Sec 4 E Maths: Equations and Inequalities, practice questions & worked solutions

$$\begin{aligned}\dfrac{3x-5}4-\dfrac x6&=3\\\dfrac{9x-15-2x}{12}&=3\\\dfrac{7x-15}{12}&=3\\7x-15&=36\\x&=7\dfrac27\end{aligned}$$

Answer: $x=7\dfrac27$

$$\frac{3x-2}{7}>1+5x$$

($\times7$ throughout)

$$\begin{aligned}3x-2&>7+35x\\-32x&>9\\x&<-\frac9{32}\end{aligned}$$

Answer: $x<-\dfrac9{32}$

Let $x$ be the number of additional girls needed to join the club.

$$\dfrac{48+x}{48+72+x}>\dfrac{45}{100}\qquad\text{or}\qquad\dfrac{48+x}{48+72+x}>\dfrac9{20}$$

$$\begin{aligned}9(120+x)&<960+20x\\1080+9x&<960+20x\\11x&>120\end{aligned}$$

$x>10.91$,  least number is 11

Answer: $11$

(a)

$$\begin{aligned}-8&\le2-3x&\text{and}\quad2-3x&<7-\frac12x\\-10&\le-3x&\text{and}\quad-5&<2\frac12x\end{aligned}$$

$$-2<x\le3\frac13.$$

(b)

$-1,0,1,2,3$.

Answer: (a) $-2<x\le3\dfrac13$ (b) $-1,0,1,2,3$

$$\begin{aligned}\frac x3&<\frac14(x+3)&\qquad&\text{and}\qquad&\frac14(x+3)&<5\\4x&<3x+9&&&x+3&<20\\x&<9&&&x&<17\end{aligned}$$

$$x<9$$

Answer: $x<9$

$$\frac{x-3}{3}-\frac{1-x^2}{2-3x}=1$$

$$(x-3)(2-3x)-3(1-x^2)=3(2-3x)$$

$$2x-3x^2-6+9x-3+3x^2=6-9x$$

$$11x-9=6-9x$$

$$20x=15$$

$$x=\frac34\quad\text{or }0.75$$

Answer: $x=\dfrac34\text{ or }0.75$

$$\begin{aligned}&2x+y=3x+5y+3\\[0pt]&x=-4y-3\qquad\quad-(1)\\[0pt]&4x+5y-7=x+y\\[0pt]&3x+4y=7\qquad\quad-(2)\end{aligned}$$

Sub (1) into (2)

$$\begin{aligned}&3(-4y-3)+4y=7\\[0pt]&{-}12y-9+4y=7\\[0pt]&{-}8y=16\\[0pt]&y=-2\end{aligned}$$

Sub $y=-2$ into (1)

$$\begin{aligned}&x=-4(-2)-3\\[0pt]&x=5\end{aligned}$$

Answer: $x=5$, $y=-2$

$$\begin{aligned}36x+20y&=5760&&\quad\text{(1)}\\36x-20y&=1800&&\quad\text{(2)}\end{aligned}$$

(1)+(2),

$$\begin{aligned}36x+20y+36x-20y&=5760+1800\\72x&=7560\\x&=105\end{aligned}$$

From (1),

$$\begin{aligned}36(105)+20y&=5760\\3780+20y&=5760\\20y&=1980\\y&=99\end{aligned}$$

The total number of dog walks $=105+99=204$.

Answer: $204$

(a)

$$\begin{aligned}5x+y&=16y-x\\5x+y&=4x+5y-3\\16y-x&=4x+5y-3\end{aligned}$$

(b)

$$\begin{aligned}5x+y&=16y-x\\6x&=15y&&-(1)\\5x+y&=4x+5y-3\\x&=4y-3&&-(2)\end{aligned}$$

sub $(2)$ into $(1)$

$$\begin{aligned}6(4y-3)&=15y\\24y-18&=15y\\9y&=18\\y&=2\end{aligned}$$

Sub $y=2$ into $(2)$

$$\begin{aligned}x&=4(2)-3\\x&=5\end{aligned}$$

$\therefore x=5,\ y=2$

$$\begin{aligned}AC&=4(5)+5(2)-3\\&=27\text{ cm}\end{aligned}$$

Perpendicular ht from $B$ to $AC$

$$\begin{aligned}&=\sqrt{27^2-\left(\frac{27}2\right)^2}\\&=23.382\text{ cm}\end{aligned}$$

$$\begin{aligned}\text{Area of }ABC&=\frac12\times27\times23.382\\&=315.657\\&=316\text{ cm}^2\ (\text{3sf})\end{aligned}$$

OR

$$\begin{aligned}\angle ABC&=60^\circ\ (\text{angles of an equilateral }\Delta)\\\text{Area of }ABC&=\frac12\times27\times27\times\sin60^\circ\\&=315.666\\&=316\text{ cm}^2\ (\text{3sf})\end{aligned}$$

Answer: (a) $5x+y=16y-x$, $5x+y=4x+5y-3$ (b) $x=5,\ y=2$; $316$ cm$^2$

(a)

20

(b)(i)

$X=pn^2+qn$

When $n=4$ and $X=2$,

$$\begin{aligned}2&=p(4)^2+q(4)\\2&=16p+4q\end{aligned}$$

When $n=5$ and $X=5$,

$$\begin{aligned}5&=p(5)^2+q(5)\\5&=25p+5q\end{aligned}$$

(b)(ii)

$$\begin{aligned}16p+4q&=2\\80p+20q&=10\quad\quad\text{(1)}\\25p+5q&=5\\100p+20q&=20\quad\quad\text{(2)}\end{aligned}$$

(2)$-$(1)

$$\begin{aligned}20p&=10\\p&=0.5\\q&=-1.5\end{aligned}$$

(b)(iii)

$$\begin{aligned}495&=0.5n^2-1.5n\\0&=-495-1.5n+0.5n^2\\0&=n^2-3n-990\\(n-33)(n+30)&=0\\n=33\quad\text{or}\quad n&=-30\ \text{(rej)}\end{aligned}$$

Since number of vertices is an integer, it is possible to have a $n$-sided polygon with 495 diagonals.

Answer: (a) $a=20$   (b)(ii) $p=0.5$, $q=-1.5$   (iii) Yes, $n=33$ is a positive integer, so a 33 sided polygon has 495 diagonals