I see that you've uploaded a set of linear equations from a worksheet. Let's solve each of these one by one.

1. $4m - 5 = 11$

Solution: $4m - 5 = 11 \quad \text{Add 5 to both sides:} \quad 4m = 16 \quad \text{Divide by 4:} \quad m = 4$

2. $-3d + 10 = 43$

Solution: $-3d + 10 = 43 \quad \text{Subtract 10 from both sides:} \quad -3d = 33 \quad \text{Divide by -3:} \quad d = -11$

3. $\frac{2(r - 3)}{4} - 8 = 50$

Solution: $\frac{2(r - 3)}{4} - 8 = 50 \quad \text{Add 8 to both sides:} \quad \frac{2(r - 3)}{4} = 58$ $\text{Multiply both sides by 4:} \quad 2(r - 3) = 232 \quad \text{Divide by 2:} \quad r - 3 = 116$ $\text{Add 3 to both sides:} \quad r = 119$

4. $5h - 13 = 12$

Solution: $5h - 13 = 12 \quad \text{Add 13 to both sides:} \quad 5h = 25 \quad \text{Divide by 5:} \quad h = 5$

5. $-8 = 3y - 2$

Solution: $-8 = 3y - 2 \quad \text{Add 2 to both sides:} \quad -6 = 3y \quad \text{Divide by 3:} \quad y = -2$

I'll continue solving the remaining problems in the next response.

Would you like to proceed with the next set of solutions?

Here are 5 related questions to practice further:

1. How would you solve an equation with a variable on both sides?
2. Can you think of an example where multiplying both sides by a negative number would reverse the inequality?
3. What are the steps to isolate the variable in an equation involving fractions?
4. How do you solve systems of linear equations with substitution?
5. What happens if a linear equation has no solution?

Tip: Always check your solution by substituting it back into the original equation.