Now that we know how to solve equations, the next step is to create our own equations to solve a particular situation or problem.
Most problems that we encounter will be presented in the form of a written description. Our first task is to translate the description into an equation. The equation is like a model for the problem. It reduces the complexity of the written problem and allows us to find a solution using equation-solving techniques.
To become familiar with this, we begin by translating some simple written statements into an equation.
Worked example
Example 1
Translate each of the following statements into an equation:
(a) The speed $P$P of an aeroplane is four times the speed $T$T of a train.
Think: The first step is to look for certain keywords that indicate which mathematical operations to use in our equation. In this case the word 'times' indicates that our equation will involve multiplication.
Do: Our equation will be $P=4T$P=4T.
(b) The sum of a number and five is twelve. Let $n$n be the number.
Think: As before, we want to identify any keywords which indicate the mathematical operation(s) to be used. This time, the word 'sum' indicates that we need to use addition.
Do: Our equation will be $n+5=12$n+5=12.
Reflect: Notice that in each description, the word 'is' corresponds to the equals sign ($=$=) in the equation.
Noticing keywords
Certain keywords in a problem description indicate the mathematical operations or concepts we might need to use in our equations. Here is a list of the more common ones:
There are many different ways the four basic operations of addition, subtraction, multiplication and division are described:
| Addition ($+$+) | add, plus, sum, total, increase, more, combined |
| Subtraction ($-$−) | subtract, minus, difference, take-away, decrease, less, deduct |
| Multiplication ($\times$×) | multiply, times, product, multiple, double, triple, twice, of, by, lots of |
| Division ($\div$÷) | divide, quotient, share, goes into, how many times, per, split, halve |
The following terms require a bit more explanation:
- The word product indicates multiplication. For example, the product of $3$3 and $x$x means $3\times x$3×x and is written as $3x$3x.
- A quotient is the name we use for the fraction we get from the division of two numbers. For example, the quotient of $x$x and $5$5 is written as $\frac{x}{5}$x5. The quotient of $5$5 and $x$x is $\frac{5}{x}$5x. Notice that the order is important.
- Consecutive numbers are numbers that follow each other in order. For example $5$5, $6$6, $7$7, $8$8 are consecutive numbers, and $45$45, $47$47, $49$49 are consecutive odd numbers, but $1$1, $3$3, $16$16, $25$25 are not consecutive.
There may also be specific words related to mathematics that indicate which concept to use. For example:
- The word average indicates that we add a series of values together, then divide by the number of values.
- The word perimeter indicates that we add side lengths together, whereas area means we multiply two lengths.
In some questions, the variable used to represent the unknown quantity will not be specified. In this case we choose a letter of the alphabet and state which variable it represents - this is called defining the variable. Note that it doesn't matter which letter we use, but it is important that we define what the letter represents.
In the following example questions, we choose $x$x to represent the unknown number.
Worked example
Example 2
Translate each of the following statements into an equation, then solve the equation to find the number.
(a) Eleven is added to the quotient of a number and three. The result is nineteen.
Think: To make it easier to write an equation from this description, let's highlight the keywords:
Eleven is added to the quotient of a number and three. The result is nineteen.
Do:
| $\frac{x}{3}+11$x3+11 | $=$= | $19$19 | (Translate the problem description into an equation) | |
| (Solve the equation) | ||||
| $\frac{x}{3}+11-11$x3+11−11 | $=$= | $19-11$19−11 | (Subtract $11$11 from both sides) | |
| $\frac{x}{3}$x3 | $=$= | $8$8 | ||
| $\frac{x}{3}\times3$x3×3 | $=$= | $8\times3$8×3 | (Multiply both sides by $3$3) | |
| $x$x | $=$= | $24$24 |
Reflect: We can check that $x=24$x=24 is correct by using $24$24 as the number in the original problem description. The quotient of $24$24 and $3$3 is $8$8. If we add $11$11 to $8$8 we get $19$19.
(b) Twenty-one less than the product of a number and nine, is double the number.
Think: Again, lets highlight the keywords:
Twenty-one less than the product of a number and nine, is double the number.
Do:
| $9x-21$9x−21 | $=$= | $2x$2x | (Translate the problem description into an equation) | |
| (Solve the equation) | ||||
| $9x-21-2x$9x−21−2x | $=$= | $2x-2x$2x−2x | (Subtract $2x$2x from both sides) | |
| $7x-21$7x−21 | $=$= | $0$0 | ||
| $7x-21+21$7x−21+21 | $=$= | $0+21$0+21 | (Add $21$21 to both sides) | |
| $7x$7x | $=$= | $21$21 | ||
| $\frac{7x}{7}$7x7 | $=$= | $\frac{21}{7}$217 | (Divide both sides by $7$7) | |
| $x$x | $=$= | $3$3 |
Reflect: We can check that $x=3$x=3 is correct by using $3$3 as the number in the original problem description. The product of $3$3 and $9$9 is $27$27. Subtracting $21$21 gives us $6$6, which is double the value of $3$3.
(c) Two thirds of a number that has been decreased by four, is equal to eight.
Think: Highlighting keywords, we have:
Two thirds of a number that has been decreased by four, is equal to eight.
Do:
| $\frac{2}{3}\times\left(x-4\right)$23×(x−4) | $=$= | $8$8 | (Translate the problem description into an equation) | |
| $\frac{2\left(x-4\right)}{3}$2(x−4)3 | $=$= | $8$8 | ||
| (Solve the equation) | ||||
| $\frac{2\left(x-4\right)}{3}\times3$2(x−4)3×3 | $=$= | $8\times3$8×3 | (Multiply both sides by $3$3) | |
| $2\left(x-4\right)$2(x−4) | $=$= | $24$24 | ||
| $\frac{2\left(x-4\right)}{2}$2(x−4)2 | $=$= | $\frac{24}{2}$242 | (Divide both sides by $2$2) | |
| $x-4$x−4 | $=$= | $12$12 | ||
| $x-4+4$x−4+4 | $=$= | $12+4$12+4 | (Add $4$4 to both sides) | |
| $x$x | $=$= | $16$16 |
Again, we can check that our solution of $x=16$x=16 is correct by using $16$16 as the number in the original problem description.
(d) The sum of three consecutive numbers is $60$60. What are the numbers?
Think: Consecutive numbers go up by $1$1 each time. So if we let the first number be $x$x, then the next consecutive number will be $x+1$x+1, the one after that will be $x+2$x+2, and so on. We will use brackets in our working to distinguish between the three numbers.
Here is the question again, with the keywords highlighted:
The sum of three consecutive numbers is $60$60. What are the numbers?
Do:
| $x+\left(x+1\right)+\left(x+2\right)$x+(x+1)+(x+2) | $=$= | $60$60 | (Translate the problem description into an equation) | |
| $3x+3$3x+3 | $=$= | $60$60 | (Simplify by collecting like terms) | |
| (Solve the equation) | ||||
| $3x+3-3$3x+3−3 | $=$= | $60-3$60−3 | (Subtract $3$3 from both sides) | |
| $3x$3x | $=$= | $57$57 | ||
| $\frac{3x}{3}$3x3 | $=$= | $\frac{57}{3}$573 | (Divide both sides by $3$3) | |
| $x$x | $=$= | $19$19 |
Remember that $x=19$x=19 is just our first number, so the three consecutive numbers that sum to $60$60 are:
$19$19, $20$20 and $21$21
Solving word problems
So far we have gained practice in translating a simple word statement into an equation. We can now apply these techniques to solving more involved problems, often based on real world situations.
Keep in mind that it can be very helpful to use a sketch or diagram as part of the problem-solving process. A sketch can help organise the problem, to see how all of the different elements fit together.
Worked examples
example 3
Sally, Kate and Emma are doing some fundraising for their sporting team. Together they raised $\$1600$$1600. Sally raised $\$272$$272 more than Emma, and Kate raised twice the amount that Emma did.
(a) Let $\$p$$p be the amount of money that Emma raised. Write an equation and solve it for $p$p.
Think: Let's start by writing expression for the amounts raised by each person:
- Emma raised $p$p
- Kate raised $2p$2p
- Sally raised $p+272$p+272
Do:
Together they raised $\$1600$$1600, so the equation becomes:
| $p+\left(2p\right)+\left(p+272\right)$p+(2p)+(p+272) | $=$= | $1600$1600 | (Translate the problem description into an equation) | |
| $4p+272$4p+272 | $=$= | $1600$1600 | (Simplify by collecting like terms) | |
| (Solve the equation) | ||||
| $4p+272-272$4p+272−272 | $=$= | $1600-272$1600−272 | (Subtract $272$272 from both sides) | |
| $4p$4p | $=$= | $1328$1328 | ||
| $\frac{4p}{4}$4p4 | $=$= | $\frac{1328}{4}$13284 | (Divide both sides by $4$4) | |
| $p$p | $=$= | $332$332 |
Therefore Emma raised $\$332$$332.
(b) How much money did Kate and Sally raise?
Solution:
Kate raised $2p$2p$=$=$2\times332$2×332$=$=$\$664$$664,
Sally raised $p+272$p+272$=$=$332+272$332+272$=$=$\$604$$604.
As a final check, we can add the three amounts to ensure they make the correct total: $332+664+604=1600$332+664+604=1600.
example 4
Tim is building a rectangular yard for his chickens. The yard is to have a perimeter of $72$72 metres, and its length is to be $6$6 metres more than double its width.
Let $w$w be the width of the yard.
(a) Write an expression for the length of the yard.
Solution:
The length of the yard is $6$6 metres more than double the width. Therefore the length can be written as $2w+6$2w+6.
(b) Write an equation in terms of $w$w and solve it to find the width of the yard.
Solution:
The yard is a rectangle with a perimeter equal to $72$72 metres. The perimeter is equal to two times the length added to two times the width.
| Perimeter | $=$= | $2\times\text{length }+2\times\text{width }$2×length +2×width | ||
| $72$72 | $=$= | $2\times\left(2w+6\right)+2\times w$2×(2w+6)+2×w | (Substitute values and expressions) | |
| $72$72 | $=$= | $2\left(2w+6\right)+2w$2(2w+6)+2w | ||
| $72$72 | $=$= | $4w+12+2w$4w+12+2w | (Expand brackets) | |
| $72$72 | $=$= | $6w+12$6w+12 | (Simplify by collecting like terms) | |
| (Solve equation) | ||||
| $72-12$72−12 | $=$= | $6w+12-12$6w+12−12 | (Subtract $12$12 from both sides) | |
| $60$60 | $=$= | $6w$6w | ||
| $\frac{60}{6}$606 | $=$= | $\frac{6w}{6}$6w6 | (Divide both sides by $6$6) | |
| $10$10 | $=$= | $w$w | ||
| $w$w | $=$= | $10$10 |
So the width of the yard is $10$10 metres.
(c) Hence, state the length of the yard.
Think: Here we substitute our value for $w$w into our expression for the length of the yard, $2w+6$2w+6.
Do:
| $2w+6$2w+6 | $=$= | $2\times10+6$2×10+6 |
| $=$= | $26$26 |
So the length of the yard is $26$26 metres.
Reflect: As a final check, we can work out the perimeter of the yard using our values for length and width:
Perimeter$=$=$2\times26+2+10$2×26+2+10$=$=$72$72 m
example 5
Skye, a squash player has won $48$48 out of $63$63 matches in her career. Find $x$x, the number of matches she must win in a row to raise her win percentage to $80%$80%.
Solution:
First, let's work out what percentage she has won so far:
| $\frac{48}{63}$4863 | $=$= | $0.7619\ldots$0.7619… |
| $=$= | $76.19%$76.19% (2 d.p.) |
If she wins the next match, her percentage would change to:
| $\frac{48+1}{63+1}$48+163+1 | $=$= | $\frac{49}{64}$4964 |
| $=$= | $0.7656\ldots$0.7656… | |
| $=$= | $76.56%$76.56% (2 d.p.) |
If she wins the next $x$x number of matches:
$\frac{48+x}{63+x}=80%$48+x63+x=80%
This is our equation. Now we need to solve it.
| $\frac{48+x}{63+x}$48+x63+x | $=$= | $0.8$0.8 | (Write the percentage as a decimal) | |
| $\frac{48+x}{63+x}\times\left(63+x\right)$48+x63+x×(63+x) | $=$= | $0.8\times\left(63+x\right)$0.8×(63+x) | (Multiply both sides by $63+x$63+x) | |
| $48+x$48+x | $=$= | $0.8\left(63+x\right)$0.8(63+x) | ||
| $48+x$48+x | $=$= | $50.4+0.8x$50.4+0.8x | (Expand brackets) | |
| $48+x-0.8x$48+x−0.8x | $=$= | $50.4+0.8x-0.8x$50.4+0.8x−0.8x | (Subtract $0.8x$0.8x from both sides) | |
| $48+0.2x$48+0.2x | $=$= | $50.4$50.4 | ||
| $48+0.2x-48$48+0.2x−48 | $=$= | $50.4-48$50.4−48 | (Subtract $48$48 from both sides) | |
| $0.2x$0.2x | $=$= | $2.4$2.4 | ||
| $\frac{0.2x}{0.2}$0.2x0.2 | $=$= | $\frac{2.4}{0.2}$2.40.2 | (Divide both sides by $0.2$0.2) | |
| $x$x | $=$= | $12$12 |
So Skye must win the next $12$12 matches in a row, to raise her win percentage to $80%$80%.
Once again, we can do a final check of our solution by adding $12$12 to the number of matches Skye has already won and the number of matches she has already played. We then convert this to a percentage:
$\frac{48+12}{63+12}$48+1263+12$=$=$\frac{60}{75}$6075$=$=$0.8$0.8$=$=$80%$80%
- First, we identify the unknown variable in the problem and represent it with a letter. This is called defining the variable. In some questions, the variable may already be chosen for us. If it isn't, we should state clearly what our chosen letter represents:
i.e. $p=\text{cost of one item}$p=cost of one item
- Look for any keywords and mathematical concepts in the problem description. We try to relate the unknown variable to other given variables in the problem. We might form some simple algebraic expressions first before combining them into an equation.
- A simple sketch or diagram can often be useful for organising all of the elements of a problem and seeing how they fit together.
- Form an equation using the unknown variable, and solve it using familiar equation-solving techniques.
- After solving our equation, it is important to re-read the question to see if we have actually answered the question or whether additional steps are required. For example, the unknown variable may have represented just one dimension of a shape, but the question may ask for all dimensions.
- It is important to state clearly what our final answer represents and include any units if necessary.
- We can test our solution using the original problem description.
Practice questions
Question 1
When a number is added to both the numerator and denominator of $\frac{1}{5}$15, the result is $\frac{3}{7}$37.
Let $n$n represent the number. Solve for $n$n.
Question 2
To manufacture sofas, the manufacturer has a fixed cost of $\$27600$$27600 plus a variable cost of $\$170$$170 per sofa. Find $n$n, the number of sofas that need to be produced so that the average cost per sofa is $\$290$$290.
Question 3
A commercial airplane has a total mass at take off of $51000$51000 kg. The luggage and fuel are $\frac{1}{3}$13 the mass of the unloaded plane, and the crew and passengers are $\frac{1}{4}$14 the mass of the fuel and luggage. Solve for $p$p, the mass of the unloaded plane.