Question 1 :
Solve for x.
x + 11 = 17
Question 2 :
Solve for p.
3p + 4 = 25
Question 3 :
Solve for y.
5y + 3 = 17 - 2y
Question 4 :
Solve for x.
12 - x = 8
Question 5 :
Solve for s.
ˢ⁄₆ = 7
Question 6 :
Solve for a.
⁽³ᵃ ⁺ ²⁾⁄₄ = 5
Question 7 :
Solve for y.
⁴ʸ⁄₃ - 1 = ¹⁴ʸ⁄₁₅ + ¹⁹⁄₅
Question 8 :
Solve for x.
⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11
Question 9 :
Solve for x.
⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄
Question 10 :
Solve for x.
ˣ⁄₀.₅ - ¹⁄₀.₀₅ + ˣ⁄₀.₀₀₅ - ¹⁄₀.₀₀₀₅ = 0
Question 11 :
How many cookies are in a box, if the box plus three more equals 23 cookies?
Question 12 :
6 less than 7 times of a number is 57. What is the number?
1. Answer :
x + 11 = 17
Subtract 11 from both sides.
(x + 11) - 11 = 17 - 11
x + 11 - 11 = 6
x = 6
2. Answer :
3p + 4 = 25
Subtract 4 from both sides.
(3p + 4) - 4 = 25 - 4
3p + 4 - 4 = 21
3p = 21
Divide both sides by 3.
³ᵖ⁄₃ = ²¹⁄₃
p = 7
3. Answer :
5y + 3 = 17 - 2y
Add 2y to both sides.
(5y + 3) + 2y = (17 - 2y) + 2y
5y + 3 + 2y = 17 - 2y + 2y
7y + 3 = 17
Subtract 3 from both sides.
(7y + 3) - 3 = 17 - 3
7y + 3 - 3 = 14
7y = 14
Divide both sides by 7.
⁷ʸ⁄₇ = ¹⁴⁄₇
y = 2
4. Answer :
12 - x = 8
Subtract 12 from both sides.
(12 - x) - 12 = 8 - 12
12 - x - 12 = -4
-x = -4
Multiply both sides by -1.
-1(-x) = -1(-4)
x = 4
5. Answer :
ˢ⁄₆ = 7
Multiply both sides by 6.
6(ˢ⁄₆) = 6(7)
s = 42
6. Answer :
⁽³ᵃ ⁺ ²⁾⁄₄ = 5
Multiply both sides by 4.
4(⁽³ᵃ ⁺ ²⁾⁄₄) = 4(5)
3a + 2 = 20
Subtract 2 from both sides.
(3a + 2) - 2 = 20 - 2
3a + 2 - 2 = 18
3a = 18
Divide both sides by 3.
³ᵃ⁄₃ = ¹⁸⁄₃
a = 6
7. Answer :
⁴ʸ⁄₃ - 1 = ¹⁴ʸ⁄₁₅ + ¹⁹⁄₅
The least common multiple of the denominators (3, 15, 5) is 15.
Multiply both sides of the equation by 15 to get rid of the denominators.
15(⁴ʸ⁄₃ - 1) = 15(¹⁴ʸ⁄₁₅ + ¹⁹⁄₅)
Using distributive property,
15(⁴ʸ⁄₃) + 15(-1) = 15(¹⁴ʸ⁄₁₅) + 15(¹⁹⁄₅)
5(4y) - 15 = 1(14y) + 3(19)
20y - 15 = 14y + 57
Subtract 14y from both sides.
(20y - 15) - 14y = (14y + 57) - 14y
20y - 15 - 14y = 14y + 57 - 14y
6y - 15 = 57
Add 15 to both sides.
(6y - 15) + 15 = 57 + 15
6y - 15 + 15 = 72
6y = 72
Divide both sides by 6.
⁶ʸ⁄₆ = ⁷²⁄₆
y = 12
8. Answer :
⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11
The least common multiple of the denominators (4, 3) is 12.
Multiply both sides of the equation by 12 to get rid of the denominators.
12(⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃) = 12(11)
Using distributive property,
12(⁽ˣ ⁺ ⁴⁾⁄₄) + 12(⁽ˣ ⁻ ⁵⁾⁄₃) = 132
3(x + 4) + 4(x - 5) = 132
Using distributive property,
3x + 12 + 4x - 20 = 132
7x - 8 = 132
Add 8 to both sides.
(7x - 8) + 8 = 132 + 8
7x - 8 + 8 = 140
7x = 140
Divide both sides by 7.
⁷ˣ⁄₇ = ¹⁴⁰⁄₇
x = 20
9. Answer :
⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄
The least common multiple of the denominators (5, 4) is 20.
Multiply both sides of the equation by 20 to get rid of the denominators.
20(⁽ˣ ⁺ ²⁴⁾⁄₅) = 20(4 + ˣ⁄₄)
4(x + 24) = 20(4) + 20(ˣ⁄₄)
4x + 96 = 80 + 5x
Subtract 5x from both sides.
(4x + 96) - 5x = (80 + 5x) - 5x
4x + 96 - 5x = 80 + 5x - 5x
-x + 96 = 80
Subtract 96 from both sides.
(-x + 96) - 96 = 80 - 96
-x + 96 - 96 = -16
-x = -16
Multiply both sides by -1.
x = 16
10. Answer :
ˣ⁄₀.₅ - ¹⁄₀.₀₅ + ˣ⁄₀.₀₀₅ - ¹⁄₀.₀₀₀₅ = 0
ˣ/⁽⁵⁄₁₀₎ - ¹/⁽⁵⁄₁₀₀₎ + ˣ/⁽⁵⁄₁₀₀₀₎ - ¹/⁽⁵⁄₁₀₀₀₀₎ = 0
x(¹⁰⁄₅) - 1(¹⁰⁰⁄₅) + x(¹⁰⁰⁰⁄₅) - 1(¹⁰⁰⁰⁰⁄₅) = 0
¹⁰ˣ⁄₅ - ¹⁰⁰⁄₅ + ¹⁰⁰⁰ˣ⁄₅ - ¹⁰⁰⁰⁰⁄₅ = 0
5(¹⁰ˣ⁄₅ - ¹⁰⁰⁄₅ + ¹⁰⁰⁰ˣ⁄₅ - ¹⁰⁰⁰⁰⁄₅) = 5(0)
5(¹⁰ˣ⁄₅) + 5(-¹⁰⁰⁄₅) + 5(¹⁰⁰⁰ˣ⁄₅) + 5(-¹⁰⁰⁰⁰⁄₅) = 5(0)
10x - 100 + 1000x - 10000 = 0
1010x - 10100 = 0
Add 10100 to both sides.
(1010x - 10100) + 10100 = 0 + 10100
1010x - 10100 + 10100 = 10100
1010x = 10100
Divide both sides by 1010.
¹⁰¹⁰ˣ⁄₁₀₁₀ = ¹⁰¹⁰⁰⁄₁₀₁₀
x = 10
11. Answer :
Let x be the number of cookies in the box.
It is given that the box plus three more equals 23 cookies.
x + 3 = 23
Subtract 3 from both sides.
(x + 3) - 3 = 23 - 3
x + 3 - 3 = 20
x = 20
There are 20 cookies in a box.
12. Answer :
Let x be the number.
It is given that 6 less than 7 times of a number is 57.
7x - 6 = 57
Add 6 to both sides.
(7x - 6) + 6 = 57 + 6
7x -6 + 6 = 63
7x = 63
Divide both sides by 7.
⁷ˣ⁄₇ = ⁶³⁄₇
x = 9
The number is 9.
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