Let's solve the equations one by one. I'll guide you through each.

a) $10\sqrt{x} - 4 = 7\sqrt{x} + 6$

1. Start by isolating the square root term: $10\sqrt{x} - 7\sqrt{x} = 6 + 4$ Simplify: $3\sqrt{x} = 10$

2. Divide both sides by 3: $\sqrt{x} = \frac{10}{3}$

3. Square both sides to eliminate the square root: $x = \left( \frac{10}{3} \right)^2 = \frac{100}{9}$

So, $x = \frac{100}{9}$.

b) $6\sqrt{x} + 4 = 9\sqrt{x} + 3$

1. Isolate the square root term: $6\sqrt{x} - 9\sqrt{x} = 3 - 4$ Simplify: $-3\sqrt{x} = -1$

2. Divide by -3: $\sqrt{x} = \frac{1}{3}$

3. Square both sides: $x = \left( \frac{1}{3} \right)^2 = \frac{1}{9}$

So, $x = \frac{1}{9}$.

c) $7\sqrt{x} + 9 = 10\sqrt{x} + 10$

1. Isolate the square root term: $7\sqrt{x} - 10\sqrt{x} = 10 - 9$ Simplify: $-3\sqrt{x} = 1$

2. Divide by -3: $\sqrt{x} = -\frac{1}{3}$

3. Square both sides: $x = \left( -\frac{1}{3} \right)^2 = \frac{1}{9}$

So, $x = \frac{1}{9}$.

d) $\sqrt{2x - 1} - 3 = 0$

1. Move the -3 to the other side: $\sqrt{2x - 1} = 3$

2. Square both sides: $2x - 1 = 9$

3. Solve for $x$: $2x = 10 \quad \Rightarrow \quad x = 5$

So, $x = 5$.

e) $\sqrt{2x - 3} = 6$

1. Square both sides: $2x - 3 = 36$

2. Solve for $x$: $2x = 39 \quad \Rightarrow \quad x = \frac{39}{2} = 19.5$

So, $x = 19.5$.

f) $\sqrt{5x - 1} = \sqrt{x + 7}$

1. Square both sides: $5x - 1 = x + 7$

2. Isolate the $x$-terms: $5x - x = 7 + 1$ Simplify: $4x = 8 \quad \Rightarrow \quad x = 2$

So, $x = 2$.

g) $3 + \sqrt{5x + 6} = 12$

1. Move the 3 to the other side: $\sqrt{5x + 6} = 12 - 3$ Simplify: $\sqrt{5x + 6} = 9$

2. Square both sides: $5x + 6 = 81$

3. Solve for $x$: $5x = 75 \quad \Rightarrow \quad x = 15$

So, $x = 15$.

h) $\sqrt{5x^2 - 8} = 2x$

1. Square both sides: $5x^2 - 8 = 4x^2$

2. Move terms to one side: $5x^2 - 4x^2 = 8$ Simplify: $x^2 = 8 \quad \Rightarrow \quad x = \pm \sqrt{8} = \pm 2\sqrt{2}$

So, $x = \pm 2\sqrt{2}$.

i) $12 - \sqrt{x + 5} = 7$

1. Isolate the square root: $\sqrt{x + 5} = 12 - 7$ Simplify: $\sqrt{x + 5} = 5$

2. Square both sides: $x + 5 = 25$

3. Solve for $x$: $x = 25 - 5 = 20$

So, $x = 20$.

If you have any questions or need more details on any step, feel free to ask!

Related questions:

1. What are some common mistakes when solving square root equations?
2. How do you check if the solution is correct in radical equations?
3. How can you solve equations with nested square roots?
4. Why do square root equations sometimes lead to extraneous solutions?
5. How do you solve equations with fractional exponents instead of square roots?

Tip:

When solving radical equations, always check for extraneous solutions by substituting the result back into the original equation.