Show graphically that each one of the following systems of equations is inconsistent (i.e. has no solution):
$3x\ –\ 4y\ –\ 1\ =\ 0$
$2x\ –\ \left(\frac{8}{3}\right) y\ +\ 5\ =\ 0$

Given:

The given system of equations is:

$3x\ –\ 4y\ –\ 1\ =\ 0$

$2x\ –\ \left(\frac{8}{3}\right) y\ +\ 5\ =\ 0$

To do:

We have to show that the above system of equations is inconsistent.

Solution:

The given pair of equations are:

$3x\ -\ 4y\ -\ 1\ =\ 0$....(i)

$4y=3x-1$

$y=\frac{3x-1}{4}$

$2x\ –\ \left(\frac{8}{3}\right) y\ +\ 5\ =\ 0$

Multiplying both sides of the equation by 3, we get,

$3(2x)-3(\frac{8}{3})y+3(5)=3(0)$

$6x-8y+15=0$.....(ii)

$8y=6x+15$

$y=\frac{6x+15}{8}$

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation (i),

If $x=-1$ then $y=\frac{3(-1)-1}{4}=\frac{-4}{4}=-1$

If $x=3$ then $y=\frac{3(3)-1}{4}=\frac{8}{4}=2$

For equation (ii),

If $x=0$ then $y=\frac{0}{2}=0$

If $x=2$ then $y=\frac{2}{2}=1$

The above situation can be plotted graphically as below:

The lines AB and PQ represent the equations $3x\ –\ 4y\ –\ 1\ =\ 0$ and $2x\ –\ \left(\frac{8}{3}\right) y\ +\ 5\ =\ 0$.

As we can see, there is no common point between the two lines.

Hence, the given system of equations is inconsistent. 

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Updated on: 10-Oct-2022

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