55–70 ■ solving equations find all real solutions of the equation. 55. $7x - 6 = 4x + 9$ 56. $8 - 2x = 14 + x$ 57. $\frac{x + 1}{x - 1} = \frac{3x}{3x - 6}$ 58. $(x + 2)^2 = (x - 4)^2$ 59. $x^2 - 9x + 14 = 0$ 60. $x^2 + 24x + 144 = 0$ 61. $2x^2 + x = 1$ 62. $3x^2 + 5x - 2 = 0$ 63. $4x^3 - 25x = 0$ 64. $x^3 - 2x^2 - 5x + 10 = 0$ 65. $3x^2 + 4x - 1 = 0$ 66. $\frac{1}{x} + \frac{2}{x - 1} = 3$ 67. $\frac{x}{x - 2} + \frac{1}{x + 2} = \frac{8}{x^2 - 4}$ 68. $x^4 - 8x^2 - 9 = 0$ 69. $|x - 7| = 4$ 70. $|2x - 5| = 9$

Question

Accepted Answer

Let's solve these equations one by one. We'll start with problem 55:

### Problem 55: $ 7x - 6 = 4x + 9 $
# Explanation:
## Step 1: Subtract $ 4x $ from both sides
$ 7x - 4x - 6 = 4x - 4x + 9 $
$ 3x - 6 = 9 $

## Step 2: Add 6 to both sides
$ 3x - 6 + 6 = 9 + 6 $
$ 3x = 15 $

## Step 3: Divide both sides by 3
$ \frac{3x}{3} = \frac{15}{3} $
$ x = 5 $

# Answer: $ x = 5 $

### Problem 56: $ 8 - 2x = 14 + x $
# Explanation:
## Step 1: Add $ 2x $ to both sides
$ 8 - 2x + 2x = 14 + x + 2x $
$ 8 = 14 + 3x $

## Step 2: Subtract 14 from both sides
$ 8 - 14 = 14 - 14 + 3x $
$ -6 = 3x $

## Step 3: Divide both sides by 3
$ \frac{-6}{3} = \frac{3x}{3} $
$ x = -2 $

# Answer: $ x = -2 $

### Problem 57: $ \frac{x + 1}{x - 1} = \frac{3x}{3x - 6} $
First, simplify the right-hand side denominator: $ 3x - 6 = 3(x - 2) $, so the equation becomes $ \frac{x + 1}{x - 1} = \frac{3x}{3(x - 2)} = \frac{x}{x - 2} $ (assuming $ x
eq 2 $)

# Explanation:
## Step 1: Cross-multiply
$ (x + 1)(x - 2) = x(x - 1) $

## Step 2: Expand both sides
$ x^2 - 2x + x - 2 = x^2 - x $
$ x^2 - x - 2 = x^2 - x $

## Step 3: Subtract $ x^2 - x $ from both sides
$ x^2 - x - 2 - (x^2 - x) = x^2 - x - (x^2 - x) $
$ -2 = 0 $

This is a contradiction, so there is no solution. However, we must also check the domain: $ x
eq 1 $ and $ x
eq 2 $ (since denominators cannot be zero). Since the equation leads to a contradiction, there are no real solutions.

# Answer: No solution

### Problem 58: $ (x + 2)^2 = (x - 4)^2 $
# Explanation:
## Step 1: Expand both sides
$ x^2 + 4x + 4 = x^2 - 8x + 16 $

## Step 2: Subtract $ x^2 $ from both sides
$ 4x + 4 = -8x + 16 $

## Step 3: Add $ 8x $ to both sides
$ 12x + 4 = 16 $

## Step 4: Subtract 4 from both sides
$ 12x = 12 $

## Step 5: Divide both sides by 12
$ x = 1 $

# Answer: $ x = 1 $

### Problem 59: $ x^2 - 9x + 14 = 0 $
# Explanation:
## Step 1: Factor the quadratic
We need two numbers that multiply to 14 and add to -9. Those numbers are -2 and -7.
$ (x - 2)(x - 7) = 0 $

## Step 2: Set each factor equal to zero
$ x - 2 = 0 $ or $ x - 7 = 0 $
$ x = 2 $ or $ x = 7 $

# Answer: $ x = 2 $ or $ x = 7 $

### Problem 60: $ x^2 + 24x + 144 = 0 $
# Explanation:
## Step 1: Factor the quadratic
This is a perfect square trinomial: $ (x + 12)^2 = 0 $

## Step 2: Solve for $ x $
$ x + 12 = 0 $
$ x = -12 $ (double root)

# Answer: $ x = -12 $ (double root)

### Problem 61: $ 2x^2 + x = 1 $
# Explanation:
## Step 1: Rewrite in standard form
$ 2x^2 + x - 1 = 0 $

## Step 2: Factor the quadratic
We need two numbers that multiply to $ 2 \times (-1) = -2 $ and add to 1. Those numbers are 2 and -1.
$ 2x^2 + 2x - x - 1 = 0 $
$ 2x(x + 1) - 1(x + 1) = 0 $
$ (2x - 1)(x + 1) = 0 $

## Step 3: Set each factor equal to zero
$ 2x - 1 = 0 $ or $ x + 1 = 0 $
$ x = \frac{1}{2} $ or $ x = -1 $

# Answer: $ x = \frac{1}{2} $ or $ x = -1 $

### Problem 62: $ 3x^2 + 5x - 2 = 0 $
# Explanation:
## Step 1: Factor the quadratic
We need two numbers that multiply to $ 3 \times (-2) = -6 $ and add to 5. Those numbers are 6 and -1.
$ 3x^2 + 6x - x - 2 = 0 $
$ 3x(x + 2) - 1(x + 2) = 0 $
$ (3x - 1)(x + 2) = 0 $

## Step 2: Set each factor equal to zero
$ 3x - 1 = 0 $ or $ x + 2 = 0 $
$ x = \frac{1}{3} $ or $ x = -2 $

# Answer: $ x = \frac{1}{3} $ or $ x = -2 $

### Problem 63: $ 4x^3 - 25x = 0 $
# Explanation:
## Step 1: Factor out $ x $
$ x(4x^2 - 25) = 0 $

## Step 2: Recognize $ 4x^2 - 25 $ as a difference of squares
$ x(2x - 5)(2x + 5) = 0 $

## Step 3: Set each factor equal to zero
$ x = 0 $ or $ 2x - 5 = 0 $ or $ 2x + 5 = 0 $
$ x = 0 $ or $ x = \frac{5}{2} $ or $ x = -\frac{5}{2} $

# Answer: $ x = 0 $, $ x = \frac{5}{2} $, $ x = -\frac{5}{2} $

### Problem 64: $ x^3 - 2x^2 - 5x + 10 = 0 $
# Explanation:
## Step 1: Factor by grouping
$ (x^3 - 2x^2) + (-5x + 10) = 0 $
$ x^2(x - 2) - 5(x - 2) = 0 $
$ (x^2 - 5)(x - 2) = 0 $

## Step 2: Set each factor equal to zero
$ x^2 - 5 = 0 $ or $ x - 2 = 0 $
$ x = \pm \sqrt{5} $ or $ x = 2 $

# Answer: $ x = 2 $, $ x = \sqrt{5} $, $ x = -\sqrt{5} $

### Problem 65: $ 3x^2 + 4x - 1 = 0 $
We'll use the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 3 $, $ b = 4 $, $ c = -1 $

# Explanation:
## Step 1: Calculate the discriminant
$ D = b^2 - 4ac = 4^2 - 4(3)(-1) = 16 + 12 = 28 $

## Step 2: Apply the quadratic formula
$ x = \frac{-4 \pm \sqrt{28}}{2(3)} = \frac{-4 \pm 2\sqrt{7}}{6} = \frac{-2 \pm \sqrt{7}}{3} $

# Answer: $ x = \frac{-2 + \sqrt{7}}{3} $ or $ x = \frac{-2 - \sqrt{7}}{3} $

### Problem 66: $ \frac{1}{x} + \frac{2}{x - 1} = 3 $
# Explanation:
## Step 1: Multiply both sides by $ x(x - 1) $ (domain: $ x
eq 0 $, $ x
eq 1 $)
$ (x - 1) + 2x = 3x(x - 1) $

## Step 2: Simplify left-hand side
$ 3x - 1 = 3x^2 - 3x $

## Step 3: Rewrite in standard form
$ 3x^2 - 6x + 1 = 0 $

## Step 4: Use quadratic formula
$ a = 3 $, $ b = -6 $, $ c = 1 $
$ D = (-6)^2 - 4(3)(1) = 36 - 12 = 24 $
$ x = \frac{6 \pm \sqrt{24}}{2(3)} = \frac{6 \pm 2\sqrt{6}}{6} = \frac{3 \pm \sqrt{6}}{3} $

## Step 5: Check domain
Both solutions are not 0 or 1, so they are valid.

# Answer: $ x = \frac{3 + \sqrt{6}}{3} $ or $ x = \frac{3 - \sqrt{6}}{3} $

### Problem 67: $ \frac{x}{x - 2} + \frac{1}{x + 2} = \frac{8}{x^2 - 4} $
Note that $ x^2 - 4 = (x - 2)(x + 2) $, so the domain is $ x
eq 2 $, $ x
eq -2 $

# Explanation:
## Step 1: Multiply both sides by $ (x - 2)(x + 2) $
$ x(x + 2) + 1(x - 2) = 8 $

## Step 2: Expand and simplify
$ x^2 + 2x + x - 2 = 8 $
$ x^2 + 3x - 10 = 0 $

## Step 3: Factor the quadratic
$ (x + 5)(x - 2) = 0 $

## Step 4: Solve for $ x $
$ x = -5 $ or $ x = 2 $

## Step 5: Check domain
$ x = 2 $ is not in the domain, so we discard it. Thus, $ x = -5 $ is the solution.

# Answer: $ x = -5 $

### Problem 68: $ x^4 - 8x^2 - 9 = 0 $
Let $ y = x^2 $, so the equation becomes $ y^2 - 8y - 9 = 0 $

# Explanation:
## Step 1: Factor the quadratic in $ y $
$ (y - 9)(y + 1) = 0 $

## Step 2: Solve for $ y $
$ y = 9 $ or $ y = -1 $

## Step 3: Substitute back $ y = x^2 $
For $ y = 9 $: $ x^2 = 9 $ ⇒ $ x = \pm 3 $
For $ y = -1 $: $ x^2 = -1 $ ⇒ no real solutions

# Answer: $ x = 3 $ or $ x = -3 $

### Problem 69: $ |x - 7| = 4 $
# Explanation:
## Step 1: Split into two cases
Case 1: $ x - 7 = 4 $ ⇒ $ x = 11 $
Case 2: $ x - 7 = -4 $ ⇒ $ x = 3 $

# Answer: $ x = 11 $ or $ x = 3 $

### Problem 70: $ |2x - 5| = 9 $
# Explanation:
## Step 1: Split into two cases
Case 1: $ 2x - 5 = 9 $ ⇒ $ 2x = 14 $ ⇒ $ x = 7 $
Case 2: $ 2x - 5 = -9 $ ⇒ $ 2x = -4 $ ⇒ $ x = -2 $

# Answer: $ x = 7 $ or $ x = -2 $

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QUESTION IMAGE

Question

Explanation:

Problem 55: \( 7x - 6 = 4x + 9 \)

Step 1: Subtract \( 4x \) from both sides

Step 2: Add 6 to both sides

Step 3: Divide both sides by 3

Step 1: Add \( 2x \) to both sides

Step 2: Subtract 14 from both sides

Step 3: Divide both sides by 3

Step 1: Cross-multiply

Step 2: Expand both sides

Step 3: Subtract \( x^2 - x \) from both sides

Answer:

Problem 56: \( 8 - 2x = 14 + x \)