Activity 5: Find Those Missing!
Solve each problem algebraically. Explain how you arrived at your answers and what
method did you use.
1. The sum of the number and its reciprocal is 13/6. Find the number.
2. The sum of the number and three times its reciprocal is 79/10. Find the
number.
3. The sum of two numbers is 20, and their product is 96. Find the two numbers.
4. One number is 5 more than 3 times a second number. If the reciprocal is -2,
what are the numbers?
5. The length of a rectangle is three more than twice its width, and its area is 90
square meters. Find its dimensions.
6. The width of a rectangle in one-third its length. If the area of the rectangle is
20 square inches, what are the dimensions of the rectangle?
thanks
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Verified answer
1.x + 1/x = 13/6
x^2 + 1 = 13/6x
6x^2 + 6 = 13x
6x^2 - 13x + 6 = 0
6x^2 - 4x - 9x + 6 = 0
2x(3x - 2) - 3(3x - 2) = 0
(2x - 3)(3x - 2) = 0
x_1 = 3/2
x_2 = 2/3
2.
x + 3/x = 79/10
10x^2 - 79x + 30 = 0
10x^2 - 4x - 75x + 30 = 0
2x(5x - 2) - 15(5x - 2) = 0
(2x - 15)(5x - 2) = 0
x_1 = 15/2
x_2 = 2/5
3.
x + y = 20
xy = 96
xy = 96
y = 96/x
x + y = 20
x + 96/x = 20
x^2 - 20x + 96 = 0
(x - 8)(x- 12) = 0
x_1 = 8
x_2 = 12
4. sorry i’m not sure how to solve
5.
length: 2w + 3
width: w
area: 90
A = lw
90 = (2w + 3)(w)
2w^2 + 3w - 90 = 0
(2x + 15)(x - 6) = 0
x_1 = -15/2
x_2 = 6
// no measurements are negative so disregard x_1
length: 18m
width: 6
6.
width: x/3
length: x
area: 20
A = lw
20 = x(x/3)
20 = x^2/3
60 = x^2
sqrt 60 = x
x_1 = -2 sqrt 15
x_2 = 2 sqrt 15
// no measurements are negative so disregard x_1
length: 2 sqrt 15
width: (2 sqrt 15)/3