Give the unknown a name
Two numbers add up to 17, and one of them is 3 more than the other. What are they?
Neither number is known, but they are linked: once you know the smaller one, you know the larger one too. So give the smaller number a letter. Call it n. The larger number is 3 more than n, so it is n + 3.
Now both numbers are written using one letter. That is the first step in forming an equation: choose one unknown, name it with a letter, and write every other quantity in terms of it.
The larger number is the smaller one, n, with 3 more added.
Say the same amount two ways
An equation says that two expressions are the same amount. The question gives that amount: the two numbers together make 17. Written with the letter, together they make n + (n + 3). So n + (n + 3) = 17.
Now solve it. Collect like terms: 2n + 3 = 17. Subtract 3 from both sides: 2n = 14. Divide both sides by 2: n = 7.
n was the smaller number, so the smaller number is 7 and the larger is 7 + 3 = 10. Check against the words of the question: 7 + 10 = 17, and 10 is 3 more than 7.
With n = 7 the bars are 7 and 10 long, and together they make 17.
The steps, every time
Forming an equation from words always follows the same steps. First, choose the unknown and name it with a letter, saying what the letter stands for. Next, write every other quantity using that letter. Then find two ways of writing the same amount, and set them equal. Solve the equation, and answer the question that was asked, in words, with units.
Here is a second example. A rectangle is 4 cm longer than it is wide, and its perimeter is 28 cm. Let the width be w cm. Then the length is (w + 4) cm. The perimeter is two widths and two lengths: w + w + (w + 4) + (w + 4) = 28. Collect like terms: 4w + 8 = 28. Subtract 8: 4w = 20. Divide by 4: w = 5. The rectangle is 5 cm wide and 9 cm long. Check: 5 + 5 + 9 + 9 = 28.
The usual mistakes
Using two letters for quantities that are linked. If the smaller number is n, the larger is n + 3, not a new letter m. One letter for each thing you do not know, and here there is only one.
Stopping at the letter. n = 7 answers only part of the question. The question asked for both numbers, so finish with 7 and 10.
Equations from real situations
The same steps turn a real problem into an equation. In the first problem below, two ways of buying coffee give two totals, and setting them equal gives an equation with the letter on both sides. In the second, a mean is a total divided by a count, so the equation has a fraction.
Worked example: A Coffee Machine at Home or Coffee Bought at a Cafe, Month by Month
Question Nadia spends $66 a month on coffee bought at a cafe. She could instead buy a coffee machine for $180 and then spend $30 a month on coffee pods. (a) After how many months would the two ways have cost her the same in total? (b) Over the first year, which way is cheaper, and by how much?
1.Let m be the number of months. Buying at the cafe costs 66m dollars in total. The machine costs 180 + 30m dollars in total: $180 once and $30 each month.
For m months the cafe costs 66m dollars and the machine costs 180 + 30m dollars. 2.The totals are the same when 66m = 180 + 30m. Subtract 30m, the smaller letter term, from both sides: 36m = 180.
The totals are equal when 66m = 180 + 30m. Subtract 30m from both sides: 36m = 180. 3.Divide both sides by 36: m = 5. (a) The two ways have cost the same after 5 months, $330 each. Check: 66 × 5 = 330 and 180 + 30 × 5 = 330.
(a) Divide both sides by 36: m = 5. After 5 months each way has cost $330. 4.For the first year put m = 12. The cafe costs 66 × 12 = $792 and the machine costs 180 + 30 × 12 = 180 + 360 = $540.
For the first year, m = 12: the cafe costs $792 and the machine costs $540. 5.(b) The machine is cheaper over the first year, by 792 − 540 = $252. This agrees with (a): after month 5 the cafe costs 66 − 30 = $36 a month more, for 7 more months, and 36 × 7 = 252.
(b) The machine is cheaper over the first year, by 792 − 540 = $252.
Answer: (a) 5 months; (b) the machine, by $252
Common mistakes
- Adding 30m to the left side instead of subtracting it from both sides, which gives 96m = 180. Whatever is done to one side is done to the other, so the 30m comes off both sides: 36m = 180.
- Comparing only the monthly amounts and saying that the machine saves 36 × 12 = $432 over the year. That leaves out the $180 the machine costs at the start, which is why the saving is only $252.
More equations and inequalities problems, worked step by step →
Worked example: The Mark Needed on the Next Test to Reach a Mean
Question Priya scored 68, 77 and 62 marks on her first three science tests. Each test is marked out of 100. (a) After the fourth test her mean mark for the four tests is 72. Find her mark on the fourth test. (b) After a fifth test her mean mark for the five tests is 75. Find her mark on the fifth test.
1.Let the fourth mark be x. The first three marks add up to 68 + 77 + 62 = 207, so the mean of the four marks is 207 + x4. The mean is 72, so 207 + x4 = 72.
The three marks add up to 207, so the mean of the four marks is 207 + x4 = 72. 2.Multiply both sides by 4 to remove the fraction: 207 + x = 288.
Multiply both sides by 4: 207 + x = 288. 3.Subtract 207 from both sides: x = 81. (a) Her fourth mark is 81. Check: 68 + 77 + 62 + 81 = 288 and 288 ÷ 4 = 72.
(a) Subtract 207 from both sides: x = 81 marks. 4.The four marks add up to 288. Let the fifth mark be y. The mean of the five marks is 75, so 288 + y5 = 75.
The four marks add up to 288, so the mean of the five marks is 288 + y5 = 75. 5.Multiply both sides by 5: 288 + y = 375. Subtract 288 from both sides: y = 87.
Multiply both sides by 5, then subtract 288 from both sides: y = 87. 6.(b) Her fifth mark is 87. Check: 375 ÷ 5 = 75, and 87 is not more than 100, so it is a possible mark.
(b) Her fifth mark is 87.
Answer: (a) 81 marks; (b) 87 marks
Common mistakes
- Writing 207 + x4 = 72. All four marks are added before the sum is divided by 4, so the whole sum 207 + x is the numerator.
- Dividing by 4 again in part (b). There are now five tests, so the sum 288 + y is divided by 5.
More equations and inequalities problems, worked step by step →