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Solve the following system of equations graphically:
$2x\ +\ y\ β\ 3\ =\ 0$
$2x\ β\ 3y\ β\ 7\ =\ 0$
Given:
The given system of equations is:
$2x\ +\ y\ –\ 3\ =\ 0$
$2x\ –\ 3y\ –\ 7\ =\ 0$
To do:
We have to represent the above system of equations graphically.
Solution:
The given pair of equations are:
$2x\ +\ y\ -\ 3\ =\ 0$....(i)
$y=3-2x$
$2x\ -\ 3y\ -\ 7\ =\ 0$....(ii)
$3y=2x-7$
$y=\frac{2x-7}{3}$
To represent the above equations graphically we need at least two solutions for each of the equations.
For equation (i),
If $x=0$ then $y=3-2(0)=3-0=3$
If $x=1$ then $y=3-2(1)=3-2=1$
$x$ | $0$ | $1$ |
$y=3-2x$ | $3$ | $1$ |
For equation (ii),
If $x=2$ then $y=\frac{2(2)-7}{3}=\frac{4-7}{3}=\frac{-3}{3}=-1$
If $x=5$ then $y=\frac{2(5)-7}{3}=\frac{10-7}{3}=\frac{3}{3}=1$
$x$ | $2$ | $5$ |
$y=\frac{2x-7}{3}$ | $-1$ | $1$ |
The above situation can be plotted graphically as below:
The line AB represents the equation $2x+y-3=0$ and the line PQ represents the equation $2x-3y-7=0$.
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 $x=2$ and $y=-1$.