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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.
Λ’ββ = 7
Question 6 :
Solve for a.
β½Β³α΅ βΊ Β²βΎββ = 5
Question 7 :
Solve for y.
β΄ΚΈββ - 1 = ΒΉβ΄ΚΈβββ + ΒΉβΉββ
Question 8 :
Solve for x.
β½Λ£ βΊ β΄βΎββ + β½Λ£ β» β΅βΎββ = 11
Question 9 :
Solve for x.
β½Λ£ βΊ Β²β΄βΎββ = 4 + Λ£ββ
Question 10 :
Solve for x.
Λ£ββ.β - ΒΉββ.ββ + Λ£ββ.βββ - ΒΉββ.ββββ = 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.
Β³α΅ββ = Β²ΒΉββ
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.
β·ΚΈββ = ΒΉβ΄ββ
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 :
Λ’ββ = 7
Multiply both sides by 6.
6(Λ’ββ) = 6(7)
s = 42
6. Answer :
β½Β³α΅ βΊ Β²βΎββ = 5
Multiply both sides by 4.
4(β½Β³α΅ βΊ Β²βΎββ) = 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.
Β³α΅ββ = ΒΉβΈββ
a = 6
7. Answer :
β΄ΚΈββ - 1 = ΒΉβ΄ΚΈβββ + ΒΉβΉββ
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(β΄ΚΈββ - 1) = 15(ΒΉβ΄ΚΈβββ + ΒΉβΉββ )
Using distributive property,
15(β΄ΚΈββ) + 15(-1) = 15(ΒΉβ΄ΚΈβββ ) + 15(ΒΉβΉββ )
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.
βΆΚΈββ = β·Β²ββ
y = 12
8. Answer :
β½Λ£ βΊ β΄βΎββ + β½Λ£ β» β΅βΎββ = 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(β½Λ£ βΊ β΄βΎββ + β½Λ£ β» β΅βΎββ) = 12(11)
Using distributive property,
12(β½Λ£ βΊ β΄βΎββ) + 12(β½Λ£ β» β΅βΎββ) = 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.
β·Λ£ββ = ΒΉβ΄β°ββ
x = 20
9. Answer :
β½Λ£ βΊ Β²β΄βΎββ = 4 + Λ£ββ
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(β½Λ£ βΊ Β²β΄βΎββ ) = 20(4 + Λ£ββ)
4(x + 24) = 20(4) + 20(Λ£ββ)
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 :
Λ£ββ.β - ΒΉββ.ββ + Λ£ββ.βββ - ΒΉββ.ββββ = 0
Λ£/β½β΅ββββ - ΒΉ/β½β΅βββββ + Λ£/β½β΅ββββββ - ΒΉ/β½β΅βββββββ = 0
x(ΒΉβ°ββ ) - 1(ΒΉβ°β°ββ ) + x(ΒΉβ°β°β°ββ ) - 1(ΒΉβ°β°β°β°ββ ) = 0
ΒΉβ°Λ£ββ - ΒΉβ°β°ββ + ΒΉβ°β°β°Λ£ββ - ΒΉβ°β°β°β°ββ = 0
5(ΒΉβ°Λ£ββ - ΒΉβ°β°ββ + ΒΉβ°β°β°Λ£ββ - ΒΉβ°β°β°β°ββ ) = 5(0)
5(ΒΉβ°Λ£ββ ) + 5(-ΒΉβ°β°ββ ) + 5(ΒΉβ°β°β°Λ£ββ ) + 5(-ΒΉβ°β°β°β°ββ ) = 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.
ΒΉβ°ΒΉβ°Λ£βββββ = ΒΉβ°ΒΉβ°β°βββββ
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.
β·Λ£ββ = βΆΒ³ββ
x = 9
The number is 9.
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