Question

题1：\frac{1}{2}(x + 24) = 21\
题2：\frac{1}{3}(x + 6) = \frac{1}{2}(x - 3)

Explanation:

Step1: Solve first equation: eliminate fraction

Multiply both sides by 2:
$$\frac{1}{2}(x+24) \times 2 = 21 \times 2$$
$$x + 24 = 42$$

Step2: Isolate x for first equation

Subtract 24 from both sides:
$$x = 42 - 24$$

Step3: Solve second equation: eliminate fractions

Multiply both sides by 6 (LCM of 3 and 2):
$$\frac{1}{3}(x+6) \times 6 = \frac{1}{2}(x-3) \times 6$$
$$2(x+6) = 3(x-3)$$

Step4: Expand both sides

$$2x + 12 = 3x - 9$$

Step5: Isolate x for second equation

Subtract $2x$ and add 9 to both sides:
$$12 + 9 = 3x - 2x$$
$$x = 21$$

Answer:

For Bonus #1: $x = 18$
For Bonus #2: $x = 21$