Sec 1 Maths: Linear Equations & Inequalities — practice questions & worked solutions

“Half the sum of $3x$ and $5$” is $\tfrac{1}{2}(3x+5)$; “$5$ subtracted from $4x$” is $4x-5$:

$$\frac{1}{2}(3x+5) = 4x-5$$

A pen costs $\tfrac{1}{4}x$ and a ruler costs $(x-3)$:

$$\begin{aligned}3x+4\left(\frac{1}{4}x\right)+2(x-3) &= 18 \\ 3x+x+2x-6 &= 18 \\ 6x &= 18+6 \\ x &= \frac{24}{6} \\ x &= 4\end{aligned}$$

Cost of one ruler $= x-3 = 4-3 = 1$.

$$\begin{aligned}\text{Spent on 2 rulers} &= 2 \times 1 \\ &= \$2\end{aligned}$$

(i)

$$\begin{aligned}0.70x+1.05y&=62.30\\ 70x+105y&=6230\\ 2x+3y&=178\end{aligned}$$

(ii) Second equation:

$$x+y=73$$

Multiply by $2$:

$$2x+2y=146$$

Subtract from $2x+3y=178$:

$$y=32$$

Then $x=73-32=41$. So $41$ cups of Refreshing and $32$ cups of Cheerful.

(i)

$$\frac{m+8}{n} = \frac{1}{2}$$
$$\frac{1}{5}(m+n) = 5$$

(ii) From the first equation:

$$2(m+8) = n$$
$$n = 2m+16$$

From the second equation:

$$m+n = 25$$

Substitute:

$$\begin{aligned}m+(2m+16) &= 25 \\ 3m &= 9 \\ m &= 3 \\ n &= 22\end{aligned}$$

The original fraction is $\dfrac{3}{22}$.

(a)(i) Substitute $x=12$:

$$\begin{aligned}p &= 4+0.55(12) \\ &= 10.6\end{aligned}$$

(iii) The number $4$ is the fixed starting (flag-down) charge of $\$$4 before any distance is travelled.

(b)(i) The number $0.3$ is the rate of charge: the fare increases by $\$$0.30 for every additional $1$ km travelled.

(ii) Solving the two equations:

$$\begin{aligned}4+0.55x &= 6+0.3x \\ 0.25x &= 2 \\ x &= 8 \\ y &= 4+0.55(8) \\ &= 8.4\end{aligned}$$

The lines intersect at $(8,\,8.40)$.

(iii) For journeys shorter than $8$ km, Company A is cheaper; for journeys longer than $8$ km, Company B is cheaper; at exactly $8$ km both cost $\$$8.40.

Rearrange the second equation:

$$\begin{aligned}8(y + 3) &= 4x – 9 \\ 8y + 24 &= 4x – 9 \\ 4x – 8y &= 33\end{aligned}$$

Label the equations:

$$\begin{aligned}2x + 8y &= 21 \quad (1) \\ 4x – 8y &= 33 \quad (2)\end{aligned}$$

Add $(1)$ and $(2)$:

$$\begin{aligned}6x &= 54 \\ x &= 9\end{aligned}$$

Substitute into $(1)$:

$$\begin{aligned}2(9) + 8y &= 21 \\ 8y &= 3 \\ y &= \frac{3}{8}\end{aligned}$$

(i)(a) Reading the graph at $x = 5.5$:

$$y = 26.50$$

The manufacturing cost is $\$$26.50.

(i)(b) Reading the graph at $y = 34$:

$$x = 8 \text{ kg}$$

(i)(c) The line passes through $(0, 10)$ and $(10, 40)$:

$$\begin{aligned}a &= \frac{40 – 10}{10 – 0} \\ &= 3\end{aligned}$$

(i)(d) $a$ represents the manufacturing cost per kilogram of Spice P, that is $\$$3 per kg.

(ii)(a) For $y = x + 20$: when $x = 7$, $y = 27$; when $x = 10$, $y = 30$. The missing $x$ value (where $y = 27$) is $7$.

(ii)(b) Plot $(0, 20)$, $(7, 27)$, $(10, 30)$ and join with a straight line labelled $y = x + 20$.

(iii)(a) Solving $3x + 10 = x + 20$ gives $2x = 10$, so $x = 5$, $y = 25$. The solution is $x = 5$, $y = 25$.

(iii)(b) The point of intersection $(5, 25)$ is the mass of Spice P ($5$ kg) for which both companies charge the same cost ($\$$25).

(i) Multiply throughout by $6$:

$$\begin{aligned}3(2x – 7) + 2(10 – x) &= 30 – (4x – 1) \\ 6x – 21 + 20 – 2x &= 30 – 4x + 1 \\ 4x – 1 &= 31 – 4x \\ 8x &= 32 \\ x &= 4\end{aligned}$$

(ii) The second equation is the first with $x$ replaced by $2y$, so

$$\begin{aligned}2y &= 4 \\ y &= 2\end{aligned}$$

(i)(a) Senior served $(x + 50)$ customers.

$$\text{Senior} = 200 + 0.16(x + 50) = \$(208 + 0.16x)$$

(i)(b)

$$\text{Junior} = \$(150 + 0.1x)$$

(ii)

$$\begin{aligned}(208 + 0.16x) – (150 + 0.1x) &= 76 \\ 58 + 0.06x &= 76 \\ 0.06x &= 18 \\ x &= 300\end{aligned}$$

Senior served $300 + 50 = 350$ customers.

(i) Reading from the graph at $y = 110$:

$$x = 400 \text{ units}$$

(ii)(a) $c$ is the $y$-intercept.

$$c = 10$$

It represents the basic/fixed cost incurred even if no electricity is used.

(ii)(b) Gradient using two points, e.g. $(0, 10)$ and $(500, 135)$:

$$m = \frac{135 – 10}{500 – 0} = \frac{125}{500} = 0.25$$

It represents the cost of $\$$0.25 for each unit of electricity used.

(i) Table of values for $y = 5x – 2$:

$$\begin{aligned}x = -2 &: y = -12 \\ x = -1 &: y = -7 \\ x = 0 &: y = -2 \\ x = 1 &: y = 3 \\ x = 2 &: y = 8\end{aligned}$$

Plot these points and join with a straight line.

(ii) Solving $5x – 2 = 5.4$:

$$\begin{aligned}5x &= 7.4 \\ x &= 1.48 \\ A &= (1.475, 5.4) \text{ (or values read from the graph)}\end{aligned}$$

(iii) Plot $B(1, 1)$ and $C(2, 1)$.

(iv) $BC$ is horizontal of length $2 – 1 = 1$ unit (base). Height is the vertical distance from $A$ to line $BC$ ($y = 1$): $5.4 – 1 = 4.4$.

$$\text{Area} = \frac{1}{2} \times 1 \times 4.4 = 2.2 \text{ units}^{2}$$

Multiply throughout by $6$:

$$\begin{aligned}3(3x+2)-(2-x)&=2(2x+3) \\ 9x+6-2+x&=4x+6 \\ 10x+4&=4x+6 \\ 6x&=2 \\ x&=\frac{1}{3}\end{aligned}$$

Both expressions equal Bruno’s total money:

$$\begin{aligned}12x+2&=16(x-2)+6 \\ 12x+2&=16x-32+6 \\ 12x+2&=16x-26 \\ 28&=4x \\ x&=7\end{aligned}$$

Amount of money:

$$12(7)+2=\$86$$

Money remaining after 2 textbooks:

$$\begin{aligned}86-2\times12&=86-24 \\ &=\$62\end{aligned}$$

Number of workbooks:

$$\frac{62}{16}=3.875$$

Since only whole workbooks can be bought, he can buy $3$ workbooks.

From the graph, $L_1$ cuts the $y$-axis at $3$ and passes through $(2,0)$. Gradient:

$$\begin{aligned}m&=\frac{0-3}{2-0} \\ &=-\frac{3}{2}\end{aligned}$$

Equation:

$$y=-\frac{3}{2}x+3$$

Substitute $x=3$ into $y=0.5x+1$:

$$\begin{aligned}y&=0.5(3)+1 \\ &=2.5\end{aligned}$$

Since the computed $y$-value equals $2.5$, the point $(3,\,2.5)$ lies on $L_2$. (Yes)

Plot $L_2:y=0.5x+1$ through the endpoints:

$$\begin{aligned}x=-4&:\quad y=0.5(-4)+1=-1 \\ x=4&:\quad y=0.5(4)+1=3\end{aligned}$$

Draw the straight line joining $(-4,\,-1)$ and $(4,\,3)$.

$L_1$ meets the $x$-axis at $(2,\,0)$ and $L_2$ meets the $x$-axis at $(-2,\,0)$, giving a base of $4$ units. The two lines intersect where $-\tfrac{3}{2}x+3=0.5x+1$, i.e. $x=1$, $y=1.5$, so the height is $1.5$ units.

$$\begin{aligned}\text{Area}&=\frac{1}{2}\times4\times1.5 \\ &=3\text{ units}^2\end{aligned}$$

(a) Multiply throughout by 30:

$$\begin{aligned}10(2m – 3) &= -6(1 – 2m) + 15 \\ 20m – 30 &= -6 + 12m + 15 \\ 20m – 30 &= 12m + 9 \\ 8m &= 39 \\ m &= \frac{39}{8}\end{aligned}$$

(b)

$$\begin{aligned}3(3x – 4y) &= -2(7x + 2y) \\ 9x – 12y &= -14x – 4y \\ 23x &= 8y \\ \frac{x}{y} &= \frac{8}{23}\end{aligned}$$

(i) Reading off the graph at $d = 5$:

$$C = \$275$$

(ii) Reading off the graph at $C = 140$ gives $d \approx 2$. The maximum whole number of days is

$$d = 2 \text{ days}$$

(iii) The graph is a straight line. The $C$-intercept (fixed charge) is

$$p = 50$$

Gradient (charge per day): using $(0, 50)$ and $(5, 275)$,

$$\begin{aligned}q &= \frac{275 – 50}{5} \\ &= 45\end{aligned}$$

So $p = 50$ and $q = 45$.

(iv) Company A: $C = 45d + 50$; Company B: $C = 51d + 20$.

$$\begin{aligned}45d + 50 &< 51d + 20 \\ 30 &< 6d \\ d &> 5\end{aligned}$$

Company A is cheaper only when $d > 5$ (i.e. renting for 6 or more days). For 5 days or fewer, Company B is cheaper or equal, so Lewis is wrong; A is not always cheaper.

(i) From the graph, $L_1$ has $y$-intercept 1 and passes through $A(-1,-2)$. Gradient:

$$\begin{aligned}m &= \frac{1 – (-2)}{0 – (-1)} \\ &= 3\end{aligned}$$
$$y = 3x + 1$$

(ii) $L_2$: $y = -x – 3$ has $x$-intercept $(-3, 0)$ and $y$-intercept $(0, -3)$. $L_3$: $x = 1.5$ is a vertical line. Draw both on the grid (intercepts labelled).

(iii) The region bounded by $L_1$, $L_2$, $L_3$ and the $y$-axis is a

$$\text{trapezium}$$

(i) Combine the left side over $3x-1$:

$$\begin{aligned}\frac{-2(3x-1)+5+6x}{3x-1} &= -\frac{4}{x+5}\\ \frac{-6x+2+5+6x}{3x-1} &= -\frac{4}{x+5}\\ \frac{7}{3x-1} &= -\frac{4}{x+5}\end{aligned}$$

Cross-multiply:

$$\begin{aligned}7(x+5) &= -4(3x-1)\\ 7x+35 &= -12x+4\\ 19x &= -31\\ x &= -\frac{31}{19}\end{aligned}$$

(ii) Replacing $x$ with $y^2$ gives the second equation, so by part (i):

$$y^2 = -\frac{31}{19}$$

Since a square can never be negative, there is no real value of $y$. Hence the equation has no real solutions.

(i) Shop A line passes through $(0, 30)$ and $(50, 190)$.

$$\begin{aligned}\text{gradient} &= \frac{190 – 30}{50 – 0}\\ &= \frac{160}{50}\\ &= 3.2\end{aligned}$$

(ii) The cost of each metre of poplin cloth from Shop A is $\$$3.20.

(iii) With $y$-intercept (delivery fee) $30$:

$$C = 3.2x + 30$$

(iv) Reading from the graphs at $x = 44$, Shop B costs about $\$$30 more than Shop A. Additional cost $\approx \$30$.

(v) Not accurate. The two lines intersect at about $x = 14$; for lengths shorter than the intersection point, Shop B is cheaper, but for longer lengths Shop A is cheaper. Hence Shop B is not always cheaper.

(i) Rewrite in the form $y = mx + c$:

$$\begin{aligned}2y &= 10 – 2x\\ y &= 5 – x\end{aligned}$$

Gradient $= -1$, $y$-intercept $= 5$.

(ii) Substitute $x = 1$:

$$\begin{aligned}y &= 5 – 1\\ y &= 4\end{aligned}$$

Since $4 \neq 3$, the point $(1, 3)$ does not lie on the line.

(iii) $A$ is the $y$-intercept, so $A = (0, 5)$ and $O = (0, 0)$; $OA = 5$ lies along the $y$-axis. For isosceles triangle $OAB$ with $OB = AB$, $B$ is level with the midpoint of $OA$, so $k = 2.5$. Area $= \frac{1}{2} \times OA \times h$:

$$\begin{aligned}10 &= \frac{1}{2} \times 5 \times h\\ 10 &= 2.5h\\ h &= 4\end{aligned}$$

Hence $h = 4$ and $k = \frac{5}{2}$.

(a) Left part, $6x – 40 < -2x – 7$:

$$\begin{aligned}8x &< 33 \\ x &< 4\tfrac{1}{8}\end{aligned}$$

Right part, $-2x – 7 \le 4x + 2$:

$$\begin{aligned}-9 &\le 6x \\ x &\ge -1\tfrac{1}{2}\end{aligned}$$

Combined:

$$-1\tfrac{1}{2} \le x < 4\tfrac{1}{8}$$

(b) A number line with a filled circle at $-1\tfrac{1}{2}$, an open circle at $4\tfrac{1}{8}$, and the segment between them shaded.

(c) The primes in $-1\tfrac{1}{2} \le x < 4\tfrac{1}{8}$ are $2$ and $3$, so the smallest prime value is $2$.

(a) Cost of one pencil $= \dfrac{x}{40}$ cents.

(b) Number of erasers $= 40 + 10 = 50$, so cost of one eraser $= \dfrac{x}{50}$ cents.

(c)(i) An eraser costs $5$ cents less than a pencil:

$$\frac{x}{40} – \frac{x}{50} = 5$$

(c)(ii) Multiply through by $200$:

$$\begin{aligned}5x – 4x &= 1000 \\ x &= 1000\end{aligned}$$

Cost of one pencil $= \dfrac{1000}{40} = 25$ cents.

(a) Plot $(-1,8)$, $(0,5)$, $(1,2)$, $(2,-1)$ and join with a straight line.

(b) Reading from the line at $x = -0.5$:

$$d = 6.5$$

(c)(i) Using points $(0,5)$ and $(1,2)$:

$$\begin{aligned}\text{gradient} &= \frac{2 – 5}{1 – 0} \\ &= -3\end{aligned}$$

(c)(ii) The line cuts the $y$-axis at $5$:

$$y = -3x + 5$$

(a) From $1-2x\le 5+x$:

$$\begin{aligned}1-5 &\le x+2x\\ -4 &\le 3x\\ -\tfrac{4}{3} &\le x\end{aligned}$$

From $5+x<13-x$:

$$\begin{aligned}x+x &< 13-5\\ 2x &< 8\\ x &< 4\end{aligned}$$

Combining:

$$-1\tfrac{1}{3}\le x<4$$

(b) Smallest integer: $x=-1$.

(c) Gradient:

$$\begin{aligned}a &= \frac{11 – (-3)}{3 – (-4)} \\ &= 2\end{aligned}$$

Using $(-1, 3)$:

$$\begin{aligned}3 &= 2(-1) + b \\ b &= 5\end{aligned}$$

So $y = 2x + 5$.

(b)(i) At the $x$-intercept, $y = 0$:

$$\begin{aligned}0 &= 2x + 5 \\ x &= -2.5\end{aligned}$$