# Word Problems

## Word Problems

To solve word problems we need to write a set of equations that represent the problem mathematically. The solution of the equations is then the solution to the problem.

## Problem Solving Strategy

2. What are we asked to solve for?

3. Assign a variable to the unknown quantity, for example, $$x$$.

4. Translate the words into algebraic expressions by rewriting the given information in terms of the variable.

5. Set up an equation or system of equations to solve for the variable.

6. Solve the equation algebraically using substitution.

7. Check the solution.

The following video shows two examples of working with word problems.

## Example

### Question

A shop sells bicycles and tricycles. In total there are $$\text{7}$$ cycles (cycles include both bicycles and tricycles) and $$\text{19}$$ wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels.

### Assign variables to the unknown quantities

Let $$b$$ be the number of bicycles and let $$t$$ be the number of tricycles.

### Set up the equations

\begin{align*} b + t & = 7 \qquad \ldots (1) \\ 2b + 3t & = 19 \qquad \ldots (2) \end{align*}

### Rearrange equation $$(1)$$ and substitute into equation $$(2)$$

\begin{align*} t & = 7 – b \\ \therefore 2b + 21 – 3b & = 19 \\ -b & = -2 \\ \therefore b & = 2 \end{align*}

### Calculate the number of tricycles $$t$$

\begin{align*} t & = 7 – b \\ & = 7 – 2 \\ & = 5 \end{align*}

There are $$\text{5}$$ tricycles and $$\text{2}$$ bicycles.

## Example

### Question

Bongani and Jane are friends. Bongani takes Jane’s maths test paper and will not tell her what her mark is. He knows that Jane dislikes word problems so he decides to tease her. Bongani says: “I have $$\text{2}$$ marks more than you do and the sum of both our marks is equal to $$\text{14}$$. What are our marks?”

### Assign variables to the unknown quantities

We have two unknown quantities, Bongani’s mark and Jane’s mark. Let Bongani’s mark be $$b$$ and Jane’s mark be $$j$$.

### Set up a system of equations

Bongani has $$\text{2}$$ more marks than Jane.

$b = j + 2 \qquad \ldots (1)$

Both marks add up to $$\text{14}$$.

$b + j = 14 \qquad \ldots (2)$

### Use equation $$(1)$$ to express $$b$$ in terms of $$j$$

$b = j + 2$

### Substitute into equation $$(2)$$

\begin{align*} b + j & = 14 \\ (j + 2) + j & = 14 \end{align*}

### Rearrange and solve for $$j$$

\begin{align*} 2j & = 14 – 2 \\ & = 12 \\ \therefore j & =6 \end{align*}

### Substitute the value for $$j$$ back into equation $$(1)$$ and solve for $$b$$

\begin{align*} b & = j + 2 \\ & = 6 + 2 \\ & = 8 \end{align*}

### Check that the solution satisfies both original equations

Bongani got $$\text{8}$$ for his test and Jane got $$\text{6}$$.

## Example

### Question

A fruitshake costs $$\text{R}\,\text{2.00}$$ more than a chocolate milkshake. If $$\text{3}$$ fruitshakes and $$\text{5}$$ chocolate milkshakes cost $$\text{R}\,\text{78.00}$$, determine the individual prices.

### Assign variables to the unknown quantities

Let the price of a chocolate milkshake be $$x$$ and let the price of a fruitshake be $$y$$.

### Set up a system of equations

\begin{align*} y & = x + 2 \qquad \ldots (1) \\ 3y + 5x & = 78 \qquad \ldots (2) \end{align*}

### Substitute equation $$(1)$$ into $$(2)$$

$3(x + 2) + 5x = 78$

### Rearrange and solve for $$x$$

\begin{align*} 3x + 6 + 5x & = 78 \\ 8x & = 72 \\ \therefore x & = 9 \end{align*}

### Substitute the value of $$x$$ back into equation $$(1)$$ and solve for $$y$$

\begin{align*} y & = x + 2 \\ & = 9 + 2 \\ & = 11 \end{align*}

### Check that the solution satisfies both original equations

One chocolate milkshake costs $$\text{R}\,\text{9.00}$$ and one fruitshake costs $$\text{R}\,\text{11.00}$$.

## Example

### Question

The product of two consecutive negative integers is $$\text{1 122}$$. Find the two integers.

### Assign variables to the unknown quantities

Let the first integer be $$n$$ and let the second integer be $$n + 1$$

### Set up an equation

$n(n + 1) = \text{1 122}$

### Expand and solve for $$n$$

\begin{align*} {n}^{2} + n & = \text{1 122} \\ {n}^{2} + n – \text{1 122}& =0 \\ (n + 34)(n – 33) & =0 \\ \therefore n & = -34 \\ \text{ or } n& = 33 \end{align*}

### Find the sign of the integers

It is given that both integers must be negative.

\begin{align*} \therefore n & =-34 \\ n + 1 & = -34 + 1 \\ & = -33 \end{align*}

The two consecutive negative integers are $$-\text{34}$$ and $$-\text{33}$$.