Chapter 16.9, Problem 18WE

To determine

To prove: For the system of equations

  $\begin{array}{c}y=m_{1}x+b_1\\ y=m_{2}x+b_2\end{array}$

Thecondition $D=0\text{ and }{D}_y\ne 0$ if and only if $m_1=m_2\ne 0\text{ and }{b}_1\ne b_2$ holds true.

Explanation of Solution

Given information:

The system of equations

  $\begin{array}{c}y=m_{1}x+b_1\\ y=m_{2}x+b_2\end{array}$

Proof:

Consider system of equations

  $\begin{array}{c}y=m_{1}x+b_1\\ y=m_{2}x+b_2\end{array}$

Write the above system in standard form,

  $\begin{array}{c}-m_{1}x+y=b_1\\ -m_{2}x+y=b_2\end{array}$

The determinant of above system of equations is,

  $\begin{array}{c}D=\left|\begin{array}{cc}-m_1& 1\\ -m_2& 1\end{array}\right|\\ =-m_1+m_2\end{array}$

And,

  $\begin{array}{c}{D}_y=\left|\begin{array}{cc}-m_1& b_1\\ -m_2& b_2\end{array}\right|\\ =-b_2{m}_1+b_1{m}_2\end{array}$

If suppose $D=0\text{ and }{D}_y\ne 0$ then,

  $\begin{array}{c}-m_1+m_2=0\\ m_1=m_2\end{array}$

That is, $m_1=m_2\ne 0$ for $y=\frac{D_y}{D}$ to be valid.

And,

  $\begin{array}{c}-b_2{m}_1+b_1{m}_2\ne 0\\ b_1\ne b_2\end{array}$

Because $m_1=m_2$ .

Next suppose $m_1=m_2\ne 0\text{ and }{b}_1\ne b_2$ so,

  $\begin{array}{c}{m}_1=m_2\\ -m_1+m_2=0\\ D=0\end{array}$

And,

  $\begin{array}{c}{b}_1\ne b_2\\ -b_2{m}_1+b_1{m}_2\ne 0\\ D_y\ne 0\end{array}$

Hence, the condition $D=0\text{ and }{D}_y\ne 0$ if and only if $m_1=m_2\ne 0\text{ and }{b}_1\ne b_2$ holds true for given system of equations.