Answer:
- Step 1: identify the constant on the side of the equal sign that has the variable term. Subtract that constant from both sides of the equation.
- Step 2: identify the coefficient of the variable. Divide both sides of the equation by that coefficient.
Explanation:
Example:
2x + 3 = 7
Step 1. The variable term (containing x) is on the left. The constant term on the left side of the equal sign is +3. We add the opposite of that to both sides of the equation:
2x +3 -3 = 7 -3 . . . . . showing -3 being added (or 3 being subtracted)
2x = 4 . . . . . . . . . . . . result of carrying out the addition/subtraction
Step 2. The variable term (on the left) has 2 as its coefficient. We divide both sides of the equation by 2 (or, equivalently, multiply by the reciprocal of 2: 1/2).
2x/2 = 4/2 . . . . . . . . showing both sides of the equation being divided by 2
x = 2 . . . . . . . . . . . . . result of carrying out the division
The solution is x=2.
_____
The problems I expect to see presented to a 2nd grader will have a "blank" instead of a variable. In that case, one would first solve the addition/subtraction problem to identify the value of the term containing the "blank". That is, the above would look like ...
2×___ + 3 = 7
This is effectively equivalent to ...
_._._ + 3 = 7 . . . . . . . . we're using _._._ to stand for 2×___
Most 2nd graders can solve this because they are familiar with their addition facts. So, they would know ...
_._._ = 4 . . . . . . or . . . . 4 + 3 = 7 . . . . . (4 goes in the _._._ blank)
Now, the second step is to solve the multiplication equation ...
2×___ = 4
When I was in 2nd grade, we hadn't learned multiplication facts, so this would not be a problem we would be expected to solve. If your second grader knows multiplication facts, then they know ...
2 × 2 = 4
so 2 goes in the blank.