An equation is a statement that two things are equal. To solve it is to discover the one value that makes the statement true — and the whole craft rests on a single idea: balance.
Equation versus expression
An expression like \( 3x + 2 \) is just a recipe — a set of instructions for combining numbers and a variable. It has no equals sign, so there is nothing to solve; the best you can do is simplify it or plug in a value. An equation, on the other hand, makes a claim. When we write \( 3x + 2 = 11 \), we are asserting that the two sides are equal, and asking a real question: what value of \( x \) makes that true?
That equals sign changes everything. An expression describes; an equation demands an answer. The rest of this lesson is about finding it.
The balance idea
Picture the equals sign as the pivot of a balance scale. The two sides of the equation weigh exactly the same — that is what "equal" means. As long as you do the same thing to both sides, the scale stays perfectly level and the equation stays true. Add the same weight to each pan, or remove it from each pan, and balance is preserved.
This single rule is the engine behind every solution. Work the buttons below to undo each operation and watch \( x \) come free; this scale solves \( 2x + 3 = 11 \).
Inverse operations
To free \( x \), you undo whatever was done to it, using the opposite operation. Subtraction undoes addition. Division undoes multiplication. Each operation has a partner that reverses it, and reaching for that partner is how you peel the equation apart.
There is also an order to the peeling. You undo in the reverse order to how the expression was built: deal with the \( + \) or \( - \) first, then the \( \times \) or \( \div \). It is like taking off your shoes before your socks — the last thing added is the first thing removed.
Solving \( 2x + 3 = 11 \)
Let's solve the equation from the scale. The variable has been multiplied by \( 2 \) and then had \( 3 \) added, so we undo the \( +3 \) first. Subtract \( 3 \) from both sides:
\[ 2x = 8 \]Now \( x \) is only multiplied by \( 2 \). Undo that by dividing both sides by \( 2 \):
\[ x = 4. \]Check by substituting back: \( 2 \times 4 + 3 = 11 \) ✓. The statement holds, so \( x = 4 \) is the value that makes it true.
Always check your answer
Once you have a solution, put it back into the original equation and see whether both sides come out equal. If they do, you're right. Checking costs only seconds, yet it catches almost every slip — a dropped sign, an arithmetic stumble, a step undone in the wrong order. Make it a habit, and your answers will rarely betray you.
One with a fraction
The same two ideas — balance and inverse operations — handle fractions just as easily. Solve \( \dfrac{x}{3} - 1 = 4 \). Here \( x \) was divided by \( 3 \) and then had \( 1 \) subtracted, so undo the \( -1 \) first by adding \( 1 \) to both sides:
\[ \frac{x}{3} = 5 \]Now \( x \) is only divided by \( 3 \). Undo that by multiplying both sides by \( 3 \):
\[ x = 15. \]Check: \( 15 \div 3 - 1 = 4 \) ✓.
The golden rule Whatever you do to one side of an equation, you must do to the other. That keeps the balance, and keeps the statement true. Every legitimate step in solving an equation is just this rule applied again and again.
- The \( -2 \) was added last, so undo it first. Add \( 2 \) to both sides: \( 5x = 15 \).
- Now \( x \) is only multiplied by \( 5 \). Divide both sides by \( 5 \): \( x = \mathbf{3} \).
Check: \( 5 \times 3 - 2 = 13 \) ✓.
- Undo the \( +2 \) first. Subtract \( 2 \) from both sides: \( \dfrac{x}{4} = 4 \).
- Now \( x \) is only divided by \( 4 \). Multiply both sides by \( 4 \): \( x = \mathbf{16} \).
Check: \( 16 \div 4 + 2 = 6 \) ✓.
Tip Solving an equation undoes the building of an expression, so work in the reverse order of operations: undo \( + \) and \( - \) first, then \( \times \) and \( \div \). The last thing done to the variable is the first thing you take away.
Practice
Try each one yourself, then reveal the full solution.
1. Solve \( x + 7 = 12 \).
Only one operation has touched \( x \): adding \( 7 \). Undo it by subtracting \( 7 \) from both sides.
So \( x = 12 - 7 = \mathbf{5} \).
Check: \( 5 + 7 = 12 \) ✓.
2. Solve \( 3x = 21 \).
Here \( x \) is multiplied by \( 3 \). Undo multiplication with division: divide both sides by \( 3 \).
So \( x = 21 \div 3 = \mathbf{7} \).
Check: \( 3 \times 7 = 21 \) ✓.
3. Solve \( 2x - 5 = 9 \).
The \( -5 \) was added last, so undo it first. Add \( 5 \) to both sides: \( 2x = 14 \).
Now \( x \) is only multiplied by \( 2 \). Divide both sides by \( 2 \): \( x = \mathbf{7} \).
Check: \( 2 \times 7 - 5 = 9 \) ✓.