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1.10 Word Problems Basic Algebra: One Step at a Time
Page 65-92: #3, 6, 7, 19, 20, 29, 31, 32, 35, 37, 38, 46, 47, 48, 49, 53, 57, 59, 61, 62, 63, 64, 65 Extra Problem, 69, 70, 71, 74, 81, Extra Problem.
Dr. Robert J. Rapalje Seminole Community College Sanford, FL 32773
3. Three times a number plus 8 is equal to five times the number. Find the number. SOLUTION: STEP 1: Let x =_____. Let x = the number (since this is the only thing you need to find!) STEP 2: Write the equation (just translate the following sentence into math symbols). Three times a number plus 8 is equal to five times the number. 3 * ( x ) + (8) = 5 * ( x ) STEP 3: Solve the equation. 3x + 8 = 5x Subtract 3x from each side: 3x − 3x + 8 = 5x − 3x 8 = 2x Divide both sides by 2: 4 = x STEP 4: Answer the question. The unknown number is 4. STEP 5: Check. Check: 3 * ( x ) + (8) = 5 * ( x ) 3 * ( 4 ) + (8) = 5 * ( 4 ) 12 + 8 = 20 It checks!!
6. Five more than four times a number is equal to 35 less than twice a number. Find the number. SOLUTION: STEP 1: Let x =_____. Let x = the number (since this is the only thing you need to find!) STEP 2: Write the equation (just translate the following sentence into math symbols). Five more than four times a number is equal to 35 less than twice a number 4 * ( x ) + 5 = 2 * ( x ) − 35 STEP 3: Solve the equation. 4x + 5 = 2x − 35 Subtract 2x from each side: 4x − 2x + 5 = 2x − 2x − 35 2x + 5 = − 35 Subtract 5 from each side: 2x + 5 – 5 = – 35 – 5 2x = – 40 Divide both sides by – 2: x = – 20 STEP 4: Answer the question. The unknown number is – 20. STEP 5: Check. Check: 4 * ( x ) + 5 = 2 * ( x ) − 35 4 * (– 20 ) + 5 = 2 * (– 20 ) − 35 − 80 + 5 = − 40 − 35 −75 = −75 It checks!!
7. Five more than four times a number is equal to 35 less twice a number. Find the number. [NOTE: This looks like # 6, but read the problem carefully. One word makes all the difference in these two problems! SOLUTION: STEP 1: Let x =_____. Let x = the number (since this is the only thing you need to find!) STEP 2: Write the equation (just translate the following sentence into math symbols). Five more than four times a number is equal to 35 less twice a number 4 * ( x ) + 5 = 35 − 2 * ( x ) STEP 3: Solve the equation. 4x + 5 = 35 − 2x Add +2x to each side: 4x + 5 + 2x = 35 − 2x + 2x 6x + 5 = 35 Subtract 5 from each side: 6x + 5 – 5 = 35 – 5 6x = 30 Divide both sides by 6: x = 5 STEP 4: Answer the question. The unknown number is 5. STEP 5: Check. Check: 4 * ( x ) + 5 = 35 − 2 * ( x ) 4 * ( 5 ) + 5 = 35 − 2 * ( 5 ) 20 + 5 = 35 − 10 25 = 25 It checks!!
19. Three numbers are such that the second number is 4 more than the first, and the third number is equal to the sum of the first two numbers. The sum of the three numbers is 256. Find the numbers. SOLUTION: STEP 1: Let x =_____. Let x = first number (since this is at the end of the first phrase) x+4 = second number 2x+4 = third number (sum of the first two numbers)
STEP 2: Write the equation. First + Second + Third = 256 ( x ) + (x+4) + (2x+4) = 256
STEP 3: Solve the equation. x + x +4 + 2x+4 = 256 4x + 8 = 256 Subtract 8 from each side: 4x + 8 – 8 = 256 – 8 4x = 248 Divide both sides by 4: x = 62
STEP 4: Answer the question. x = 62 First number x + 4 = 62 + 4 = 66 Second number 2x+4 = 2(62)+4 = 128 Third number 256
STEP 5: Check. Check: Sum of the numbers = 256 See above! It checks!!
20. Three numbers are such that the second number is 6 less than twice the first, and the third number is 5 more than the sum of the first two numbers. The sum of the numbers is 293. Find the numbers. SOLUTION: STEP 1: Let x =_____. Let x = first number (since this is at the end of the first phrase) 2x–6 = second number 3x–6+5 = third number (sum of the first two numbers plus 5 more) 3x – 1 = third number
STEP 2: Write the equation. First + Second + Third = 293 ( x ) + (2x – 6) + (3x – 1) = 293
STEP 3: Solve the equation. x + 2x – 6 + 3x – 1 = 293 6x – 7 = 293 Add +7 to each side: 6x – 7 + 7 = 293 + 7 6x = 300 Divide both sides by 6: x = 50
STEP 4: Answer the question. (YOU FINISH IT!!) x = 50 First number ________ = ____ Second number ________ = ____ Third number 293
STEP 5: Check. Check: Sum of the numbers = 293 See above! Does it check?
29. Two consecutive even integers are such that twice the first, plus three times the second, is equal to 156. Find the numbers. SOLUTION: STEP 1: Let x =_____. Let x = first number x+2 = second number (since they are consecutive EVEN numbers!) STEP 2: Write the equation. 2* First + 3 * Second = 156 2*( x ) + 3 * (x+2) = 156 STEP 3: Solve the equation. 2 x + 3 x + 6 = 156 5x + 6 = 156 Subtract 6 from each side: 5x + 6 – 6 = 156 – 6 5x = 150 Divide both sides by 5: x = 30 STEP 4: Answer the question. x = 30 First number x + 2 = 32 Second number STEP 5: Check. Check: 2*30 + 3* 32 60 + 96 = 156 It checks!!
31. Three consecutive odd integers are such that the sum of the integers is 7 less than four times the smallest. Find the numbers. SOLUTION: STEP 1: Let x =_____. Let x = first integer x+2 = second integer (since these are consecutive ODD integers!) x+4 = third integer 3x+6 = the sum STEP 2: Write the equation. The sum of the integers is 7 less than four times the smallest 3x+6 = 4 * x − 7 STEP 3: Solve the equation. 3x+6 = 4x − 7 Subtract 3x from each side: 3x+6 − 3x = 4x − 7 − 3x 6 = x – 7 Add +7 to each side: 6 + 7 = x – 7 + 7 13 = x STEP 4: Answer the question. x = 13 First number x + 2 = 15 Second number x + 4 = 17 Third number STEP 5: Check. Check: Sum of the numbers equals 7 less than 4 times smallest number. 45 = 4 * 13 – 7 45 = 52 – 7 It checks!!
32. Three consecutive integers are such that the first, plus twice the second, plus three times the third is equal to 200. Find the numbers. SOLUTION: STEP 1: Let x =_____. Let x = first number x+1 = second number (since these are consecutive integers!) x+2 = third number STEP 2: Write the equation. First + 2 * Second + 3 * Third = 200 ( x ) + 2 * (x+1) + 3 * (x+2) = 200 STEP 3: Solve the equation. x + 2x +2 + 3x+6 = 200 6x + 8 = 200 Subtract 8 from each side: 6x + 8 – 8 = 200 – 8 6x = 192 Divide both sides by 6: x = 192/6 = 32 STEP 4: Answer the question. x = 32 First number x + 1 = 33 Second number x + 2 = 34 Third number STEP 5: Check. Check: ( x ) + 2 * (x+1) + 3 * (x+2) = 200 32 + 2 * 33 + 3 * 34 32 + 66 + 102 = 200 It checks!!
35. The second side of a triangle is four more the first side, and the third side is 10 less than the second side. The perimeter of the triangle is 70 feet. Find the sides of the triangle. SOLUTION: STEP 1: Let x =_____.
Let x = first side (since this is at the end of the first phrase) x+4 = second side x+4 –10 = third side x – 6 = third side
STEP 2: Write the equation.
First + Second + Third = Perimeter ( x ) + (x+4) + (x – 6) = 70
STEP 3: Solve the equation.
x + x+4 + x – 6 = 70 3x – 2 = 70
Add +2 to each side: 3x – 2 + 2 = 70 + 2 3x = 72 Divide both sides by 3: x = 24
STEP 4: Answer the question. x = 24 First side x+4 = 28 Second side x+4 – 10 = 18_ Third side
STEP 5: Check. 70 It does check!!
37. The third side of a triangle is 6 less the first side, and the second side is twice the third side. The perimeter of the triangle is 82 feet. Find the sides of the triangle.
SOLUTION:
STEP 1: Let x =_____.
Let x = first side (since this is at the end of the first phrase) x – 6 = third side 2(x –6) = second side
STEP 2: Write the equation.
First + Third + Second = Perimeter ( x ) + (x– 6) + 2(x– 6) = 82
STEP 3: Solve the equation.
x + x– 6 + 2x – 12 = 82 4x – 18 = 82
Add +18 to each side: 4x – 18 + 18 = 82 + 18 4x = 100 Divide both sides by 4: x = 25
STEP 4: Answer the question.
x = 25 First side x-6 = 19 Third side 2(x-6) = 38_ Second side
STEP 5: Check. 82 It does check!!
38. The third side of a triangle is twice the first side, and the second side is 10 less than the third side. The perimeter of the triangle is 70 feet. Find the sides of the triangle. SOLUTION: STEP 1: Let x =_____. Let x = first side (since this is at the end of the first phrase) 2x = third side 2x–10 = second side STEP 2: Write the equation. First + Second + Third = Perimeter ( x ) + (2x– 10) + (2x) = 70
STEP 3: Solve the equation. x + 2x – 10 + 2x = 70 5x – 10 = 70 Add +10 to each side: 5x – 10 + 10 = 70 + 10 5x = 80 Divide both sides by 5: x = 16 STEP 4: Answer the question. x = 16 First side 2x = 32 Second side 2x – 10 = 22_ Third side STEP 5: Check. 70 It does check!!
46. The width of a rectangle is 50 feet less than the length. If the perimeter is 400 feet, find the length and width of the rectangle. SOLUTION: STEP 1: Let x =_____. Let x = length of the rectangle x − 50 = width of the rectangle STEP 2: Write the equation. 2( ) + 2( ) = Perimeter 2( x ) + 2( x–50) = 400 or 2(x–50 ) + 2(x) = 400 STEP 3: Solve the equation. 2x + 2x – 100 = 400 4x – 100 = 400 Add +100 to each side: 4x – 100 + 100 = 400 + 100 4x = 500 Divide both sides by 4: x = 125 STEP 4: Answer the question. x = 125 Length x – 50 = 75 Width STEP 5: Check. Check: 2W + 2 L = P 2(75) + 2(125) = 400 150 + 250 = 400 It checks!!
47. The length of a rectangle is 3 more than twice the width. The perimeter is 56 meters. Find the dimensions of the rectangle. (Note: the length and/or width do not have to come out even! Express the answer in fractional form or as a repeating decimal.) SOLUTION: STEP 1: Let x =_____. Let x = width of the rectangle 2x+3 = length of the rectangle 56 = Perimeter STEP 2: Write the equation. 2( ) + 2( ) = Perimeter 2( x ) + 2( 2x+3) = 56 STEP 3: Solve the equation. 2x + 4x + 6 = 56 6x + 6 = 56 Subtract 6 from each side: 6x +6 – 6 = 56 – 6 6x = 50 Divide both sides by 6:
STEP 4: Answer the question.
2x + 3 = 2(
=
STEP 5: Check. Check: 2(W + L) = P
2( 2( 28 ) = 56 It checks!!
48. The length of a rectangle is three less than five times the width. The perimeter is ten times the width. Find the dimensions and perimeter of the rectangle. SOLUTION: STEP 1: Let x =_____. Let x = width of the rectangle 5x-3 = length of the rectangle 10x = Perimeter
STEP 2: Write the equation. 2( ) + 2( ) = Perimeter 2( x ) + 2( 5x-3) = 10x
STEP 3: Solve the equation. 2x + 10x – 6 = 10x 12x – 6 = 10x Subtract 12x from each side: 12x – 12x – 6 = 10x -12x – 6 = –2x Divide both sides by -2: x = 3 STEP 4: Answer the question. x = 3 Width 5x – 3 = 5(3) – 3 = 12 Length 10x = 30 Perimeter
STEP 5: Check. Check: 2W + 2 L = P 2(3) + 2(12) = 30 6 + 24 = 30 It checks!!
49. The perimeter of a rectangle is 46. Twice the length is 4 more than five times the width. Find the length and width of the rectangle. SOLUTION: STEP 1: Let x =_____. Let x = width of the rectangle 5x+4 = two lengths of the rectangle STEP 2: Write the equation. 2( W ) + 2( L ) = Perimeter 2( x ) + 5x+4 = 46 STEP 3: Solve the equation. 2x + 5x + 4 = 46 7x + 4 = 46 Subtract 4 from each side: 7x +4 – 4 = 46 – 4 7x = 42 Divide both sides by 7: x = 6
STEP 4: Answer the question. x = 6 Width 5x + 4 = 5(6) + 4 = 34 = 2•Length 17 = Length
STEP 5: Check. Check: 2W + 2 L = P 2(6) + 2(17) = 46 12 + 34 = 46 It checks!!
53. A certain number of nickels and some dimes are worth $7.20. The number of dimes is three less than twice the number of nickels. How many of each are there? SOLUTION: STEP 1: Let x =_____. Let x = number of nickels 2x – 3 = number of dimes
STEP 2: Write the equation from the last column of the chart above. 5(x) + 10(2x – 3) = 720 STEP 3: Solve the equation. 5 x + 20x – 30 = 720 25x – 30 = 720 25x = 750 x = 30 STEP 4: Answer the question. x = 30 Nickels 2x – 3 = 2(30) – 3 = 57 Dimes STEP 5: Check. Check: 30 Nickels $1.50 57 Dimes 5.70 TOTAL: $7.20 It checks!!
57. A certain number of quarters, four times as many pennies as quarters, and 6 more dimes than pennies are worth $3.36. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of quarters 4x = number of pennies 4x+6 = number of dimes
STEP 2: Write the equation from the last column of the chart above. 25(x) + 1(4x) + 10(4x+6) = 336 STEP 3: Solve the equation.
25 x + 4x + 40x + 60 = 336 69x + 60 = 336 69x = 276 x = 4 STEP 4: Answer the question. x = 4 Quarters 4x =4(4) = 16 Pennies 4x+6 = 16+6= 22 Dimes STEP 5: Check. Check: 4 Quarters $1.00 16 Pennies .16 22 Dimes 2.20 TOTAL: $3.36 It checks!! 59. A box contains $6.60 in nickels, dimes, and quarters. There are three times as many nickels as quarters, and the number of dimes is 4 less than the number of nickels. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of quarters 3x = number of nickels 3x-4 = number of dimes
STEP 2: Write the equation from the last column of the chart above. 25(x) + 5(3x) + 10(3x-4) = 660 STEP 3: Solve the equation. 25 x + 15x + 30x - 40 = 660 70x - 40 = 660 70x = 700 x = 10 STEP 4: Answer the question. x = 10 Quarters 3x =3(10) = 30 Nickels 3x-4 = 30-4= 26 Dimes
STEP 5: Check. Check: 10 Quarters $2.50 30 Nickels 1.50 26 Dimes 2.60 TOTAL: $6.60 It checks!!
61. A certain number of pennies, four times as many dimes as pennies, and a number of quarters which is 16 less than twice the number of dimes, are worth $24.92. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of pennies 4x = number of dimes 2(4x)-16 = number of quarters 8x - 16 = number of quarters
STEP 2: Write the equation from the last column of the chart above. 1(x) + 10(4x) + 25(8x-16) = 2492 STEP 3: Solve the equation. 1 x + 40x + 200x - 400 = 2492 241x - 400 = 2492
x = 12 STEP 4: Answer the question. x = 12 Pennies 4x =4(12) = 48 Dimes 8x-16 = 96-16 = 80 Quarters
STEP 5: Check. Check: 12 Pennies $ 0.12 48 Dimes 4.80 80 Quarters 20.00 TOTAL: $24.92 It checks!!
62. A sum of money consists of nickels, dimes, and quarters amounting to $1.90. If there are twice as many nickels as quarters and three less dimes than nickels, how many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of quarters 2x = number of nickels 2x-3 = number of dimes
STEP 2: Write the equation from the last column of the chart above. 25(x) + 5(2x) + 10(2x-3) = 190 STEP 3: Solve the equation. 25 x + 10x + 20x - 30 = 190 55x - 30 = 190 55x = 220
x = 4 STEP 4: Answer the question. x = 4 Quarters 2x =2(4) = 8 Nickels 2x-3 = 8 - 3 = 5 Dimes STEP 5: Check. Check: 4 Quarters $ 1.00 8 Nickels .40 5 Dimes .50 TOTAL: $ 1.90 It checks!!
63. A box contains nickels, dimes, and quarters worth $12.60. The number of dimes is 2 less than three times the number of nickels, and the number of quarters is 4 less than twice the number of dimes. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of nickels 3x− 2 = number of dimes 2(3x− 2) − 4 = number of quarters 6x − 4 − 4 = number of quarters 6x − 8 = number of quarters
STEP 2: Write the equation from the last column of the chart above. 5(x) + 10(3x−2) + 25(6x−8) = 1260 STEP 3: Solve the equation. 5 x + 30x - 20 + 150x - 200 = 1260 185x - 220 = 1260 185x = 1480
x = 8 STEP 4: Answer the question. x = 8 Nickels 3x-2 = 3(8) -2 = 22 Dimes 2(3x− 2) − 4= 2(22) − 4 = 40 Quarters STEP 5: Check. Check: 8 Nickels $ .40 22 Dimes 2.20 40 Quarters 10.00 TOTAL: $12.60 It checks!!
64. A box contains nickels, dimes, and quarters worth $69.50. The number of nickels is 10 more than twice the number of dimes. There are as many quarters as nickels and dimes combined. How many of each coin are there? SOLUTION: STEP 1: Let x =_____. Let x = number of dimes 2x+10 = number of nickels x + 2x+10 = number of quarters 3x + 10 = number of quarters
STEP 2: Write the equation from the last column of the chart above. 10(x) + 5(2x+10) + 25(3x+10) = 6950 STEP 3: Solve the equation. 10 x + 10x + 50 + 75x + 250 = 6950 95x + 300 = 6950 95x = 6650
x = 70 STEP 4: Answer the question. x = 70 Dimes 2x+10 =2(70) +10 = 150 Nickels x+2x+10 = 70 + 150 = 220 Quarters
STEP 5: Check. Check: 70 Dimes $ 7.00 150 Nickels 7.50 220 Quarters 55.00 TOTAL: $69.50 It checks!!
65. A merchant mixes some candy worth $3.50 per pound with cheap stuff worth $1.00 per pound. There are 10 more pounds of the cheap stuff than the more expensive candy. If the total value of the mixture is $28, how many pounds of each are there? SOLUTION: STEP 1: Let x =_____. Let x = number of pounds of expensive candy x+10 = number of cheap stuff
STEP 2: Write the equation from the last column of the chart above. 3.50(x) + 1(x+10) = 28 STEP 3: Solve the equation. 3.5 x + 1.0x + 10 = 28 4.5x + 10 = 28 4.5x = 18
x = 4
STEP 4: Answer the question. x = 4 pounds Expensive Candy x+10 = 4 +10 = 14 pounds Cheap Stuff STEP 5: Check. Check: 4 Expensive @ $3.50 = $ 14.00 14 Cheap Stuff @ $1.00 = 14.00 TOTAL: $28.00 It checks!!
EXTRA PROBLEM from Daniel in Kyrgyzstan: How many pounds of hamburger that cost $1.00 per pound must be mixed with 50 pounds of hamburger that cost 2.00 per pound to make a mixture that cost $1.20 per pound. SOLUTION: STEP 1: Let x =_____. Let x = number of pounds of hamburger at $1.00 per pound 50 = number of pounds of hamburger at $2.00 per pound x+50 = number of pounds of hamburger at $1.20 per pound
STEP 2: Write the equation from the last column of the chart above. 1.00(x) + 2.00(50) = 1.20(x+50) STEP 3: Solve the equation. 1.0 x + 100 = 1.2x + 60 Clear the decimal by multiplying both sides times 10: 10 x + 1000 = 12x + 600 400 = 2x x = 200 STEP 4: Answer the question. x = 200 pounds $1 hamburger
STEP 5: Check. Check: 200 pounds @ $1.00 = $ 200 50 pounds @ $2.00 = $ 100 TOTAL: 250 pounds @ $1.20 = $ 300 It checks!!
69. A sum of money was invested at 8% simple interest, and three times this much at 10%. The total interest earned for the year was $190. How much was invested at each rate. SOLUTION: STEP 1: Let x =_____. Let x = principle invested 8% 3x = principle invested at 10%
STEP 2: Write the equation from the last column of the chart above. .08(x) + .10(3x) = 190 STEP 3: Solve the equation. .08 x + .30x = 190 .38x = 190
STEP 4: Answer the question. x = $ 500 at 8% 3x = $1500 at 10% STEP 5: Check. Check: $ 500 @ 8% = $ 40 $1500 @ 10% = $150 TOTAL: $190 It checks!!
70. A sum of money was invested at 12% simple interest, and $1000 less than this at 10%. The total interest earned for the year was $1000. How much was invested at each rate. SOLUTION: STEP 1: Let x =_____. Let x = principle invested 12% x−1000 = principle invested at 10%
STEP 2: Write the equation from the last column of the chart above. .12(x) + .10(x−1000) = 1000 STEP 3: Solve the equation. .12 x + .10x − 100 = 1000 .22x − 100 = 1000 .22x = 1100
STEP 4: Answer the question. x = $5000 at 12% x−1000 = $5000-$1000 = $4000 at 10% STEP 5: Check. Check: $5000 @ 12% = $ 600 $4000 @ 10% = $ 400 TOTAL: $1000 It checks!!
71. A sum of money was invested at 5% annual interest, and $500 less than twice this amount was invested at 12%. If the total interest earned for the year was $375, how much was invested at each rate? SOLUTION: STEP 1: Let x =_____. Let x = principle invested 5% 2x−500 = principle invested at 12%
STEP 2: Write the equation from the last column of the chart above. .05(x) + .12(2x−500) = 375 STEP 3: Solve the equation. .05 x + .24x − 60 = 375 .29x − 60 = 375 .29x = 435
STEP 4: Answer the question. x = $1500 at 5% 2x−500 = $3000-$500 = $2500 at 12% STEP 5: Check. Check: $1500 @ 5% = $ 75 $2500 @ 12% = $ 300 TOTAL: $ 375 It checks!!
74. A man has $10,000 to invest, some in a relatively safe account earning 5% interest per year, and the rest in more speculative investments earning 12% per year. If the total interest earned for the year was $955, how much was invested at each rate? SOLUTION: STEP 1: Let x =_____. Let x = principle invested 5% 10000− x = principle invested at 12%
STEP 2: Write the equation from the last column of the chart above. .05(x) + .12(10000−x) = 955 STEP 3: Solve the equation. .05 x + 1200 −.12x = 955 −.07x + 1200 = 955 −.07x = −245
STEP 4: Answer the question. x = $3500 at 5% 10000−x = 10000-3500 = $6500 at 12% STEP 5: Check. Check: $3500 @ 5% = $ 175 $6500 @ 12% = $ 780 TOTAL: $ 955 It checks!!
81. How much water must be added to 50% solution to obtain 100 liters of 10% solution? SOLUTION: STEP 1: Let x =_____. Let x = number of liters of water (which is 0% alcohol!)
STEP 2: Write the equation from the last column of the chart above. 0 + .50(100-x) = .10 (100) STEP 3: Solve the equation. 50 - .50 x = 10 -.50 x = - 40
STEP 4: Answer the question. x = 80 liters of water 100 – x = 20 liters of 50% alcohol STEP 5: Check. Check: 20 liters @ 50% solution = 10 liters 100 liters @ 10% solution = 10 liters . It checks!!
Age Problem #1. Caylyne is 5 years older than Cassie. The sum of their ages is 17. How old is Cassie, and how old is Caylyne? SOLUTION: STEP 1: Let x =_____. Let x = Cassie’s age x+5 = Caylyne’s age
STEP 2: Write the equation from the last sentence in the problem above. “ The SUM of their ages is 17.” (That is, you have to ADD the ages together!) x + x+5 = 17 STEP 3: Solve the equation. 2 x + 5 = 17 Subtract 5 from each side of the equation: 2 x + 5 = 17 −5 −5
Divide both sides by 2: x = 6 STEP 4: Answer the question. x = 6 Cassie’s Age x+5 = 6 + 5 = 11 Caylyne’s Age STEP 5: Check. The sum of their ages is 6 + 11 = 17. It checks!!
Age Problem #2. The age of Caylyne is one year less than twice the age of Cassie. Daddy’s age is three times the age of Caylyne. The sum (click here!!) of all of their ages is 50. How old is each? SOLUTION: STEP 1: Let x =_____. Let x = Cassie’s age 2x−1 = Caylyne’s age 3(2x−1) = Daddy’s age STEP 2: Write the equation from the last sentence in the problem above. “The SUM of all their ages is 50.” (That is, you have to ADD all the ages together!) Cassie + Caylyne + Daddy = 50 x + 2x−1 + 3(2x−1) = 50 STEP 3: Solve the equation. x + 2x−1 + 3(2x−1) = 50 x + 2x−1 + 6x−3 = 50 9 x − 4 = 50 Add + 4 to each side of the equation: 9 x − 4 = 50 + 4 + 4
Divide both sides by 9:
x = 6 STEP 4: Answer the question. Let x = 6 Cassie’s age 2x−1 = 12 − 1 = 11 Caylyne’s age 3(2x−1) = 3*11 = 33 Daddy’s age STEP 5: Check. The sum of their ages is 6 + 11 + 33 = 50. It checks!!
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