## Problem Solving Strategy

Contents

- Problem Solving Strategy
- Optional Investigation: Simple word problems
- Example
- Example
- Question
- Method 1: identify the unknown quantities and assign two variables
- Use the given information to set up a system of equations
- Solve the equations by substituting the second equation into the first equation
- Check that the solution satisfies both original equations
- Write the final answer
- Method 2: identify the unknown quantities and assign one variable
- Use the given information to set up an equation
- Solve for \(x\)
- Write the final answer

- Example
- Question
- Identify the unknown quantities and assign variables
- Convert all units of time to be the same
- Use the given information to set up a system of equations
- Multiply the equation through by the lowest common denominator and simplify
- Check that the solution satisfies the original equation
- Write the final answer

Read the problem carefully.

What is the question and what do we need to solve for?

Assign variables to the unknown quantities, for example, \(x\) and \(y\).

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

Set up a system of equations.

Solve for the variables using substitution.

Check the solution.

Write the final answer.

## Optional Investigation: Simple word problems

Write an equation that describes the following real-world situations mathematically:

Mohato and Lindiwe both have colds. Mohato sneezes twice for each sneeze of Lindiwe’s. If Lindiwe sneezes \(x\) times, write an equation describing how many times they both sneezed.

The difference of two numbers is \(\text{10}\) and the sum of their squares is \(\text{50}\). Find the two numbers.

Liboko builds a rectangular storeroom. If the diagonal of the room is \(\sqrt{ \text{1 312}}\) \(\text{m}\) and the perimeter is \(\text{80}\) \(\text{m}\), determine the dimensions of the room.

It rains half as much in July as it does in December. If it rains \(y\) mm in July, write an expression relating the rainfall in July and December.

Zane can paint a room in \(\text{4}\) hours. Tlali can paint a room in \(\text{2}\) hours. How long will it take both of them to paint a room together?

\(\text{25}\) years ago, Arthur was \(\text{5}\) years more than a third of Bongani’s age. Today, Bongani is \(\text{26}\) years less than twice Arthur’s age. How old is Bongani?

The product of two integers is \(\text{95}\). Find the integers if their total is \(\text{24}\).

## Example

### Question

The annual gym subscription for a single member is \(\text{R}\,\text{1 000}\), while an annual family membership is \(\text{R}\,\text{1 500}\). The gym is considering increasing all membership fees by the same amount. If this is done then a single membership would cost \(\dfrac{5}{7}\) of a family membership. Determine the amount of the proposed increase.

### Identify the unknown quantity and assign a variable

Let the amount of the proposed increase be \(x\).

### Use the given information to complete a table

now | after increase | |

single | \(\text{1 000}\) | \(\text{1 000} + x\) |

family | \(\text{1 500}\) | \(\text{1 500} + x\) |

### Set up an equation

\[\text{1 000} + x = \cfrac{5}{7}(\text{1 500} + x)\]

### Solve for \(x\)

\begin{align*} \text{7 000} + 7x &= \text{7 500} + 5x \\ 2x &= 500 \\ x &= 250 \end{align*}

### Write the final answer

The proposed increase is \(\text{R}\,\text{250}\).

## Example

### Question

Erica has decided to treat her friends to coffee at the Corner Coffee House. Erica paid \(\text{R}\,\text{54.00}\) for four cups of cappuccino and three cups of filter coffee. If a cup of cappuccino costs \(\text{R}\,\text{3.00}\) more than a cup of filter coffee, calculate how much a cup of each type of coffee costs?

### Method 1: identify the unknown quantities and assign two variables

Let the cost of a cappuccino be \(x\) and the cost of a filter coffee be \(y\).

### Use the given information to set up a system of equations

\begin{align*} 4x + 3y &= 54 \qquad \ldots (1) \\ x &= y + 3 \qquad \ldots (2) \end{align*}

### Solve the equations by substituting the second equation into the first equation

\begin{align*} 4(y+3) + 3y &= 54 \\ 4y+12 + 3y &= 54 \\ 7y &= 42 \\ y &= 6 \end{align*}

If \(y=6\), then using the second equation we have \begin{align*} x &= y + 3 \\ &= 6 + 3 \\ &= 9 \end{align*}

### Check that the solution satisfies both original equations

### Write the final answer

A cup of cappuccino costs \(\text{R}\,\text{9}\) and a cup of filter coffee costs \(\text{R}\,\text{6}\).

### Method 2: identify the unknown quantities and assign one variable

Let the cost of a cappuccino be \(x\) and the cost of a filter coffee be \(x-3\).

### Use the given information to set up an equation

\[4x + 3(x-3) = 54\]

### Solve for \(x\)

\begin{align*} 4x + 3(x-3) &= 54 \\ 4x + 3x-9 &= 54 \\ 7x &= 63 \\ x &= 9 \end{align*}

### Write the final answer

A cup of cappuccino costs \(\text{R}\,\text{9}\) and a cup of filter coffee costs \(\text{R}\,\text{6}\).

## Example

### Question

Two taps, one more powerful than the other, are used to fill a container. Working on its own, the less powerful tap takes \(\text{2}\) hours longer than the other tap to fill the container. If both taps are opened, it takes \(\text{1}\) hour, \(\text{52}\) minutes and \(\text{30}\) seconds to fill the container. Determine how long it takes the less powerful tap to fill the container on its own.

### Identify the unknown quantities and assign variables

Let the time taken for the less powerful tap to fill the container be \(x\) and let the time taken for the more powerful tap be \(x – 2\).

### Convert all units of time to be the same

First we must convert \(\text{1}\) hour, \(\text{52}\) minutes and \(\text{30}\) seconds to hours: \[1 + \cfrac{52}{60} + \cfrac{30}{(60)^2} = \text{1.875}\text{ hours}\]

### Use the given information to set up a system of equations

Write an equation describing the two taps working together to fill the container: \begin{align*} \cfrac{1}{x} + \cfrac{1}{x-2} &= \cfrac{1}{\text{1.875}} \end{align*}

### Multiply the equation through by the lowest common denominator and simplify

\begin{align*} \text{1.875}(x-2) + \text{1.875}x &= x(x-2) \\ \text{1.875}x – \text{3.75} + \text{1.875}x &= x^2 – 2x \\ 0 &= x^2 – \text{5.75}x + \text{3.75} \end{align*}

Multiply the equation through by \(\text{4}\) to make it easier to factorise (or use the quadratic formula) \begin{align*} 0 &= 4x^2 – 23x + 15 \\ 0 &= (4x-3)(x-5) \end{align*} Therefore \(x = \cfrac{3}{4}\) or \(x = 5\).

We have calculated that the less powerful tap takes \(\cfrac{3}{4}\) hours or \(\text{5}\) hours to fill the container, but we know that when both taps are opened it takes \(\text{1.875}\) hours. We can therefore discard the first solution \(x = \dfrac{3}{4}\) hours.

So the less powerful tap fills the container in \(\text{5}\) hours and the more powerful tap takes \(\text{3}\) hours.

### Check that the solution satisfies the original equation

### Write the final answer

The less powerful tap fills the container in \(\text{5}\) hours and the more powerful tap takes \(\text{3}\) hours.