Equations from proportions and percents

When you reach for it

Solving a proportion finds a plain unknown such as the x in \displaystyle \frac{x}{5} = \frac{3}{4}: one multiplication and one division and it is done. Very often, though, the unknown is not alone. It sits inside an expression in a numerator,

or in a denominator,

and no single operation sets it free. The move in this lesson turns such a proportion into a two-step linear equation, which you can already solve.

Percent problems lead to the same place. “A number increased by 20\% is 54” describes an unknown that has been changed and asks you to undo the change. Writing the change as a multiplier turns the sentence into 1.2n = 54, again a linear equation.

So reach for this whenever two ratios are set equal and the unknown is tangled up in one of them, or whenever a percent has been added to or taken from an amount you do not know. The lesson runs over two parts: Part 1 is proportions with the unknown inside an expression, Part 2 is percent equations.

The procedure

Part 1: a proportion with the unknown inside an expression

Cross multiplying. For any numbers with b \ne 0 and d \ne 0,

This is not a new rule. Multiplying both sides of \displaystyle \frac{a}{b} = \frac{c}{d} by bd is allowed by the multiplication property of equality, and it cancels each denominator: the left side becomes ad and the right side becomes bc. Since bd is not zero, the step can be undone by dividing by bd, so the two equations have exactly the same solutions. (What a proportion says has the same fact from the side of equal ratios.)

Keep the expression whole. In \displaystyle \frac{x + 2}{5} = \frac{3}{4} the numerator x + 2 is one number, the top of the fraction. Multiplying it by 4 multiplies all of it, so the cross product is 4(x + 2), with brackets. The distributive property then opens the brackets:

From here it is a two-step equation: subtract 8 from both sides, then divide both sides by 4, and \displaystyle x = \frac{7}{4}. Instead of distributing you may divide both sides by 4 first, getting \displaystyle x + 2 = \frac{15}{4}; both routes are allowed by the properties of equality and give the same answer.

No denominator may be zero. When the unknown is in a denominator, some values of it make that denominator zero, and for those values the fraction does not exist. In \displaystyle \frac{6}{x - 1} = \frac{2}{3} the value x = 1 is ruled out before any work starts: write x \ne 1 beside the equation. The reason this matters is in the justification above: cross multiplying is multiplying by bd, and that step can be undone only when bd \ne 0. So after solving, check that the answer does not make any denominator zero, and put it back into the original proportion to check that it really is a solution.

Part 2: percent equations

A percent of an amount is that decimal times the amount, as in finding a percent of a quantity. If the unknown amount is n and it grows by 20\%, the growth is 0.2n and the new amount is

The middle step is the distributive property, read from right to left. If instead n falls by 15\%, what is left is n - 0.15n = 0.85n. In general an increase of r (written as a decimal) multiplies by 1 + r and a decrease multiplies by 1 - r; the lesson on the multiplier for a percent change builds this in full.

Once the change is written as a multiplier, the sentence is an equation. “A number increased by 20\% is 54” becomes

and dividing both sides by 1.2 undoes the change. A decimal coefficient is easier to divide by after multiplying both sides by 10 or 100, which is again allowed by the properties of equality: 12n = 540, so n = 45.

The same sentence can be written as a proportion, which joins the two parts of the lesson. The new amount is to the original as 120 is to 100:

Cross multiplying gives 120n = 5400, and dividing both sides by 100 gives 1.2n = 54, the same equation. When the percent is applied to an expression rather than to a plain number, as in “35\% of x - 2”, the brackets matter exactly as they did in Part 1: 0.35(x - 2).

Worked examples

Examples 1 to 4 belong to Part 1 and Examples 5 to 7 to Part 2.

Boundaries

A denominator may not be zero. Cross multiplying replaces \displaystyle \frac{a}{b} = \frac{c}{d} by ad = bc, and the two say the same thing only when b \ne 0 and d \ne 0. When the unknown sits in a denominator, the values that make it zero are excluded, and an answer must be checked against them. In \displaystyle \frac{2x - 6}{x - 3} = \frac{2}{1}, cross multiplying gives 2x - 6 = 2x - 6, which is true for every x, yet x = 3 is not a solution: there the fraction is \displaystyle \frac{0}{0}, which is not a number. (Exercise 25 asks for the full answer.)

Cross multiplying needs one fraction on each side, and nothing else. In \displaystyle \frac{x + 1}{2} + 3 = \frac{3}{4} the left side is not a single fraction, so it is not a proportion, and multiplying x + 1 by 4 and 2 by 3 does not give an equivalent equation. Either first collect the left side into one fraction, or multiply every term by a common denominator.

One unknown, on one side, to the first power. This lesson stays with proportions whose cross products leave the unknown on one side of a linear equation. If the unknown sits in more than one place, as in \displaystyle \frac{x - 12}{x} = \frac{5}{7}, cross multiplying leaves it on both sides, and that is the subject of equations with variables on both sides. If it sits in a numerator on one side and a denominator on the other, as in \displaystyle \frac{x + 1}{3} = \frac{4}{x - 2}, cross multiplying gives (x + 1)(x - 2) = 12, which has an x^2 in it once multiplied out, and those equations belong to solving a rational equation.

A percent is a percent of a particular amount. “Increased by 20\%” means 20\% of the starting amount, so the starting amount is what gets multiplied by 1.2. “54 is 20\% more than n” gives 1.2n = 54 and n = 45, while “n is 20\% less than 54” gives n = 0.8 \times 54 = 43.2. The two sentences sound alike but take the percent of different amounts.

A decrease of 100\% or more cannot be undone. A decrease of 100\% multiplies by 1 - 1 = 0, so “a number decreased by 100\% is 5” becomes 0 \cdot n = 5, which no number satisfies. Every amount decreased by 100\% is 0, and the original cannot be recovered from that. A decrease of more than 100\% gives a negative multiplier, which makes sense for a number but not for a count of things.

Where this goes next

Proportions with an expression inside are the first step toward rational equations, where unknowns sit in several denominators and the same two ideas, clear the denominators and check that none is zero, carry the whole method. They are also how missing lengths are found in figures with parallel lines, as in the triangle proportionality theorem, where a side is often written as x - 1 or x + 4.

On the percent side, writing a change as a multiplier is what makes two percent changes one after the other simple: the multipliers multiply. Choosing which quantity goes on top when a proportion comes from words is the subject of setting up proportions from word problems, and the plain part-whole percent equation is in solving percent problems with an equation and finding the whole from a part and a percent.

Exercises

Part 1: the unknown in a numerator

Solve each proportion, and check your answer.

1. \dfrac{x + 4}{6} = \dfrac{5}{3}
2. \dfrac{x - 6}{9} = \dfrac{2}{3}
3. \dfrac{2x + 1}{7} = \dfrac{3}{5}
4. \dfrac{5 - x}{4} = -\dfrac{3}{8}

Part 1: the unknown in a denominator

State the value that x may not take, then solve.

5. \dfrac{10}{x + 4} = \dfrac{5}{3}
6. \dfrac{8}{x - 2} = \dfrac{4}{7}
7. \dfrac{3}{2x - 1} = \dfrac{1}{5}
8. \dfrac{-6}{x + 1} = \dfrac{3}{4}

Part 1: a sign or a side to watch

Solve, stating any value that x may not take.

9. \dfrac{12}{5 - 2x} = -\dfrac{4}{3}
10. \dfrac{4}{9} = \dfrac{2x + 6}{15}

Part 1: in context

11. Iolanthe mixes packets of wildflower seed and packets of grass seed in the ratio 2 : 7. She has some wildflower packets, and after buying 3 more she has exactly the right number to go with 35 grass packets. How many wildflower packets did she have to begin with?
12. Jovan cuts 18 cm off a rope. The length that is left and the piece cut off are in the ratio 7 : 3. How long was the rope?
13. Thessaly’s class has 12 girls. After 4 new boys join, the girls are 3 out of every 7 students in the class. How many students were in the class before the new boys joined?

Part 2: write the equation and solve

Write each sentence as an equation, then solve it.

14. A number increased by 30\% is 117.
15. A number decreased by 45\% is 33.
16. A number increased by 125\% is 63.
17. A number decreased by 8\% is 46.
18. 35\% of x - 2 is 14.
19. When 12 is added to a number and the sum is decreased by 25\%, the result is 18.

Part 2: in context

20. Isaura’s shell collection grew by 15\% to 138 shells. How many shells did she have before?
21. A population of bacteria in a dish fell by 60\% overnight, to 1{,}800 cells. How many cells were there the evening before?
22. Hesper sold 20\% of her comic books in spring, then bought 9 more. She now has 41 comic books. How many did she have at the start?

Reasoning

23. Show that “a number increased by 20\% is 54” leads to the same equation whether it is written as 1.2n = 54 or as the proportion \displaystyle \frac{54}{n} = \frac{120}{100}.
24. Explain why cross multiplying \displaystyle \frac{x + 2}{5} = \frac{3}{4} gives 4(x + 2) = 15 and not 4x + 2 = 15. Solve 4x + 2 = 15 and show that its answer does not satisfy the proportion.
25. Cross multiplying \displaystyle \frac{2x - 6}{x - 3} = \frac{2}{1} gives an equation that is true for every x. Which numbers are solutions of the original proportion, and why?