Equations of the form p(x + q) = r, from words

When you reach for it

Dorothea and three friends go to a film. Each of the four buys a ticket and a drink, and a drink costs 5 dollars. Together they pay 48 dollars. What does one ticket cost?

Call the price of one ticket x dollars. One person’s bundle, a ticket and a drink, costs x + 5 dollars. There are four bundles, all the same, so the four together cost 4 times that whole amount:

The brackets carry the meaning: the drink is in every bundle, so the 4 multiplies it as well as the ticket. Without brackets, 4x + 5 means four tickets and a single drink, because by the order of operations only the x is multiplied.

Reach for this whenever a situation is made of equal groups, each holding the unknown together with the same fixed amount, and the total is known: bundles bought by each person, teams that each gain the same number of players, the four sides of a square with a border, the sections of a fence. Every such equation has the shape

where p is the number of groups, q the fixed amount in each group, and r the total. The numbers may be negative or fractions; only p = 0 is ruled out. An equation with no brackets, such as 4x + 20 = 48, is a two-step equation, and one of the two ways of solving below turns this equation into exactly that.

The procedure

There are two honest ways to solve p(x + q) = r: undo the multiplication by p first, or open the brackets first. Both are always allowed, both give the same answer, and the skill is in seeing which one the numbers favor. Over two days, Part 1 writes the equation from words and solves it both ways with positive numbers; Part 2 brings in negatives and fractions and asks which way to choose.

Part 1: from words to an equation, and two ways to solve it

Writing the equation. Say what the letter stands for, with its unit: “let x be the price of one ticket, in dollars”. Then build one group, here a ticket and a drink, x + 5. Only then multiply by the number of groups. Building one group first is what puts the fixed amount inside the brackets.

Dividing first. The left side of 4(x + 5) = 48 is 4 times a single quantity, the bracket. Dividing both sides by the same non-zero number keeps the equation true for exactly the same values of x; that is the division property of equality. Divide by 4:

This middle equation has a meaning of its own: one bundle costs 12 dollars, as the figure shows. Subtract 5 from both sides: x = 7.

Distributing first. By the distributive property, 4(x + 5) = 4x + 20 for every x, so swapping one for the other does not change the solutions:

Four tickets cost 4x and four drinks cost 20. This is a two-step equation. Subtract 20 from both sides, then divide both sides by 4:

Why the two ways must agree. Every step on either path can be undone (multiply by 4 again, add 5 back, put the brackets back), so each new equation is true for exactly the same values of x as the one before. Both paths start from the same equation and end at “x = a number”, so the number is the same. With letters: dividing first gives \displaystyle x + q = \frac{r}{p}, so \displaystyle x = \frac{r}{p} - q; distributing first gives px = r - pq, so \displaystyle x = \frac{r - pq}{p}, which is the same number.

Checking. In the equation: 4(7 + 5) = 4 \cdot 12 = 48. Better, in the words: a 7 dollar ticket and a 5 dollar drink make a 12 dollar bundle, and four bundles cost 48 dollars. A check in the words also catches an equation that was written wrongly, which a check in the equation cannot.

Part 2: signed and fraction values, and choosing a way

Signs. Dividing by a negative p is allowed (only 0 is not), and the signs follow the rules for dividing integers. Distributing a negative p carries its sign into both terms. A bracket such as x - 2 is x + (-2), so q = -2: in words, each group has lost a fixed amount. A negative r changes nothing in the algebra, though in a story about counts or lengths it cannot happen. Example 4 works all three.

Fractions. Dividing by a fraction p is multiplying by its reciprocal, which undoes it in one step; distributing would spread the fraction onto q as well. A fraction q is often cleared by distributing, when p \cdot q comes out whole. Example 5 compares the two.

Choosing. Divide first when r \div p comes out neatly, or when p is a fraction: the bracket is then one step from finished, and the middle equation says what one group is worth. Distribute first when p \cdot q is neat but r \div p is not. Neither way is ever wrong; a poor choice only means more fractions along the way.

Worked examples

Examples 1 to 3 go with Part 1, and Examples 4 to 6 with Part 2.

Boundaries

p must not be 0. Dividing by 0 is undefined, and when p = 0 the left side is 0 for every x. Then 0(x + q) = r is false for every x if r \neq 0, so there is no solution, and true for every x if r = 0. The two cases are taken up in no solution, one solution, or every number.

Each move acts on the whole of a side. Dividing 4(x + 5) by 4 leaves the whole bracket, x + 5, and the whole right side is divided too. Distributing multiplies every term inside the bracket, q as well as x.

The shape needs equal groups. Every group must hold the unknown and the same fixed amount, and there must be p of them. A single amount outside the groups, like the shared snack in Example 3(b), gives a different equation; if the groups are not all alike, the situation is not p(x + q) = r at all.

The answer must fit what x counts. The algebra returns a number; the story decides whether it can be true. A count must be a whole number and a length must be positive. A frame with a 3 cm border and a perimeter of 20 cm gives 4(x + 6) = 20 and x = -1: no photo has that side, so those numbers cannot describe a real frame.

The unknown appears once, inside the brackets. If x also appears outside the brackets or on the other side, the equation needs collecting like terms or gathering the variable on one side first.

Where this goes next

Longer equations with brackets, such as 5(x - 1) + 2x = 30, are solved by distributing first and then collecting like terms on each side; dividing first no longer works, because the 5 does not multiply the whole left side. Brackets on both sides, such as 2(x + 1) = 5(x - 4), lead to equations with the variable on both sides.

The same shape with an inequality sign, p(x + q) > r, is solved with the same two moves, with one difference: dividing both sides by a negative p reverses the inequality, as in two-step inequalities. A proportion such as \displaystyle \frac{x - 2}{3} = \frac{4}{5} turns into this shape after cross multiplying, giving 5(x - 2) = 12. And solving p(x + q) = r with letters, which gave \displaystyle x = \frac{r}{p} - q in the procedure, is a first example of solving for a variable.

Exercises

Part 1: solve both ways

Solve each equation by dividing first, then again by distributing first.

1. 5(x + 3) = 45
2. 7(n - 2) = 56
3. 3(m + 5) = 20

Part 1: from words

Write an equation, solve it, and check the answer in the words.

4. Six friends each buy a ticket and a snack that costs 4 dollars. Together they pay 90 dollars. What does one ticket cost?
5. Florian adds a border 2 cm wide on every side of a square photo. The outer edge of the framed photo measures 72 cm all the way round. How long is a side of the photo?
6. Leocadia pays 52 dollars. Find the price of one ticket if (a) she buys 4 bundles, each a ticket and a snack that costs 2 dollars, and (b) she buys 4 tickets and one snack that costs 2 dollars.
7. Write a word problem about friends buying bundles that leads to 5(x + 6) = 85, then solve it.

Part 2: negative values

8. -4(x + 2) = 20
9. -6(y - 3) = -42
10. 5(x - 8) = -15
11. -(n + 9) = 4

Part 2: fractions and decimals

Choose the shorter way for each, and say why.

12. \displaystyle \frac{3}{4}(x + 8) = 18
13. \displaystyle \frac{1}{2}(y - 6) = -5
14. \displaystyle 3\left(x - \frac{2}{3}\right) = 7
15. \displaystyle -\frac{2}{5}(x - 10) = 6
16. 0.5(x + 1.2) = 2.4

Part 2: from words

17. A club has 7 equal teams. Then 2 players leave each team, and 63 players are left. How many players were on each team at first?
18. Ottilie trims a strip 1.5 cm wide from every edge of a square photo. The trimmed photo measures 48 cm all the way round. How long was a side of the photo before trimming?
19. A fence is made of 10 sections, each a panel and a post 0.15 m wide. The fence is 22 m long. How wide is one panel?
20. Each of three friends buys a ticket and a drink that costs 2.75 dollars. Together they pay 39.75 dollars. What does one ticket cost?
21. A coach says: “I had 4 equal teams, then 3 new players joined each team, and now there are 50 players.” Show that the coach has made a mistake somewhere.

Reasoning

22. Explain why dividing first and distributing first must give the same solution of p(x + q) = r.
23. Suppose p, q and r are integers and p \neq 0. Show that p(x + q) = r has an integer solution exactly when p divides r.
24. What happens when you try to solve 0(x + 3) = 5 and 0(x + 3) = 0? Explain why dividing first cannot be used.
25. Write one story that leads to 4(x + 3) = 40 and another that leads to 4x + 3 = 40, solve both, and explain which words make the brackets.