Linearly Related Variables, Linear Equations and Inequations - Questions
The solution to \(\dfrac{{x + 2}}{3} - \dfrac{x}{2} + \dfrac{{x + 1}}{4} = 1\) is?
\begin{aligned}
\frac{x+2}{3}-\frac{x}{2}+\frac{x+1}{4}&=1 \\
\textcolor{red}{12} \times \frac{x+2}{3}-\textcolor{red}{12} \times \frac{x}{2}+\textcolor{red}{12} \times \frac{x+1}{4}&=\textcolor{red}{12} \times 1 \\
4(x+2)-6 x+3(x+1)&=12 \\
4 x+8-6 x+3 x+3&=12 \\
x+11&=12 \\
x&=1
\end{aligned}
Given \(\dfrac{{x + a}}{b} - 1 = \dfrac{{x - b}}{a}\) then \(x = \)?
\begin{aligned}
\frac{x+a}{b}-\frac{b}{b} &=\frac{x-b}{a} \\
\frac{x+a-b}{b} &=\frac{x-b}{a} \\
a(x+a-b) &=b(x-b)\\
ax+a(a-b)&=b x-b^{2} \\
ax-b x&=-b^{2}-a(a-b) \\
x(a-b)&=-b^{2}-a^{2}+a b \\
\therefore x&=\frac{-a^{2}-b^{2}+a b}{a-b} \\
\text { or } x&=\frac{a^{2}+b^{2}-a b}{b-a}
\end{aligned}
The solutions to the equations \(4x - 3y = - 38\) and \(5x + 2y = - 13\) are?
\begin{align}
&\begin{aligned}
4x-3y&=-38 \cdots (1) \\
5x+2y&=-13 \cdots (2) \\
(1) \times 2 \quad \;\; 8x-6y&=-76 \cdots (3) \\
(2) \times 3 \quad 15x+6y&=-39 \cdots (4)\\
\end{aligned}\\
&\begin{aligned}
(3)+(4) \quad 23x&=-115 \\
x&=-5 \\
\end{aligned}\\
&\begin{aligned}
\text {In }(2) \quad -25+2y &=-13 \\
2y &=12 \\
y &=6 \\
\end{aligned}\\
&\therefore x=-5, \; y=6
\end{align}
A cyclist travels at an average speed of \({\text{15 km/h}}\). Twenty minutes later, a motorist starting at the same place travels along the same road at an average speed of \({\text{45 km/h}}\). How many minutes would the cyclist have travelled before being overtaken by the motorist?
\begin{align}
&\text{Let }d= \text{the distance travelled by the cyclist and the motorist}\\
&\begin{aligned}
t+\frac{1}{3}&= \text{time taken for the cyclist } (20 \mathrm{~min}=\frac{1}{3} \mathrm{~hour}) \\
t&=\text{time taken for the motorist}\\
\end{aligned}\\
&\text{For cyclist } \;\;\;d=15\left(t+\frac{1}{3}\right)\\
&\text{For motorist } d=45t\\
&\qquad\qquad \quad \begin{aligned}
45t&=15\left(t+\frac{1}{3}\right)\\
45t&=15 t+5 \\
30t&=5 \\
t&=\frac{5}{30}=\frac{1}{6} \text { hour }
\end{aligned}\\
& \begin{aligned}
\therefore \text {time for cyclist }&=\frac{1}{6}+\frac{1}{3} \\
&=\frac{1}{2}=30 \mathrm{~mins.}
\end{aligned}
\end{align}