Question
Solve the following system of equations graphically:
$2x + 3y = 8,$
$x - 2y + 3 = 0$
$2x + 3y = 8,$
$x - 2y + 3 = 0$
✓
Answer
On a graph paper, draw a horizontal line $X'OX$ and a vertical line $YOY'$ representing the $x-$axis and $y-$axis, respectively.
Given equations are $2x + 3y = 8$
and $x - 2y + 3 = 0$
Graph of $2x + 3y = 8:$
$2x + 3y = 8$
$\Rightarrow\text{y}=\frac{8-\text{2x}}{3}\ \dots(1)$
Thus we have the following table for $2x + 3y = 8$
On the graph paper plot the points $A (1,2), B (-5,6)$ and $C (7,-2)$.
Join $A B$ and $A C$ to get the graph line $B C$.
Thus, the line $A C$ is the equation of $2 x+3 y=8$.
Graph of $x-2 y+3=0$ :
For graph of $x-2 y+3=0$
$\Rightarrow\text{y}=\frac{\text{x}+3}{2}\ \dots(2)$
Thus, we have the following table for$ x - 2y + 3 = 0$
Now, on the same graph paper plot the points $P(3,3)$ and $Q(-3,0)$.
The point $A(1,2)$ has already been plotted.
Join PA and $QA$ to get the line $PQ.$
Thus, line $P Q$ is the graph of the equation $x-2 y+3=0$.
The two graph lines intersect at $A(1, 2).$
$\therefore x = 1, y = -2$ is the solution of the given system of equations.
Given equations are $2x + 3y = 8$
and $x - 2y + 3 = 0$
Graph of $2x + 3y = 8:$
$2x + 3y = 8$
$\Rightarrow\text{y}=\frac{8-\text{2x}}{3}\ \dots(1)$
Thus we have the following table for $2x + 3y = 8$
|
$x:$
|
$1$ |
$-5$
|
$7$
|
|
$y:$
|
$2$
|
$6$
|
$-2$
|
Join $A B$ and $A C$ to get the graph line $B C$.
Thus, the line $A C$ is the equation of $2 x+3 y=8$.
Graph of $x-2 y+3=0$ :
For graph of $x-2 y+3=0$
$\Rightarrow\text{y}=\frac{\text{x}+3}{2}\ \dots(2)$
Thus, we have the following table for$ x - 2y + 3 = 0$
|
$x:$
|
$1$
|
$3$
|
$-3$
|
|
$y:$
|
$2$
|
$3$
|
$0$
|
The point $A(1,2)$ has already been plotted.
Join PA and $QA$ to get the line $PQ.$
Thus, line $P Q$ is the graph of the equation $x-2 y+3=0$.
The two graph lines intersect at $A(1, 2).$
$\therefore x = 1, y = -2$ is the solution of the given system of equations.
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