11. Solve: [tex]2x - 3 = 7[/tex]

12. Solve: [tex]\frac{x}{3} + \frac{5}{2} = \frac{-3}{2}[/tex]

13. Solve: [tex]\frac{x + 6}{4} + \frac{x - 3}{5} = \frac{5x - 4}{8}[/tex]

14. Solve: [tex]\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}[/tex]

15. Find [tex]x[/tex] if [tex]2x + 5 = x + 25[/tex]



Answer :

Of course! Let's break down and solve each equation step-by-step:

### Problem 11:
Solve [tex]\(2x - 3 = 7\)[/tex].

1. Add 3 to both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 2x - 3 + 3 = 7 + 3 \][/tex]
[tex]\[ 2x = 10 \][/tex]

2. Divide both sides by 2:
[tex]\[ x = \frac{10}{2} \][/tex]
[tex]\[ x = 5 \][/tex]

### Problem 12:
Solve [tex]\(\frac{x}{3} + \frac{5}{2} = \frac{-3}{2}\)[/tex].

1. Multiply every term by the least common multiple of the denominators (which is 6) to clear the fractions:
[tex]\[ 6 \left(\frac{x}{3}\right) + 6 \left(\frac{5}{2}\right) = 6 \left(\frac{-3}{2}\right) \][/tex]
[tex]\[ 2x + 15 = -9 \][/tex]

2. Subtract 15 from both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 2x + 15 - 15 = -9 - 15 \][/tex]
[tex]\[ 2x = -24 \][/tex]

3. Divide both sides by 2:
[tex]\[ x = \frac{-24}{2} \][/tex]
[tex]\[ x = -12 \][/tex]

### Problem 13:
Solve [tex]\(\frac{x + 6}{4} + \frac{x - 3}{5} = \frac{5x - 4}{8}\)[/tex].

1. Multiply every term by the least common multiple of the denominators (which is 40) to clear the fractions:
[tex]\[ 40 \left(\frac{x + 6}{4}\right) + 40 \left(\frac{x - 3}{5}\right) = 40 \left(\frac{5x - 4}{8}\right) \][/tex]
[tex]\[ 10(x + 6) + 8(x - 3) = 5(5x - 4) \][/tex]

2. Simplify and combine like terms:
[tex]\[ 10x + 60 + 8x - 24 = 25x - 20 \][/tex]
[tex]\[ 18x + 36 = 25x - 20 \][/tex]

3. Subtract 18x from both sides:
[tex]\[ 18x + 36 - 18x = 25x - 20 - 18x \][/tex]
[tex]\[ 36 = 7x - 20 \][/tex]

4. Add 20 to both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 36 + 20 = 7x - 20 + 20 \][/tex]
[tex]\[ 56 = 7x \][/tex]

5. Divide both sides by 7:
[tex]\[ x = \frac{56}{7} \][/tex]
[tex]\[ x = 8 \][/tex]

### Problem 14:
Solve [tex]\(\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}\)[/tex].

1. Multiply every term by the least common multiple of the denominators (which is 6) to clear the fractions:
[tex]\[ 6 \left(\frac{6x + 1}{3}\right) + 6 \cdot 1 = 6 \left(\frac{x - 3}{6}\right) \][/tex]
[tex]\[ 2(6x + 1) + 6 = x - 3 \][/tex]
[tex]\[ 12x + 2 + 6 = x - 3 \][/tex]

2. Simplify and combine like terms:
[tex]\[ 12x + 8 = x - 3 \][/tex]

3. Subtract [tex]\(x\)[/tex] from both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 12x + 8 - x = x - 3 - x \][/tex]
[tex]\[ 11x + 8 = -3 \][/tex]

4. Subtract 8 from both sides:
[tex]\[ 11x + 8 - 8 = -3 - 8 \][/tex]
[tex]\[ 11x = -11 \][/tex]

5. Divide both sides by 11:
[tex]\[ x = \frac{-11}{11} \][/tex]
[tex]\[ x = -1 \][/tex]

### Problem 15:
Solve [tex]\(2x + 5 = x + 25\)[/tex].

1. Subtract [tex]\(x\)[/tex] from both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 2x + 5 - x = x + 25 - x \][/tex]
[tex]\[ x + 5 = 25 \][/tex]

2. Subtract 5 from both sides:
[tex]\[ x + 5 - 5 = 25 - 5 \][/tex]
[tex]\[ x = 20 \][/tex]

So the solutions are:
1. [tex]\(x = 5\)[/tex]
2. [tex]\(x = -12\)[/tex]
3. [tex]\(x = 8\)[/tex]
4. [tex]\(x = -1\)[/tex]
5. [tex]\(x = 20\)[/tex]