Chapter 6.1, Problem 40PPE

a.

To determine

To write: the second equation so that the system has one solution

Answer to Problem 40PPE

  $2x-5y+6=0$

Explanation of Solution

Given information: One equation in a system is $y=\frac{x}{2}-1$

One equation in a system is $y=\frac{x}{2}-1$

This equation can be written as follows:

  $\begin{array}{l}y=\frac{x}{2}-1\\ \Rightarrow y=\frac{x-2}{2}\\ \Rightarrow 2y=x-2\\ \Rightarrow x-2y-2=0\end{array}$

The system of equations: $a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$ has one solution if $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

Take the second equation as $2x-5y+6=0$

Here,

  $\begin{array}{l}{a}_1=1,b_1=-2,c_1=-2\\ a_2=2,b_2=-5,c_2=6\end{array}$

  $\begin{array}{l}\frac{a_1}{a_2}=\frac{1}{2}\\ \frac{b_1}{b_2}=\frac{-2}{-5}=\frac{2}{5}\end{array}$

So,

  $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

The system of equations: $y=\frac{x}{2}-1,2x-5y+6=0$ has one solution.

b.

To determine

To write:the second equation so that the system has no solution

Answer to Problem 40PPE

  $2x-4y+6=0$

Explanation of Solution

Given information:One equation in a system is $y=\frac{x}{2}-1$

One equation in a system is $y=\frac{x}{2}-1$

This equation can be written as follows:

  $\begin{array}{l}y=\frac{x}{2}-1\\ \Rightarrow y=\frac{x-2}{2}\\ \Rightarrow 2y=x-2\\ \Rightarrow x-2y-2=0\end{array}$

The system of equations: $a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$ has no solution if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Take the second equation as $2x-4y+6=0$

Here,

  $\begin{array}{l}{a}_1=1,b_1=-2,c_1=-2\\ a_2=2,b_2=-4,c_2=6\end{array}$

  $\begin{array}{l}\frac{a_1}{a_2}=\frac{1}{2}\\ \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\\ \frac{c_1}{c_2}=\frac{-2}{6}=\frac{-1}{3}\end{array}$

So,

  $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

The system of equations: $y=\frac{x}{2}-1,2x-4y+6=0$ has no solution.

c.

To determine

To write: the second equation so that the system has infinitely many solutions

Answer to Problem 40PPE

  $2x-4y+6=0$

Explanation of Solution

Given information: One equation in a system is $y=\frac{x}{2}-1$

One equation in a system is $y=\frac{x}{2}-1$

This equation can be written as follows:

  $\begin{array}{l}y=\frac{x}{2}-1\\ \Rightarrow y=\frac{x-2}{2}\\ \Rightarrow 2y=x-2\\ \Rightarrow x-2y-2=0\end{array}$

The system of equations: $a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$ has infinitely many solutions if $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Take the second equation as $2x-4y-4=0$

Here,

  $\begin{array}{l}{a}_1=1,b_1=-2,c_1=-2\\ a_2=2,b_2=-4,c_2=-4\end{array}$

  $\begin{array}{l}\frac{a_1}{a_2}=\frac{1}{2}\\ \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\\ \frac{c_1}{c_2}=\frac{-2}{-4}=\frac{1}{2}\end{array}$

So,

  $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

The system of equations: $y=\frac{x}{2}-1,2x-4y-4=0$ has infinitely many solutions.