Question

1. select all of the equations that have ( x = 7 ) as a solution

( square , x cdot 3 = 28 )
( square , 3 + x = 12 )
( \boxtimes , 8 cdot x = 56 )
( \boxtimes , x + 4 = 11 )

Explanation:

Step1: Test $x=7$ in first equation

Substitute $x=7$ into $x \cdot 3 = 28$: $7 \cdot 3 = 21
eq 28$

Step2: Test $x=7$ in second equation

Substitute $x=7$ into $8 \cdot x = 56$: $8 \cdot 7 = 56$

Step3: Test $x=7$ in third equation

Substitute $x=7$ into $3 + x = 12$: $3 + 7 = 10
eq 12$

Step4: Test $x=7$ in fourth equation

Substitute $x=7$ into $x + 4 = 11$: $7 + 4 = 11$

Answer:

$\square 8 \cdot x = 56$
$\square x + 4 = 11$