Question

Select all equations that have 3
as a solution.
A
x+7=10x+7=10x+7=10
B
3+x=33+x=33+x=3
C
x⋅3=1x\cdot 3=1x⋅3=1
D
4⋅x=124\cdot x=124⋅x=12

Answer 1

To determine which equations have 3 as a solution, let's solve each equation for x.

**Option A:**
\[ x + 7 = 10 \]
Subtract 7 from both sides:
\[ x = 10 - 7 \]
\[ x = 3 \]
So, option A has 3 as a solution.

**Option B:**
\[ 3 + x = 3 \]
Subtract 3 from both sides:
\[ x = 3 - 3 \]
\[ x = 0 \]
So, option B does not have 3 as a solution.

**Option C:**
\[ x \cdot 3 = 1 \]
Divide both sides by 3:
\[ x = \frac{1}{3} \]
So, option C does not have 3 as a solution.

**Option D:**
\[ 4 \cdot x = 12 \]
Divide both sides by 4:
\[ x = \frac{12}{4} \]
\[ x = 3 \]
So, option D has 3 as a solution.

Thus, the equations that have 3 as a solution are:

**A** and **D**.