I can see your uploaded image clearly, and I'll solve problems 16, 17, and 18 from it. Let's get started!

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Problem 16:

Equation: 9x² = 4 + 7x

Solution:

1. Move all terms to one side to set the equation to 0: $9x^2 - 7x - 4 = 0$

2. Use the quadratic formula: $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ For $A = 9$, $B = -7$, and $C = -4$, the formula becomes: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(9)(-4)}}{2(9)}$

3. Simplify: $x = \frac{7 \pm \sqrt{49 + 144}}{18}$ $x = \frac{7 \pm \sqrt{193}}{18}$

Thus, the two solutions are: $x = \frac{7 + \sqrt{193}}{18} \quad \text{or} \quad x = \frac{7 - \sqrt{193}}{18}$

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Problem 17:

Equation: 5x + 9x = 4

Solution:

1. Combine like terms: $14x = 4$

2. Solve for $x$: $x = \frac{4}{14} = \frac{2}{7}$

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Problem 18:

Equation: 2x² - 36 = x

Solution:

1. Move all terms to one side: $2x^2 - x - 36 = 0$

2. Use the quadratic formula: $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ For $A = 2$, $B = -1$, and $C = -36$, the formula becomes: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-36)}}{2(2)}$

3. Simplify: $x = \frac{1 \pm \sqrt{1 + 288}}{4}$ $x = \frac{1 \pm \sqrt{289}}{4}$

4. Since $\sqrt{289} = 17$, we get: $x = \frac{1 + 17}{4} \quad \text{or} \quad x = \frac{1 - 17}{4}$

5. The two solutions are: $x = \frac{18}{4} = 4.5 \quad \text{or} \quad x = \frac{-16}{4} = -4$

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Final Answers:

• Problem 16: $x = \frac{7 + \sqrt{193}}{18}$ or $x = \frac{7 - \sqrt{193}}{18}$
• Problem 17: $x = \frac{2}{7}$
• Problem 18: $x = 4.5$ or $x = -4$

Would you like more detailed steps for any part of the solutions? 😊

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Here are 5 related questions for further practice:

1. Solve $3x^2 + 5x - 2 = 0$ using the quadratic formula.
2. Solve the linear equation $6x - 4 = 10$.
3. Solve $7x^2 - 2x + 1 = 0$.
4. Factor the quadratic expression $x^2 - 5x + 6$.
5. Solve $4x + 3 = 15 - 2x$.

Tip: Always simplify your equations first before applying formulas like the quadratic formula!