Question
Solve the following systems of equations graphically:
$2x + 3y = 4$
$x - y + 3 = 0$
✓
Answer
The given equations are,
$2x + 3y = 4 ......(i)$
$x - y + 3 = 0 ........(ii)$
Putting $x = 0$ in equation $(i)$, we get,
$\Rightarrow 2 \times 0 + 3y = 4$
$\Rightarrow\text{y}=\frac{4}{3}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{4}{3}$
Putting $y = 0$ in equation $(i)$, we get,
$\Rightarrow 2x + 3 \times 0 = 4$
$\Rightarrow x = 2$
$\Rightarrow x = 2, y = 0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $\text{A}\Big(0, \frac{4}{3}\Big)$ and $B(2, 0)$ from table.
Graph of the equation,
$\Rightarrow x - y = -3 ........(ii)$
Putting $x = 0$ in equation $(ii)$, we get,
$\Rightarrow 0 - y = -3$
$\Rightarrow y = 3$
$\Rightarrow x = 0, y = 3$
Putting $y = 0$ in equation $(ii)$, we get,
$\Rightarrow x - 0 = -3$
$\Rightarrow x = -3$
$\Rightarrow x = -3, y = 0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $C(0, 3)$ and $D(-3, 0)$ from table.
The two lines intersect at points $P(-1, 2).$
Hence, $x = -1$ and $y = 2$ is the solution.
$2x + 3y = 4 ......(i)$
$x - y + 3 = 0 ........(ii)$
Putting $x = 0$ in equation $(i)$, we get,
$\Rightarrow 2 \times 0 + 3y = 4$
$\Rightarrow\text{y}=\frac{4}{3}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{4}{3}$
Putting $y = 0$ in equation $(i)$, we get,
$\Rightarrow 2x + 3 \times 0 = 4$
$\Rightarrow x = 2$
$\Rightarrow x = 2, y = 0$
Use the following table to draw the graph.
|
$X$
|
$0$
|
$2$
|
|
$Y$
|
$\frac{4}{3}$
|
$0$
|
Graph of the equation,
$\Rightarrow x - y = -3 ........(ii)$
Putting $x = 0$ in equation $(ii)$, we get,
$\Rightarrow 0 - y = -3$
$\Rightarrow y = 3$
$\Rightarrow x = 0, y = 3$
Putting $y = 0$ in equation $(ii)$, we get,
$\Rightarrow x - 0 = -3$
$\Rightarrow x = -3$
$\Rightarrow x = -3, y = 0$
Use the following table to draw the graph.
|
$x$
|
$0$
|
$-3$
|
|
$y$
|
$3$
|
$0$
|
The two lines intersect at points $P(-1, 2).$
Hence, $x = -1$ and $y = 2$ is the solution.
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