Chapter 6.2, Problem 4LC

To determine

To write: whether the system of equations has one solution, infinitely many solutions or no solution.

Answer to Problem 4LC

No solution

Explanation of Solution

Given information: Equation are $-x+\frac{y}{2}=13,x+15=\frac{y}{2}$

Consider the equation $-x+\frac{y}{2}=13$

This equation can be written as follows:

  $\begin{array}{l}-x+\frac{y}{2}=13\\ \Rightarrow \frac{-2x+y}{2}=13\\ \Rightarrow -2x+y-26=0\end{array}$

Compare this equation with $a_{1}x+b_{1}y+c_1=0$

  $a_1=-2,b_1=1,c_1=-26$

Consider the equation $x+15=\frac{y}{2}$

This equation can be written as follows:

  $\begin{array}{l}x+15=\frac{y}{2}\\ \Rightarrow 2x+30=y\\ \Rightarrow 2x-y+30=0\end{array}$

Compare this equation with $a_{2}x+b_{2}y+c_2=0$

  $a_2=2,b_2=-1,c_2=30$

Now find ratios $\frac{a_1}{a_2},\frac{b_1}{b_2},\frac{c_1}{c_2}$

  $\begin{array}{l}\frac{a_1}{a_2}=\frac{-2}{2}=-1\\ \frac{b_1}{b_2}=\frac{1}{-1}=-1\\ \frac{c_1}{c_2}=\frac{-26}{30}=\frac{-13}{15}\end{array}$

Here,

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

So, the system of equations has no solution.

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