Solving Problems with Linear Equations

Let the smallest odd integer be $n$.

The next two consecutive odd integers are $(n + 2)$ and $(n + 4)$.

Form the equation:

$$n + (n + 2) + (n + 4) = 57$$

Collect like terms:

$$3n + 6 = 57$$

Subtract $6$ from both sides:

$$3n + 6 - 6 = 57 - 6$$

$$3n = 51$$

Divide both sides by $3$:

$$n = \frac{51}{3} = 17$$

The three integers are $17$, $17 + 2 = 19$, and $17 + 4 = 21$.

Check: $17$, $19$ and $21$ are consecutive odd integers, and $17 + 19 + 21 = 57 \quad \checkmark$

Let the smaller number be $x$.

The larger number is $4x$.

Their difference is 36:

$$4x - x = 36$$

Collect like terms:

$$3x = 36$$

Divide both sides by $3$:

$$x = 12$$

The larger number is $4(12) = 48$.

Check: $48 = 4 \times 12$ and $48 - 12 = 36 \quad \checkmark$

Let the daughter's present age be $d$ years.

Mrs Goh's present age is $4d$ years.

Five years ago:

The daughter was $(d - 5)$ years old.

Mrs Goh was $(4d - 5)$ years old.

The sum of their ages then was 40:

$$(d - 5) + (4d - 5) = 40$$

Collect like terms:

$$5d - 10 = 40$$

Add $10$ to both sides:

$$5d - 10 + 10 = 40 + 10$$

$$5d = 50$$

Divide both sides by $5$:

$$d = 10$$

The daughter is 10 years old and Mrs Goh is $4(10) = 40$ years old.

Check: $40 = 4 \times 10$. Five years ago they were 5 and 35, and $5 + 35 = 40$ $\quad \checkmark$

Let Siti's present age be $s$ years.

Aisyah's present age is $(s + 6)$ years.

Five years ago:

Siti was $(s - 5)$ years old.

Aisyah was $(s + 6 - 5) = (s + 1)$ years old.

Form the equation:

$$s + 1 = 2(s - 5)$$

Expand the bracket:

$$s + 1 = 2s - 10$$

Subtract $s$ from both sides:

$$s + 1 - s = 2s - 10 - s$$

$$1 = s - 10$$

Add $10$ to both sides:

$$11 = s$$

Siti is 11 years old; Aisyah is $11 + 6 = 17$ years old.

Check: 5 years ago, Siti was 6 and Aisyah was 12, and $12 = 2 \times 6 \quad \checkmark$

Let Raju's savings be $s$ dollars.

The book cost $\frac{s}{3}$ dollars and lunch cost 12 dollars.

What is left is his savings minus what he spent:

$$s - \frac{s}{3} - 12 = 28$$

Multiply every term on both sides by $3$:

$$3s - 3\left(\frac{s}{3}\right) - 3(12) = 3(28)$$

$$3s - s - 36 = 84$$

Collect like terms:

$$2s - 36 = 84$$

Add $36$ to both sides:

$$2s - 36 + 36 = 84 + 36$$

$$2s = 120$$

Divide both sides by $2$:

$$s = 60$$

Check: the book cost $\frac{60}{3} = 20$ dollars; $60 - 20 - 12 = 28$ dollars left $\quad \checkmark$

Let the number of 50-cent coins be $n$.

The number of 20-cent coins is $2n$ and the number of 10-cent coins is $(n + 5)$.

Work in cents: $6.50 is 650 cents. The coins are worth $50n$, $20(2n)$ and $10(n + 5)$ cents.

Form the equation:

$$50n + 20(2n) + 10(n + 5) = 650$$

Expand the brackets:

$$50n + 40n + 10n + 50 = 650$$

Collect like terms:

$$100n + 50 = 650$$

Subtract $50$ from both sides:

$$100n + 50 - 50 = 650 - 50$$

$$100n = 600$$

Divide both sides by $100$:

$$n = 6$$

He has 6 fifty-cent coins, $2(6) = 12$ twenty-cent coins and $6 + 5 = 11$ ten-cent coins.

Check: $6 \times 50 + 12 \times 20 + 11 \times 10 = 300 + 240 + 110 = 650$ cents, which is $6.50 $\quad \checkmark$

Let the first angle be $x^\circ$.

The second angle is $(x + 20)^\circ$.

The third angle is $2x^\circ$.

The angles in a triangle add up to $180^\circ$:

$$x + (x + 20) + 2x = 180$$

Collect like terms:

$$4x + 20 = 180$$

Subtract $20$ from both sides:

$$4x + 20 - 20 = 180 - 20$$

$$4x = 160$$

Divide both sides by $4$:

$$x = 40$$

First angle: $40^\circ$; second angle: $40 + 20 = 60^\circ$; third angle: $2(40) = 80^\circ$.

Check: $40^\circ + 60^\circ + 80^\circ = 180^\circ \quad \checkmark$

Let the length of a side of the square be $s$ cm.

Each side of the triangle is $(s + 5)$ cm.

Both shapes use the whole wire, so their perimeters are equal:

$$4s = 3(s + 5)$$

Expand the bracket:

$$4s = 3s + 15$$

Subtract $3s$ from both sides:

$$4s - 3s = 3s + 15 - 3s$$

$$s = 15$$

A side of the square is 15 cm, and the wire is $4 \times 15 = 60$ cm long.

Check: each side of the triangle is $15 + 5 = 20$ cm, and $3 \times 20 = 60$ cm $\quad \checkmark$

Let the time be $t$ minutes.

After $t$ minutes, Tank A holds $(50 + 6t)$ litres and Tank B holds $(200 - 9t)$ litres.

The two amounts are equal:

$$50 + 6t = 200 - 9t$$

Add $9t$ to both sides:

$$50 + 6t + 9t = 200 - 9t + 9t$$

$$50 + 15t = 200$$

Subtract $50$ from both sides:

$$50 + 15t - 50 = 200 - 50$$

$$15t = 150$$

Divide both sides by $15$:

$$t = 10$$

Check: after 10 minutes, Tank A holds $50 + 6 \times 10 = 110$ litres and Tank B holds $200 - 9 \times 10 = 110$ litres $\quad \checkmark$

Number of game booklets sold: $g$.

Number of food booklets sold: $(50 - g)$.

Total money collected is $720:

$$12g + 18(50 - g) = 720$$

Expand the bracket:

$$12g + 900 - 18g = 720$$

Collect like terms:

$$-6g + 900 = 720$$

Subtract $900$ from both sides:

$$-6g + 900 - 900 = 720 - 900$$

$$-6g = -180$$

Divide both sides by $-6$:

$$g = \frac{-180}{-6} = 30$$

Game booklets: 30; food booklets: $50 - 30 = 20$.

Check: 30 game booklets bring in $30 \times 12 = 360$ dollars; 20 food booklets bring in $20 \times 18 = 360$ dollars; total is $360 + 360 = 720$ dollars $\quad \checkmark$

Number of plates of chicken rice: $c$.

Number of bowls of fishball noodles: $(14 - c)$.

Total cost is $60:

$$4.5c + 4(14 - c) = 60$$

Expand the bracket:

$$4.5c + 56 - 4c = 60$$

Collect like terms:

$$0.5c + 56 = 60$$

Subtract $56$ from both sides:

$$0.5c + 56 - 56 = 60 - 56$$

$$0.5c = 4$$

Divide both sides by $0.5$:

$$c = \frac{4}{0.5} = 8$$

Plates of chicken rice: 8; bowls of fishball noodles: $14 - 8 = 6$.

Check: 8 plates of chicken rice cost $8 \times 4.5 = 36$ dollars; 6 bowls of noodles cost $6 \times 4 = 24$ dollars; total is $36 + 24 = 60$ dollars $\quad \checkmark$

Convert 10 minutes to hours: $10 \text{ minutes} = \frac{10}{60} = \frac{1}{6} \text{ hour}$.

Journey to the airport: $t$ hours at 60 km/h, so the distance is $60t$ km.

Journey back: $\left(t + \frac{1}{6}\right)$ hours at 45 km/h, so the distance is $45\left(t + \frac{1}{6}\right)$ km.

The distance is the same both ways:

$$60t = 45\left(t + \frac{1}{6}\right)$$

Expand the bracket, using $45 \times \frac{1}{6} = \frac{45}{6} = 7.5$:

$$60t = 45t + 7.5$$

Subtract $45t$ from both sides:

$$60t - 45t = 45t + 7.5 - 45t$$

$$15t = 7.5$$

Divide both sides by $15$:

$$t = \frac{7.5}{15} = 0.5$$

The journey to the airport takes 0.5 hours (30 minutes). Distance $= 60 \times 0.5 = 30$ km.

Check: the way back takes $0.5 + \frac{1}{6} = \frac{2}{3}$ hour (40 minutes, which is 10 minutes longer), and $45 \times \frac{2}{3} = 30$ km $\quad \checkmark$