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# 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 $$310$$.

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