Algebraic equations worksheet with 18 problems involving fractions and variables.

Problem Set Solution

We will solve each equation step by step. Let's go through them one by one.

---

#### 1. \(\frac{5a}{3} - 2 = \frac{a}{4} + 15\)

1. Eliminate the fractions by finding the least common denominator (LCD), which is 12.
\[
12 \left( \frac{5a}{3} - 2 \right) = 12 \left( \frac{a}{4} + 15 \right)
\]
\[
12 \cdot \frac{5a}{3} - 12 \cdot 2 = 12 \cdot \frac{a}{4} + 12 \cdot 15
\]
\[
4 \cdot 5a - 24 = 3 \cdot a + 180
\]
\[
20a - 24 = 3a + 180
\]

2. Move all terms involving \(a\) to one side and constants to the other:
\[
20a - 3a = 180 + 24
\]
\[
17a = 204
\]

3. Solve for \(a\):
\[
a = \frac{204}{17} = 12
\]

Answer:
\[
\boxed{12}
\]

---

#### 2. \(\frac{x}{3} + \frac{x}{6} - 1 = \frac{5x}{12}\)

1. Find the LCD, which is 12.
\[
12 \left( \frac{x}{3} + \frac{x}{6} - 1 \right) = 12 \left( \frac{5x}{12} \right)
\]
\[
12 \cdot \frac{x}{3} + 12 \cdot \frac{x}{6} - 12 \cdot 1 = 12 \cdot \frac{5x}{12}
\]
\[
4x + 2x - 12 = 5x
\]
\[
6x - 12 = 5x
\]

2. Move all terms involving \(x\) to one side:
\[
6x - 5x = 12
\]
\[
x = 12
\]

Answer:
\[
\boxed{12}
\]

---

#### 3. \(\frac{2a}{5} - \frac{7}{5} = \frac{3a}{4} - \frac{5}{4}\)

1. Eliminate the fractions by finding the LCD, which is 20.
\[
20 \left( \frac{2a}{5} - \frac{7}{5} \right) = 20 \left( \frac{3a}{4} - \frac{5}{4} \right)
\]
\[
20 \cdot \frac{2a}{5} - 20 \cdot \frac{7}{5} = 20 \cdot \frac{3a}{4} - 20 \cdot \frac{5}{4}
\]
\[
4 \cdot 2a - 4 \cdot 7 = 5 \cdot 3a - 5 \cdot 5
\]
\[
8a - 28 = 15a - 25
\]

2. Move all terms involving \(a\) to one side and constants to the other:
\[
8a - 15a = -25 + 28
\]
\[
-7a = 3
\]

3. Solve for \(a\):
\[
a = -\frac{3}{7}
\]

Answer:
\[
\boxed{-\frac{3}{7}}
\]

---

#### 4. \(\frac{x}{2} - \frac{x}{3} = \frac{x}{4} - 1\)

1. Find the LCD, which is 12.
\[
12 \left( \frac{x}{2} - \frac{x}{3} \right) = 12 \left( \frac{x}{4} - 1 \right)
\]
\[
12 \cdot \frac{x}{2} - 12 \cdot \frac{x}{3} = 12 \cdot \frac{x}{4} - 12 \cdot 1
\]
\[
6x - 4x = 3x - 12
\]
\[
2x = 3x - 12
\]

2. Move all terms involving \(x\) to one side:
\[
2x - 3x = -12
\]
\[
-x = -12
\]

3. Solve for \(x\):
\[
x = 12
\]

Answer:
\[
\boxed{12}
\]

---

#### 5. \(\frac{3x}{4} + \frac{1}{2} = \frac{x}{5} + \frac{8}{5}\)

1. Find the LCD, which is 20.
\[
20 \left( \frac{3x}{4} + \frac{1}{2} \right) = 20 \left( \frac{x}{5} + \frac{8}{5} \right)
\]
\[
20 \cdot \frac{3x}{4} + 20 \cdot \frac{1}{2} = 20 \cdot \frac{x}{5} + 20 \cdot \frac{8}{5}
\]
\[
5 \cdot 3x + 10 \cdot 1 = 4 \cdot x + 4 \cdot 8
\]
\[
15x + 10 = 4x + 32
\]

2. Move all terms involving \(x\) to one side and constants to the other:
\[
15x - 4x = 32 - 10
\]
\[
11x = 22
\]

3. Solve for \(x\):
\[
x = \frac{22}{11} = 2
\]

Answer:
\[
\boxed{2}
\]

---

#### 6. \(\frac{x}{4} - \frac{5}{12} = \frac{x}{2} - \frac{x}{3}\)

1. Find the LCD, which is 12.
\[
12 \left( \frac{x}{4} - \frac{5}{12} \right) = 12 \left( \frac{x}{2} - \frac{x}{3} \right)
\]
\[
12 \cdot \frac{x}{4} - 12 \cdot \frac{5}{12} = 12 \cdot \frac{x}{2} - 12 \cdot \frac{x}{3}
\]
\[
3x - 5 = 6x - 4x
\]
\[
3x - 5 = 2x
\]

2. Move all terms involving \(x\) to one side:
\[
3x - 2x = 5
\]
\[
x = 5
\]

Answer:
\[
\boxed{5}
\]

---

#### 7. \(\frac{x}{5} = \frac{3}{35} + \frac{x+1}{7}\)

1. Find the LCD, which is 35.
\[
35 \left( \frac{x}{5} \right) = 35 \left( \frac{3}{35} + \frac{x+1}{7} \right)
\]
\[
35 \cdot \frac{x}{5} = 35 \cdot \frac{3}{35} + 35 \cdot \frac{x+1}{7}
\]
\[
7x = 3 + 5(x + 1)
\]
\[
7x = 3 + 5x + 5
\]
\[
7x = 5x + 8
\]

2. Move all terms involving \(x\) to one side:
\[
7x - 5x = 8
\]
\[
2x = 8
\]

3. Solve for \(x\):
\[
x = \frac{8}{2} = 4
\]

Answer:
\[
\boxed{4}
\]

---

#### 8. \(\frac{2y-1}{3} + 3 = y\)

1. Eliminate the fraction by multiplying through by 3:
\[
3 \left( \frac{2y-1}{3} + 3 \right) = 3 \cdot y
\]
\[
(2y - 1) + 9 = 3y
\]
\[
2y - 1 + 9 = 3y
\]
\[
2y + 8 = 3y
\]

2. Move all terms involving \(y\) to one side:
\[
2y - 3y = -8
\]
\[
-y = -8
\]

3. Solve for \(y\):
\[
y = 8
\]

Answer:
\[
\boxed{8}
\]

---

#### 9. \(\frac{5x+2}{3} + \frac{x}{5} = \frac{3x-5}{15} + x\)

1. Find the LCD, which is 15.
\[
15 \left( \frac{5x+2}{3} + \frac{x}{5} \right) = 15 \left( \frac{3x-5}{15} + x \right)
\]
\[
15 \cdot \frac{5x+2}{3} + 15 \cdot \frac{x}{5} = 15 \cdot \frac{3x-5}{15} + 15 \cdot x
\]
\[
5(5x + 2) + 3x = (3x - 5) + 15x
\]
\[
25x + 10 + 3x = 3x - 5 + 15x
\]
\[
28x + 10 = 18x - 5
\]

2. Move all terms involving \(x\) to one side and constants to the other:
\[
28x - 18x = -5 - 10
\]
\[
10x = -15
\]

3. Solve for \(x\):
\[
x = \frac{-15}{10} = -\frac{3}{2}
\]

Answer:
\[
\boxed{-\frac{3}{2}}
\]

---

#### 10. \(\frac{2y-1}{3} + 3 = y\)

This is the same as problem 8. The solution is:
\[
\boxed{8}
\]

---

#### 11. \(\frac{1}{2} + \frac{x-1}{3} = \frac{x}{2}\)

1. Find the LCD, which is 6.
\[
6 \left( \frac{1}{2} + \frac{x-1}{3} \right) = 6 \left( \frac{x}{2} \right)
\]
\[
6 \cdot \frac{1}{2} + 6 \cdot \frac{x-1}{3} = 6 \cdot \frac{x}{2}
\]
\[
3 + 2(x - 1) = 3x
\]
\[
3 + 2x - 2 = 3x
\]
\[
1 + 2x = 3x
\]

2. Move all terms involving \(x\) to one side:
\[
2x - 3x = -1
\]
\[
-x = -1
\]

3. Solve for \(x\):
\[
x = 1
\]

Answer:
\[
\boxed{1}
\]

---

#### 12. \(x - \frac{x-1}{2} = 0\)

1. Eliminate the fraction by multiplying through by 2:
\[
2 \left( x - \frac{x-1}{2} \right) = 2 \cdot 0
\]
\[
2x - (x - 1) = 0
\]
\[
2x - x + 1 = 0
\]
\[
x + 1 = 0
\]

2. Solve for \(x\):
\[
x = -1
\]

Answer:
\[
\boxed{-1}
\]

---

#### 13. \(\frac{4x}{3} - \frac{3x-4}{6} = 5 - \frac{x-2}{2}\)

1. Find the LCD, which is 6.
\[
6 \left( \frac{4x}{3} - \frac{3x-4}{6} \right) = 6 \left( 5 - \frac{x-2}{2} \right)
\]
\[
6 \cdot \frac{4x}{3} - 6 \cdot \frac{3x-4}{6} = 6 \cdot 5 - 6 \cdot \frac{x-2}{2}
\]
\[
2 \cdot 4x - (3x - 4) = 30 - 3(x - 2)
\]
\[
8x - 3x + 4 = 30 - 3x + 6
\]
\[
5x + 4 = 36 - 3x
\]

2. Move all terms involving \(x\) to one side and constants to the other:
\[
5x + 3x = 36 - 4
\]
\[
8x = 32
\]

3. Solve for \(x\):
\[
x = \frac{32}{8} = 4
\]

Answer:
\[
\boxed{4}
\]

---

#### 14. \(\frac{x+1}{2} + \frac{x+2}{3} - \frac{x+3}{4} = 2\)

1. Find the LCD, which is 12.
\[
12 \left( \frac{x+1}{2} + \frac{x+2}{3} - \frac{x+3}{4} \right) = 12 \cdot 2
\]
\[
12 \cdot \frac{x+1}{2} + 12 \cdot \frac{x+2}{3} - 12 \cdot \frac{x+3}{4} = 24
\]
\[
6(x + 1) + 4(x + 2) - 3(x + 3) = 24
\]
\[
6x + 6 + 4x + 8 - 3x - 9 = 24
\]
\[
7x + 5 = 24
\]

2. Move the constant to the other side:
\[
7x = 24 - 5
\]
\[
7x = 19
\]

3. Solve for \(x\):
\[
x = \frac{19}{7}
\]

Answer:
\[
\boxed{\frac{19}{7}}
\]

---

#### 15. \(\frac{5x-4}{3} - \frac{4x-3}{2} - \frac{3x-2}{1} = 0\)

1. Find the LCD, which is 6.
\[
6 \left( \frac{5x-4}{3} - \frac{4x-3}{2} - \frac{3x-2}{1} \right) = 6 \cdot 0
\]
\[
6 \cdot \frac{5x-4}{3} - 6 \cdot \frac{4x-3}{2} - 6 \cdot \frac{3x-2}{1} = 0
\]
\[
2(5x - 4) - 3(4x - 3) - 6(3x - 2) = 0
\]
\[
10x - 8 - 12x + 9 - 18x + 12 = 0
\]
\[
10x - 12x - 18x - 8 + 9 + 12 = 0
\]
\[
-20x + 13 = 0
\]

2. Solve for \(x\):
\[
-20x = -13
\]
\[
x = \frac{13}{20}
\]

Answer:
\[
\boxed{\frac{13}{20}}
\]

---

#### 16. \(\frac{3(2x+7)}{5} - 3 = \frac{5x-2}{3}\)

1. Eliminate the fractions by finding the LCD, which is 15.
\[
15 \left( \frac{3(2x+7)}{5} - 3 \right) = 15 \left( \frac{5x-2}{3} \right)
\]
\[
15 \cdot \frac{3(2x+7)}{5} - 15 \cdot 3 = 15 \cdot \frac{5x-2}{3}
\]
\[
3 \cdot 3(2x+7) - 45 = 5(5x-2)
\]
\[
9(2x+7) - 45 = 25x - 10
\]
\[
18x + 63 - 45 = 25x - 10
\]
\[
18x + 18 = 25x - 10
\]

2. Move all terms involving \(x\) to one side and constants to the other:
\[
18x - 25x = -10 - 18
\]
\[
-7x = -28
\]

3. Solve for \(x\):
\[
x = \frac{-28}{-7} = 4
\]

Answer:
\[
\boxed{4}
\]

---

#### 17. \(\frac{2x}{a} + \frac{b-a}{a} = \frac{2b}{a} \left( \frac{2x}{a} - \frac{b}{a} \right)\)

1. Simplify the left-hand side:
\[
\frac{2x}{a} + \frac{b-a}{a} = \frac{2x + b - a}{a}
\]

2. Simplify the right-hand side:
\[
\frac{2b}{a} \left( \frac{2x}{a} - \frac{b}{a} \right) = \frac{2b}{a} \cdot \frac{2x - b}{a} = \frac{2b(2x - b)}{a^2} = \frac{4bx - 2b^2}{a^2}
\]

3. Equate the two sides:
\[
\frac{2x + b - a}{a} = \frac{4bx - 2b^2}{a^2}
\]

4. Eliminate the denominators by multiplying through by \(a^2\):
\[
a(2x + b - a) = 4bx - 2b^2
\]
\[
2ax + ab - a^2 = 4bx - 2b^2
\]

5. Move all terms involving \(x\) to one side and constants to the other:
\[
2ax - 4bx = -2b^2 - ab + a^2
\]
\[
x(2a - 4b) = a^2 - ab - 2b^2
\]

6. Solve for \(x\):
\[
x = \frac{a^2 - ab - 2b^2}{2a - 4b}
\]
\[
x = \frac{a^2 - ab - 2b^2}{2(a - 2b)}
\]

Answer:
\[
\boxed{\frac{a^2 - ab - 2b^2}{2(a - 2b)}}
\]

---

#### 18. \(\frac{5}{6y} + \frac{2}{3y} = 1 + \frac{1}{2y}\)

1. Find the LCD, which is \(6y\).
\[
6y \left( \frac{5}{6y} + \frac{2}{3y} \right) = 6y \left( 1 + \frac{1}{2y} \right)
\]
\[
6y \cdot \frac{5}{6y} + 6y \cdot \frac{2}{3y} = 6y \cdot 1 + 6y \cdot \frac{1}{2y}
\]
\[
5 + 2 \cdot 2 = 6y + 3
\]
\[
5 + 4 = 6y + 3
\]
\[
9 = 6y + 3
\]

2. Move the constant to the other side:
\[
9 - 3 = 6y
\]
\[
6 = 6y
\]

3. Solve for \(y\):
\[
y = \frac{6}{6} = 1
\]

Answer:
\[
\boxed{1}
\]

---

Final Answers:

\[
\boxed{12}, \boxed{12}, \boxed{-\frac{3}{7}}, \boxed{12}, \boxed{2}, \boxed{5}, \boxed{4}, \boxed{8}, \boxed{-\frac{3}{2}}, \boxed{8}, \boxed{1}, \boxed{-1}, \boxed{4}, \boxed{\frac{19}{7}}, \boxed{\frac{13}{20}}, \boxed{4}, \boxed{\frac{a^2 - ab - 2b^2}{2(a - 2b)}}, \boxed{1}
\]