Chapter 9
Basic Engineering Mathematics · 18 exercises
Problem 1
Solve the following equations for \(x\) and \(y\), (a) by substitution, and (b) by elimination: $$ \begin{aligned} x+2 y &=-1 \\ 4 x-3 y &=18 \end{aligned} $$
6 step solution
Problem 2
Solve, by a substitution method, the simultaneous equations $$ \begin{aligned} 3 x-2 y &=12 \\ x+3 y &=-7 \end{aligned} $$
5 step solution
Problem 3
Use an elimination method to solve the simultaneous equations $$ \begin{aligned} &3 x+4 y=5 \\ &2 x-5 y=-12 \end{aligned} $$
7 step solution
Problem 4
Solve $$ \begin{aligned} &7 x-2 y=26 \\ &6 x+5 y=29 \end{aligned} $$
6 step solution
Problem 5
Solve $$ \begin{aligned} &3 p=2 q \\ &4 p+q+11=0 \end{aligned} $$
5 step solution
Problem 6
Solve $$ \begin{aligned} &\frac{x}{8}+\frac{5}{2}=y \\ &13-\frac{y}{3}=3 x \end{aligned} $$
5 step solution
Problem 7
Solve $$ \begin{aligned} &2.5 x+0.75-3 y=0 \\ &1.6 x=1.08-1.2 y \end{aligned} $$
6 step solution
Problem 8
Solve $$ \begin{aligned} &\frac{2}{x}+\frac{3}{y}=7 \\ &\frac{1}{x}-\frac{4}{y}=-2 \end{aligned} $$
7 step solution
Problem 9
Solve $$ \begin{gathered} \frac{1}{2 a}+\frac{3}{5 b}=4 \\ \frac{4}{a}+\frac{1}{2 b}=10.5 \end{gathered} $$
9 step solution
Problem 10
Solve $$ \begin{gathered} \frac{1}{x+y}=\frac{4}{27} \\ \frac{1}{2 x-y}=\frac{4}{33} \end{gathered} $$
7 step solution
Problem 11
Solve $$ \begin{aligned} &\frac{x-1}{3}+\frac{y+2}{5}=\frac{2}{15} \\ &\frac{1-x}{6}+\frac{5+y}{2}=\frac{5}{6} \end{aligned} $
6 step solution
Problem 12
The law connecting friction \(F\) and load \(L\) for an experiment is of the form \(F=a L+b\), where \(a\) and \(b\) are constants. When \(F=5.6, L=8.0\) and when \(F=4.4\), \(L=2.0\). Find the values of \(a\) and \(b\) and the value of \(F\) when \(L=6.5\).
4 step solution
Problem 13
The equation of a straight line, of gradient \(m\) and intercept on the \(y\)-axis \(c\), is \(y=m x+c\). If a straight line passes through the point where \(x=1\) and \(y=-2\), and also through the point where \(x=3 \frac{1}{2}\) and \(y=10 \frac{1}{2}\), find the values of the gradient and the \(y\)-axis intercept.
5 step solution
Problem 14
When Kirchhoff's laws are applied to the electrical circuit shown in Figure \(9.1\) the currents \(I_{1}\) and \(I_{2}\) are connected by the equations: $$ \begin{aligned} 27 &=1.5 I_{1}+8\left(I_{1}-I_{2}\right) \\ -26 &=2 I_{2}-8\left(I_{1}-I_{2}\right) \end{aligned} $$ Solve the equations to find the values of currents \(I_{1}\) and \(I_{2}\)
6 step solution
Problem 15
The distance \(s\) metres from a fixed point of a vehicle travelling in a straight line with constant acceleration, \(a \mathrm{~m} / \mathrm{s}^{2}\), is given by \(s=u t+\frac{1}{2} a t^{2}\), where \(u\) is the initial velocity in \(\mathrm{m} / \mathrm{s}\) and \(t\) the time in seconds. Determine the initial velocity and the acceleration given that \(s=42 \mathrm{~m}\) when \(t=2 \mathrm{~s}\) and \(s=144 \mathrm{~m}\) when \(t=4 \mathrm{~s}\). Find also the distance travelled after \(3 \mathrm{~s}\).
4 step solution
Problem 16
A craftsman and 4 labourers together earn £865 per week, whilst 4 craftsmen and 9 labourers earn £2340 basic per week. Determine the basic weekly wage of a craftsman and a labourer.
7 step solution
Problem 17
The resistance \(R \Omega\) of a length of wire at \(t^{\circ} \mathrm{C}\) is given by \(R=R_{0}(1+\alpha t)\), where \(R_{0}\) is the resistance at \(0^{\circ} \mathrm{C}\) and \(\alpha\) is the temperature coefficient of resistance in \({ }^{\circ} \mathrm{C}\). Find the values of \(\alpha\) and \(R_{0}\) if \(R=30 \Omega\) at \(50^{\circ} \mathrm{C}\) and \(R=\) \(35 \Omega\) at \(100^{\circ} \mathrm{C}\).
8 step solution
Problem 18
The molar heat capacity of a solid compound is given by the equation \(c=a+b T\), where \(a\) and \(b\) are constants. When \(c=52, T=100\) and when \(c=172, T=400\). Determine the values of \(a\) and \(b\).
4 step solution