## 6.6 Solving problems

## Worked example 6.23: Formulating expressions

Delphino has \(k\) sweets. Phindile has \(4\) times as many sweets as Delphino. Give an expression for the number of sweets that Phindile has, in terms of \(k\).

### Multiply the number of sweets that Delphino has by \(4\).

We don’t know how many sweets Delphino has. We use the variable \(k\) to stand in the place of the number of sweets that Delphino has.

Delphino has \(k\) sweets and Phindile has \(4\) times as many.

\[4 \times k = 4k\]Therefore, Phindile has \(4k\) sweets.

## Worked example 6.24: Formulating expressions

Yu Yan has \(y\) sweets. Abdoul has \(7\) fewer sweets than Yu Yan. Give an expression for the number of sweets that Abdoul has, in terms of \(y\).

### Subtract the difference from the number of sweets that Yu Yan has.

Abdoul has \(7\) fewer sweets than Yu Yan has. So we must subtract \(7\) sweets from Yu Yan’s sweets:

\[y - 7\]Therefore, Abdoul has \(y - 7\) sweets.

## Worked example 6.25: Formulating expressions

- Phindile has \(z\) sweets.
- Abdoul has \(4\) more sweets than Phindile.
- Delphino has twice as many sweets as Abdoul.

Give simplified expressions, in terms of \(z\), for:

- The number of sweets that Abdoul has.
- The number of sweets that Delphino has.

### Write an expression for Abdoul’s sweets.

We don’t know how many sweets Phindile has. We use the variable \(z\) to stand in the place of Phindile’s sweets.

Abdoul has \(4\) more sweets than Phindile. Phindile has \(z\) sweets.

Therefore, Abdoul has \(z + 4\) sweets.

### Write an expression for Delphino’s sweets.

Delphino has twice as many sweets as Abdoul. Abdoul has \(z + 4\) sweets.

\[2(z + 4) = 2k + 8\]Therefore, Delphino has \(2k + 8\) sweets.

## Worked example 6.26: Consecutive integers

Determine a simplified expression for the sum of five consecutive integers, where the smallest integer is \(x\).

Hence, determine the sum of five consecutive integers if the smallest integer is \(60\).

### Write expressions for five consecutive integers.

To get consecutive integers, we start with an integer and add \(1\) each time. For example, if we start at \(2\), the next consecutive integer is \(2 + 1 = 3\).

If we start at \(x\), the next consecutive integer is \(x + 1\).

We add \(1\) again to get the following consecutive integer: \((x + 1) + 1 = x + 2\).

The fourth consecutive integer is: \((x + 2) + 1 = x + 3\).

The fifth consecutive integer is: \((x + 3) + 1 = x + 4\).

Therefore the consecutive integers are \(x\), \(x + 1\), \(x + 2\), \(x + 3\) and \(x + 4\).

### Add the expressions together to get the sum.

\[= x + (x + 1) + (x + 2) + (x + 3) + (x + 4)\] \[= x + x + 1 + x + 2 + x + 3 + x + 4\] \[= 5x + 10\]**Hint:** Adding \((x + 1)\) is the same as adding \(x\) and then adding \(1\). That is why we
can get rid of the brackets.

### Substitute.

Substitute \(x = 60\) into \(5x + 10\).

From Step \(2\), we know that an expression for the sum of five consecutive integers is \(5x + 10\), where \(x\) stands for the smallest integer.

If \(x = 60\):

\[5x + 10 = 5(60) + 10 = 300 + 10 = 310\]Therefore, if the smallest integer is \(60\), the sum of five consecutive integers is \(310\).