Question

Select all equations that have
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 solve each equation for \( x \), we will analyze each option.

**A.** \( x + 7 = 10 \)

To solve for \( x \):
\[
x + 7 = 10 \\
x = 10 - 7 \\
x = 3
\]

**B.** \( 3 + x = 3 \)

To solve for \( x \):
\[
3 + x = 3 \\
x = 3 - 3 \\
x = 0
\]

**C.** \( x \cdot 3 = 1 \)

To solve for \( x \):
\[
x \cdot 3 = 1 \\
x = \frac{1}{3}
\]

**D.** \( 4 \cdot x = 12 \)

To solve for \( x \):
\[
4 \cdot x = 12 \\
x = \frac{12}{4} \\
x = 3
\]

Now, we summarize the solutions:
- **A:** \( x = 3 \)
- **B:** \( x = 0 \)
- **C:** \( x = \frac{1}{3} \)
- **D:** \( x = 3 \)

The equations that have \( x \) as a solution are **A** and **D** (both yield \( x = 3 \)). The other equations have different solutions: **B** gives \( x = 0 \) and **C** gives \( x = \frac{1}{3} \).

Thus, the answers are:
- A and D are solutions for \( x = 3 \).