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Solve the following system of equations graphically:
$x\ β\ 2y\ =\ 5$
$2x\ +\ 3y\ =\ 10$
Given:
The given system of equations is:
$x\ –\ 2y\ =\ 5$
$2x\ +\ 3y\ =\ 10$
To do:
We have to represent the above system of equations graphically.
Solution:
The given pair of equations are:
$x\ -\ 2y\ -\ 5\ =\ 0$....(i)
$2y=x-5$
$y=\frac{x-5}{2}$
$2x\ +\ 3y\ -\ 10\ =\ 0$....(ii)
$3y=10-2x$
$y=\frac{10-2x}{3}$
To represent the above equations graphically we need at least two solutions for each of the equations.
For equation (i),
If $x=5$ then $y=\frac{5-5}{2}=0$
If $x=1$ then $y=\frac{1-5}{2}=\frac{-4}{2}=-2$
$x$ | $5$ | $1$ |
$y=\frac{x-5}{2}$ | $0$ | $-2$ |
For equation (ii),
If $x=5$ then $y=\frac{10-2(5)}{3}=\frac{10-10}{3}=0$
If $x=2$ then $y=\frac{10-2(2)}{3}=\frac{10-4}{3}=\frac{6}{3}=2$
$x$ | $5$ | $2$ |
$y=\frac{10-2x}{3}$ | $0$ | $2$ |
The above situation can be plotted graphically as below:
The line AB represents the equation $x-2y-5=0$ and the line PQ represents the equation $2x+3y-10=0$.
The solution of the given system of equations is the intersecting point of both the lines.
Hence, the solution of the given system of equations is $x=5$ and $y=0$.