum + q is ositive, the negative sign has to be in front of the smaller number. We only need to consider 1 with 0, and with 7. Clearly with 7 work as and q. We use these these numbers to eress the second term as the sum of two terms, and then factor by grouing. Thus our equation is a + a 1 0 a {z + 7a a {z 1 0 (a (a + 7) (a + 7)) 0 (a ) (a + 7) 0 (a ) (a + 7) 0 We now aly the secial zero roerty. If this roduct is zero, then either 0 or a 0 or a We solve these equations for a. a 0 or a or 0 a or a 7 or no solution here

10 age 10 We check both solutions. If a, then LHS 8 () + () RHS X If a 7, then LHS 8 ( 7) + ( 7) 8 ( 7) RHS X Thus both solutions, 7 and are correct. (g) 8 0 ; 0 Solution: since this equation is of a higher degree than 1, our only method is to reduce one side to zero, factor, and then aly the secial zero roerty. 8 0 subtract the GCF is ( ) 0 We now aly the secial zero roerty. If this roduct is zero, then either 0 or 0. We solve these equations for. 0 or 0 0 or 0 or We check both solutions. If 0, then If, then LHS RHS LHS RHS Thus both solutions, 0 and are correct. (h) 8 0 ; 0; Solution: since this equation is of a higher degree than 1, our only method is to reduce one side to zero, factor, and then aly the secial zero roerty. 8 0 subtract the GCF is 0 () 0 factor via di erence of squares theorem ( + ) ( ) 0

11 age 11 We now aly the secial zero roerty. If this roduct is zero, then either 0 or + 0 or 0. We solve these equations for. + 0 or 0 or 0 or or 0 or We check all three solutions. If, then LHS 8 8 RHS If, then LHS 8 8 RHS and if 0; then LHS RHS Thus all three solutions, ; 0; and are correct. (i) ( ) ( + ) ( 1) ( 1) 7 Solution: We have to use the FOIL method on both sides to erform the multilications. It is very imortant, however, to kee the eressions in a arentheses since we are dealing with subtraction between algebraic eressions. ( ) ( + ) ( 1) ( 1) FOIL combine like terms erform subtraction combine like terms subtract (the equation is linear!) add 1 1 add 1 1 divide by 7

12 age 1 We check: if 7, then LHS ( 7) ( (7) + ) ( ) (1 + ) ( ) 19 ( 76) 78 RHS (7 1) ( (7) 1) 6 (1 1) Thus our solution, 7 is correct. (j) + 1: + ; Solution : We comlete the square. + {z ( ) ( ) 0 ( ) and + We check: if, then LHS RHS If +, then LHS RHS Thus our solution, + and (k) Solution: + ; is correct Half of the linear coe cient is. Thus the comlete square is

13 age 1 So we smuggle in.! + + {z + 7 +! 0 0 Although can be simli ed, we will kee it as it is, because the common denominator will be easier this way. We net aly the di erence of squares theorem. is the square of a number - its own square root. r!!! We solve for the zeroes of both linear factors: +! 0 +! 1 A 0 +! + +!

14 age 1 We check: if, then!! LHS ! RHS and if +, then LHS RHS +! + 0 +! ! Thus our solution, + and Note: since, the nal answer + is correct. and is equally correct.

15 age 1 7. Grah the straight lines + y and y 1 in the same coordinate system. Use your grah to nd the coordinates of the oint where the lines intersect. ( ; ) Solution: We can easily grah the lines by using the sloe-intercet forms: y 1 and y + 1 y Find an equation of the straight line that is arendicular to y 6 and asses through the oint ( 1; ). y 1 Solution: We obtain the sloe of the line give by solving for y. y 6 add y y 6 add y ivie by + 6 y ) y Our line is erenicular to one with sloe : thus its sloe must be the negative recirocal of ; which is. Using the sloe-oint form of the equation, we write y ( + 1). If we simlify the equation, we obtain the sloe-intercet form, y 1: 9. Find an equation of the straight line that asses through the oints (; 7) and ( ; ). y + 1 Solution 1: We obtain the sloe from the sloe formula m rise run y y ( ) ( ) 1 For b; we substitute either oint into the sloe-intercet form of the equation. y m + b we know m y + b (; 7) is on the line 7 () + b b 1 b

16 age 16 Thus the equation is y + 1. We can easily check by substituting the coordinates of the oints into the equation. Solution : We obtain the sloe from the sloe formula m rise run y y ( ) ( ) 1 and then aly the sloe-oint form (we will use (; 7)) y 7 ( ) Solution : Write y m + b and solve the system that we obtain by substituting the oints given. Thus we get the system 7 m () + b (; 7) is on the line We eliminate m if simly add the two equations m ( ) + b ( ; ) is on the line m + b 7 m + b b b 1 Now we can easily solve for m in the rst equation m m 6 m Thus the equation is y + 1. We can easily check by substituting the coordinates of the oints into the equation.

17 age Grah the arabola y 8 1. Clearly label the coordinates of ve oints on the arabola, including verte and intercets. Solution: We obtain all forms of the equation rst. y ) olynomial form y ( ) y 8 {z y ( ) 1 y ( ) + 1 ) comlete square form We factor via the di erence of squares theorem y ( ) 1 since 1 1 y ( + 1) ( 1) y ( ) ( ) ) factored form From the olynomial form we obtain the y intercet, (0; 1). From the comlete square form, the verte is (; 1). Finally, the factored form tells us that there are two intercets, (; 0) and (; 0). The few missing oints, close to the verte can be found by substituting values for into any of the three forms of the equations to nd y. This time we will work with the olynomial form. We are ready to grah: if, then y () + 8 () if 6, then y (6) + 8 (6) y

18 age 11. One side of a rectangle is ft shorter than three times the other side. Find the sides if the erimeter is 6 ft. 9 ft and ft Solution: Let us denote the shorter side by : Then the other side is : The equation eresses the erimeter of the rectangle. () + ( ) 6 multily out arentheses combine like terms add 8 7 divide by 8 9 If the shorter side was denoted by ; we now know it is 9 ft. The longer side was denoted by ; 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. 1. One side of a rectangle is ft shorter than three times the other side. Find the sides if the area is 8 ft. 6 ft and 1 ft Solution: Let us denote the shorter side by : Then the other side is : The equation eresses the area of the rectangle. ( ) 8 multily out arentheses 8 subtract Because the equation is quadratic, we need to factor the left-hand side and then aly the zero roerty. We will factor by grouing. First we conduct the "q game". The sum of and q has to be the linear coe cient (the number in front of, with its sign), so it is. The roduct of and q has to be the roduct of the other coe cients, ( 8). q + q Now we need to nd and q. Because the roduct is negative, we re looking for a ositive and a negative number. Becuse the sum is negtive, the larger number must carry the negative sign. We enter 1: into the calculator and get a decimal: 1:87:::: So we start looking for factors of, starting at 1; and moving down. We soon nd 1 and. These are our values for and q: We use these numbers to eress the linear term: 1 and factor by grouing. 8 0 {z + 1 {z 8 0 ( + 1) 6 ( + 1) 0 ( 6) ( + 1) 0

19 age 19 We now aly the zero roerty. Either for or We solve both these equations and Since distances can not be negative, the second solution for, is ruled out. Thus 6: Then the longer side is (6 ft) ft 1 ft; and so the rectangle s sides are 6 ft and 1 ft long. We check: 6 ft (1 ft) 8 ft and 1 ft (6 ft) ft. Thus our solution is correct. 1. One side of a rectangle is in shorter than times the other side. Find the sides of the rectangle if its area is 19 in. 11 in by 9 in Solution: Let us denote the shorter side by. Then the larger side is. The equation eresses the area. ( ) We will comlete the square. (It can be done by the q-game as well) Half of the linear coe cient is ; thus we work out Thus we smuggle in 9. + {z 9 9 9! 19!

20 age 0 We bring the last two numbers to the common denominator:! 19 () 9 () !! Since , we factor via the di erence of squares theorem.! ( 11) 0 This equation has two solutions, 1 9 and Since distances are ositive, is ruled out as a solution for the shorter side. The other solution is 11 in. This makes the longer side 11 9 in. We check: (11) 9 and 11 (9) 19. Thus our solution, 11 in by 9 in is correct. 1. A bank teller has more ve-dollar bills than ten-dollar bills. The total value of the money is $610: How much of each denomination of bill does he have? ten-dollar bills and 6 ve-dollar bills Solution: Let us denote the number of ten-dollar bills by : Then we have + many ve-dollar bills. The equation eresses the value of the bills ( + ) 610 distribute combine like terms subtract divide by 1 Thus we have tens and + 6 ves: We check: 6 and (10) + 6 () 610: Thus our solution; ten-dollar bills and 6 ve-dollar bills; is correct.

21 age 1 1. The oulation of a town has decreased from to What ercent of a decrease does this reresent? 1% decrease Solution 1: We subtract from to determine the change Now the question is: is what ercent of ? Then We substitute the data into the formula: Thus 0:1 0:1 1 (is) F (of) (is) F (of) :1 0:1 (100) 1 (100) % This is a 1% decrease. Solution : The question may be re-hrased as: is what ercent of ? Then We substitute the data into the formula: Thus 0:8 0:8 1 (is) F (of) (is) F (of) :8 0:8 (100) 1 (100) % Since the oulation has decreased to 8% of its revious count, this is a 1% decrease.

22 age 16. We invested $10000 into two bank accounts. One account earns 1% er year, the other account earns 8% er year. How much did we invest into each account if the combined interest from the two accounts is $ after the rst year? $ 700 at 1% and $ 700 at 8% Solution: Let us denote the amount invested at 1% by and the amount invested at 8% by y. The two equations eress that + y the amounts add u to $ :1 + 0:08y the interests earned add u to $ We solve the system of equation by elimination. But let us rst make the second equation simler: We now have 0:1 + 0:08y multily by y 00 divide by 7 + y y y We will multily the rst equation by to eliminate y. We add the equations and solve for : y y divide by 700 Thus we invested $700 at 1%. The other amount is then from the rst equation: y y 700 We invested $ 700 at 1% and $ 700 at 8%. We check: the amounts add u to $700 + $700 $ The interest from the accounts are 1% of 700 is 0:1 (700) 10 and 8% of 700 is 0:08 (700) 16 Since , our solution is correct. 17. The hyotenuse of a right triangle is 68 cm. The di erence between the other two sides is 8. cm. Find the sides of the triangle. cm and 60 cm Solution: See handout on the Pythagorean Theorem.. Find the distance between (; 8) and (8; ). 1 units Solution: See handout on the Pythagorean Theorem.