Show graphically that each one of the following systems of equation has infinitely many solution:
$x\ –\ 2y\ +\ 11\ =\ 0$
$3x\ +\ 6y\ +\ 33\ =\ 0$

Given:

The given system of equations is:

$x\ –\ 2y\ +\ 11\ =\ 0$

$3x\ -\ 6y\ +\ 33\ =\ 0$

To do:

We have to show that the above system of equations has infinitely many solutions.

Solution:

The given pair of equations are:

$x\ -\ 2y\ +\ 11\ =\ 0$....(i)

$2y=x+11$

$y=\frac{x+11}{2}$

$3x\ -\ 6y\ +\ 33\ =\ 0$....(ii)

$6y=3x+33$

$y=\frac{33+3x}{6}$

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

For equation (i),

If $x=-1$ then $y=\frac{-1+11}{2}=\frac{10}{2}=5$

If $x=-3$ then $y=\frac{-3+11}{2}=\frac{8}{2}=4$

For equation (ii),

If $x=-1$ then $y=\frac{33+3(-1)}{6}=\frac{30}{6}=5$

If $x=1$ then $y=\frac{33+3(1)}{6}=\frac{33+3}{6}=\frac{36}{6}=6$

The above situation can be plotted graphically as below:

The lines AB and PQ represent the equations $x-2y+11=0$ and $3x-6y+33=0$.

As we can see, both equations represent the same line.

Hence, the given system of equations has infinitely many solutions.

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Updated on: 10-Oct-2022

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