Problems 1-5 : In each case, solve the equation by isolating the variable.
Problem 1 :
5x - 3 = 22
Problem 2 :
²ˣ⁄₃ + 5 = 13
Problem 3 :
ˣ⁄₃ + ⁷⁄₂ = 0
Problem 4 :
2(p + 3) - 3(1 - 5p) = 3(3p + 7) - (9p - 16)
Problem 5 :
8(y - 1) + y(y - 1) = y^{2} + 2(3y + 5)
Problem 6 :
If k = -5 and ⁽²ˣ ⁻ ⁷⁾⁄₃ = k, find the value of x.
Problem 7 :
If the circumference of a circle is 24π cm, find the radius.
Problem 8 :
If the perimeter of a rectangle is 24 in., find the length of the rectangle in terms of its width.
Problem 9 :
If 6x - 7 = -5. find the value of 12x + 1 without isolating the variable x.
Problem 10 :
If 14x + 3 = 25. find the value of 21x - 7 without isolating the variable x.
Problem 11 :
If 5 is decreased from 7 times of a number, the result would be 37. Find the number.
Problem 12 :
13 is added to 5 times of a number results 3. Find the number results when 5 is added to 7 times of the same number.
Problem 13 :
It is given that 3(a + 5) = 2(b + 9) - 3. Find the value of ᵃ⁄b.
Problem 14 :
Adam is thrice as old as his son now. 5 years ago, Adam's age was 1 more than four times of his son's age. Find the present age of Adam and son.
1. Answer :
5x - 3 = 22
Add 3 to both sides.
5x = 25
Divide both sides by 5.
x = 5
2. Answer :
²ˣ⁄₃ + 5 = 13
Subtract 5 from both sides.
²ˣ⁄₃ = 8
Multiply both sides by 3.
2x = 24
Divide both sides by 2.
x = 12
3. Answer :
ˣ⁄₃ + ⁷⁄₂ = 0
Least common multiple of the decominators (3, 2) is 6.
Multiply both sides by the equation by 6 to get rid of the denominators 3 and 2.
6(ˣ⁄₃ + ⁷⁄₂) = 6(0)
6(ˣ⁄₃) + 6(⁷⁄₂) = 0
2x + 3(7) = 0
2x + 14 = 0
Subtract 14 from both sides.
2x = -14
Divide both sides by 2.
x = -7
4. Answer :
2(p + 3) - 3(1 - 5p) = 3(3p + 7) - (9p - 16)
Using Distributive Property,
2p + 6 - 3 + 15p = 9p + 21 - 9p + 16
Combine the like terms.
17p + 3 = 37
Subtract 3 from both sides.
17p = 34
Divide both sides by 17.
p = 2
5. Answer :
8(y - 1) + y(y - 1) = y^{2} + 2(3y + 5)
Using Distributive Property,
8y - 8 + y^{2} - y = y^{2} + 6y + 10
Combine the like terms.
y^{2} + 7y - 8 = y^{2} + 6y + 10
Subtract y^{2} from both sides.
7y - 8 = 6y + 10
Subtract 6y from both sides.
y - 8 = 10
Add 8 to both sides.
y = 18
6. Answer :
⁽²ˣ ⁻ ⁷⁾⁄₃ = k
Substitute k = -5.
⁽²ˣ ⁻ ⁷⁾⁄₃ = -5
Multiply both sides by 3.
2x - 7 = -15
Add 7 to both sides.
2x = -8
Divide both sides by 2.
x = -4
7. Answer :
Circumference of the circle = 24π cm
2πr = 24π
r = 12
The radius of the circle is 12 cm.
8. Answer :
Perimeter of the rectangle = 24 in.
2(l + w) = 24
Divide both sides by 2.
l + w = 12
Subtract w from both sides.
l = 12 - w
length = 12 - width
9. Answer :
6x - 7 = -5
Add 7 to both sides.
6x = 2
Multiply both sides by 2.
12x = 4
Add 1 to both sides.
12x + 1 = 5
10. Answer :
14x + 3 = 25
Subtract 3 from both sides.
14x = 22
Divide both sides by 2.
7x = 11
Multiply both sides by 3.
21x = 33
Subtract 7 from both sides.
21x - 7 = 26
11. Answer :
Let x be the required number.
Given : Decreasing 5 from 7 times of a number results 37.
7x - 5 = 37
Add 5 to both sides.
7x = 42
Divide both sides by 7.
x = 6
The required number is 2.
12. Answer :
Let y be the number.
Given : 13 is added to 5 times of a number results 12.
5y + 13 = 3
Subtract 13 from both sides.
5y = -10
Divide both sides by 5.
y = -2
Multiply both sides by 7.
7y = -14
Add 5 to both sides.
7y + 5 = -9
13. Answer :
3(a + 5) = 2(b + 9) - 3
Using Distributive Property,
3a + 15 = 2b + 18 - 3
3a + 15 = 2b + 15
Subtract 15 from both sides.
3a = 2b
Divide both sides by b.
³ᵃ⁄b = 2
Multiply both sides by ⅓.
ᵃ⁄b = ⅔
14. Answer :
Let x be the present of the son.
Then the present age of Adam is 3x.
Age of the son 5 years ago = x - 5
Age of Adam 5 years ago = 3x - 5
Given : Adam's age was 1 more than four times of his son's age.
3x - 5 = 4(x - 5) + 1
Using Distributive Property,
3x - 5 = 4x - 20 + 1
3x - 5 = 4x - 19
Subtract 4x from both sides.
-x - 5 = -19
Add 5 to both sides.
-x = -14
Multiply both sides by -1.
x = 14
3x = 3(14) = 42
The present age of Adam is 42 years and that of his son is 14 years.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Sep 29, 23 10:55 PM
Sep 29, 23 10:49 PM
Sep 29, 23 07:56 PM