|
1.10 Word Problems Basic Algebra: One Step at a Time
Page 65-92: #3, 6, 7, 19, 20, 29, 31, 32, 38, 46, 47, 48, 49, 57, 59, 61, 62, 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) 2x6 = second number 3x6+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!!
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 2x10 = third number STEP 2: Write the equation. First + Second + Third = Perimeter ( x ) + (2x) + (2x 10) = 70
STEP 3: Solve the equation. x + 2x + 2x 10 = 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 number 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( x50) = 400 or 2(x50 ) + 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 = 2Length 17 = Length
STEP 5: Check. Check: 2W + 2 L = P 2(6) + 2(17) = 46 12 + 34 = 46 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
|
