The balance method

Imagine a weighing scale that is perfectly balanced. If you add $2$ kg to one pan and $2$ kg to the other, the scale stays balanced. Take $5$ kg off each , still balanced. Double both pans , still balanced. This intuitive idea is the heart of solving equations.

Idea

The balance principle: an equation stays true if you perform the same operation on both sides. There are four allowed operations.

Goal. Use these operations to isolate the unknown letter on one side, with a number on the other.

Worked-through example: solve $5x - 4 = 7$.

We want $x$ alone. Two things stand in the way: the $-4$ and the $5$. Remove them, in that order.

Step 1. Get rid of the $-4$ by adding $4$ to both sides: $5x - 4 + 4 = 7 + 4 \ \Rightarrow\ 5x = 11.$

Step 2. Get rid of the $5$ multiplying $x$ by dividing both sides by $5$: $\frac{5x}{5} = \frac{11}{5} \ \Rightarrow\ x = \frac{11}{5}.$

Check. Substitute $x = \tfrac{11}{5}$ back into the original: LHS $= 5\left(\tfrac{11}{5}\right) - 4 = 11 - 4 = 7$. ✓ Matches RHS.

Order of operations. When undoing an equation, work in the reverse order of how it was built. If the equation was built by "multiply $x$ by $5$, then subtract $4$", we undo by "add $4$, then divide by $5$". This is the same idea as un-wrapping a parcel: outer layer first.

Multiply or divide first? Multiply or divide both sides only by non-zero numbers. Dividing by zero is forbidden , the equation collapses.

Multiple terms with the unknown. If both sides contain the variable, like $6y + 7 = 4y + 21$, first collect the variable on one side by subtracting the smaller variable term from both sides: $6y - 4y + 7 = 21 \ \Rightarrow\ 2y + 7 = 21,$ then proceed as usual.

Worked examples

Example 1. Solve $3x - 10 = 35$.

Add $10$ to both sides: $3x = 45$. Divide by $3$: $x = 15$. Check: $3(15) - 10 = 35$. ✓

Example 2. Solve $11y - 5 = 61$.

Add $5$ to both sides: $11y = 66$. Divide by $11$: $y = 6$. Check: $11(6) - 5 = 66 - 5 = 61$. ✓

Example 3. Solve $\dfrac{u}{15} = 6$.

Multiply both sides by $15$: $u = 90$. Check: $\tfrac{90}{15} = 6$. ✓

Example 4. Solve $6y + 7 = 4y + 21$.

Subtract $4y$ from both sides: $2y + 7 = 21$. Subtract $7$: $2y = 14$. Divide by $2$: $y = 7$. Check: LHS $= 6(7) + 7 = 49$; RHS $= 4(7) + 21 = 49$. ✓

Try it yourself

1. Solve $x + 9 = 22$.
2. Solve $3x = 45$.
3. Solve $2y - 5 = 19$.
4. Solve $7p + 4 = 39$.
5. Solve $\dfrac{m}{8} = 5$.
6. Solve $5s = 3s + 12$.
7. Solve $4(a + 2) = 28$. (Distribute first, then balance.)
8. Solve $\dfrac{u}{15} = 4$.

Activity

Take a clothes-hanger and hang two paper cups from the ends so it balances. Drop two identical coins in the left cup and three coins (each labelled "$x$") in the right. Add coins to the left until it balances. The number of coins added equals $3x - 2$, balanced against $0$ on the right , wait, the cups must balance. Set up the equation and solve!