Question
Solve the following system of equations graphically:
$2x + 3y = 2,$
$x - 2y = 8$
$2x + 3y = 2,$
$x - 2y = 8$
✓
Answer
On a graph paper, draw a horizontal line $X^{\prime} O X$ and a vertical line $Y O Y^{\prime}$ representing the $x$-axis and $y$-axis, respectively. Graph of $2 x+3 y=2$ :
$y=\frac{2(1-x)}{3}$
Putting $x=1$, we get $y=0$
Putting $x=-2$, we get $y=2$
Putting $x=4$, we get $y=-2$
$\therefore$ Table for $2 x+3 y=2$ is
Plot the points $A(1,0), B(-2,2)$ and $C(4,-2)$ on the graph paper.
Join $A B$ and $A C$ to get the graph line $B C$.
Extend it on both ways.
Thus, the line $B C$ is the graph of $x+3 y=2$.
Graph of $x-2 y=8$ :
$y=\frac{x-8}{2}$
Putting $x=2$, we get $y=-3$
Putting $x=4$, we get $y=-2$
Putting $x=0$, we get $y=-4$
Table for $x-2 y=8$ is
Now, on the same graph paper plot the points $P(0,-4)$ and $Q(2,-3)$. The point $C(4,-2)$ has already been plotted. Join $PQ$ and $QC$ and extend it on both ways.
Thus, line $PC$ is the graph of $x-2 y=8$.
The two graph lines intersect at $C (4,-2)$.
$\therefore x=4, y=-2$ is the solution of the given system of equations.
$y=\frac{2(1-x)}{3}$
Putting $x=1$, we get $y=0$
Putting $x=-2$, we get $y=2$
Putting $x=4$, we get $y=-2$
$\therefore$ Table for $2 x+3 y=2$ is
|
$x:$
|
$1$
|
$-2$
|
$4$
|
|
$y:$
|
$0$
|
$2$
|
$-2$
|
Join $A B$ and $A C$ to get the graph line $B C$.
Extend it on both ways.
Thus, the line $B C$ is the graph of $x+3 y=2$.
Graph of $x-2 y=8$ :
$y=\frac{x-8}{2}$
Putting $x=2$, we get $y=-3$
Putting $x=4$, we get $y=-2$
Putting $x=0$, we get $y=-4$
Table for $x-2 y=8$ is
|
$x:$
|
$2$
|
$4$
|
$0$
|
|
$y:$
|
$-3$
|
$-2$
|
$-4$
|
Thus, line $PC$ is the graph of $x-2 y=8$.
The two graph lines intersect at $C (4,-2)$.
$\therefore x=4, y=-2$ is the solution of the given system of equations.
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