1. Three values are listed below each equatior Circle the value that is a solution to the equation shown. a. $$ 2 m+4=-10 $$ $$ m=4 \quad m=3 $$ b. $$ \frac{1}{2} n-1=23 $$ $$ n=48 \quad n=12 $$

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**(a) Equation:**  
$$
2m + 4 = -10
$$

1. Subtract $$4$$ from both sides:  
   $$
   2m + 4 - 4 = -10 - 4 \quad \Rightarrow \quad 2m = -14
   $$
2. Divide by $$2$$:  
   $$
   m = \frac{-14}{2} = -7
   $$

The provided choices are $$ m = 4 $$ and $$ m = 3 $$. Since neither equals $$-7$$, neither value is a solution.

---

**(b) Equation:**  
$$
\frac{1}{2}n - 1 = 23
$$

1. Add $$1$$ to both sides:  
   $$
   \frac{1}{2}n - 1 + 1 = 23 + 1 \quad \Rightarrow \quad \frac{1}{2}n = 24
   $$
2. Multiply by $$2$$:  
   $$
   n = 24 \times 2 = 48
   $$

The provided choices are $$ n = 48 $$ and $$ n = 12 $$. Since $$ n = 48 $$ is obtained, it is the correct solution.
**(a) Equation:** $$ 2m + 4 = -10 $$ Solutions: $$ m = 4 $$ and $$ m = 3 $$ are not solutions. **(b) Equation:** $$ \frac{1}{2}n - 1 = 23 $$ Solution: $$ n = 48 $$

Three values are listed below each equatior
Circle the value that is a solution to the
equation shown.
a.

b.

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Three values are listed below each equatior Circle the value that is a solution to the equation shown. a. b.