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

1 answer

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 \( x = 3 \).