Solve the following system of linear equations graphically:
$4x\ -\ 5y\ -\ 20\ =\ 0$
$3x\ +\ 5y\ -\ 15\ =\ 0$
Determine the vertices of the triangle formed by the lines representing the above equation and the y-axis.

 Given:

The given equations are:

$4x\ -\ 5y\ -\ 20\ =\ 0$

$3x\ +\ 5y\ -\ 15\ =\ 0$

To do:

We have to solve the given system of equations and determine the vertices of the triangle formed by the lines representing the above equations and the y-axis.

Solution:

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

For equation $4x-5y-20=0$,

$5y=4x-20$

$y=\frac{4x-20}{5}$

If $x=0$ then $y=\frac{4(0)-20}{5}=\frac{-20}{5}=-4$

If $x=5$ then $y=\frac{4(5)-20}{5}=\frac{20-20}{5}=0$

For equation $3x+5y-15=0$,

$5y=15-3x$

$y=\frac{15-3x}{5}$

If $x=0$ then $y=\frac{15-3(0)}{5}=\frac{15}{5}=3$

If $x=5$ then $y=\frac{15-3(5)}{5}=\frac{15-15}{5}=0$

The equation of y-axis is $x=0$.

The above situation can be plotted graphically as below:

The lines AB, CD and AC represent the equations $4x-5y-20=0$, $3x+5y-15=0$ and y-axis respectively.

As we can see, the points of intersection of the lines AB, CD and AC taken in pairs are the vertices of the given triangle.

Hence, the vertices of the given triangle are $(0,-4), (5,0)$ and $(0,3)$.  

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Updated on: 10-Oct-2022

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