Chapter 6.1, Problem 50HP

To determine

Find a system of equation which is different from the other three and also explain the reason.

Answer to Problem 50HP

  $\begin{array}{l}4x+2y=14\\ 12x+6y=18\end{array}$ is different from the other three because the system of equation is inconsistent means the lines are parallel.

Explanation of Solution

Given information:

it is given in the question that,

the linear equations are

  $\begin{array}{l}4x-y=5\\ -2x+y=-1\end{array}$ , $\begin{array}{l}-x+4y=8\\ 3x-6y=6\end{array}$ , $\begin{array}{l}4x+2y=14\\ 12x+6y=18\end{array}$ and $\begin{array}{l}3x-2y=1\\ 2x+3y=18\end{array}$ .

Formula used:

By using the concept of consistent and inconsistent for find the nature of line that is,

  $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ (for consistent nature of line)

  $\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ (for inconsistent nature of line)

Where,

  $a_1$ , $a_2$ are the co-efficient of $x$ .

  $b_1$ ,

  $b_2$ are the co-efficient of $y$ and,

  $c_1$ , $c_2$ are the constant term.

Calculation:

  $\begin{array}{l}4x-y=5\\ -2x+y=-1\end{array}$ Here, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\Rightarrow \frac{4}{-2}\ne \frac{-1}{1}$ so lines are consistent.

  $\begin{array}{l}-x+4y=8\\ 3x-6y=6\end{array}$ Here, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\Rightarrow \frac{-1}{3}\ne \frac{4}{-6}$ so lines are consistent.

  $\begin{array}{l}4x+2y=14\iff 4x+2y-14=0\\ 12x+6y=18\iff 12x+6y-18=0\end{array}$

Here,

  $\begin{array}{l}\frac{a_1}{a_2}=\frac{b_1}{b_2}\Rightarrow \frac{4}{12}=\frac{2}{6}\\ \Rightarrow \frac{1}{3}=\frac{1}{3}\end{array}$

So, use the second formula that is $\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ ,

  $\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\Rightarrow \frac{2}{6}\ne \frac{-14}{-18}$ (so the line are inconsistent)

  $\begin{array}{l}3x-2y=1\\ 2x+3y=18\end{array}$ Here, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\Rightarrow \frac{3}{2}\ne \frac{-2}{3}$ so lines are consistent.