Question
Determine, by drawing graphs, whether the following system of linear equations has a unique solution or not:
$2x - 3y = 6, x + y = 1.$
$2x - 3y = 6, x + y = 1.$
✓
Answer
The given equations are
$2x - 3y = 6 .......(i)$
$x + y = 1 ..........(ii)$
Putting x = 0 in equation (i), we get,
$\Rightarrow 2 \times 0 - 3y = 6$
$\Rightarrow y = -2$
$\Rightarrow x = 0, y = -2$
Putting y = 0 in equation (i), we get,
$\Rightarrow 2x - 3 \times 0 = 6$
$\Rightarrow x = 3$
$\Rightarrow x = 3, y = 0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $A(0, -2), B(3, 0)$ from table.
Graph of the equation.
$x + y = 1 .......(ii)$
Putting $x = 0$ in equation $(ii)$, we get,
$\Rightarrow 0 + y = 1$
$\Rightarrow y = 1$
$\therefore x = 0, y = 1$
Putting $y = 0$ in equation $(ii)$, we get,
$\Rightarrow x + 0 =1$
$\Rightarrow x = 1$
$\Rightarrow x = 1, y = 0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $C(0, 1), D(1, 0)$ from table. The two lines intersect at point $\text{P}\Big(\frac{9}{5},\frac{-4}{5}\Big).$
Hence the equations have unique solution.
$2x - 3y = 6 .......(i)$
$x + y = 1 ..........(ii)$
Putting x = 0 in equation (i), we get,
$\Rightarrow 2 \times 0 - 3y = 6$
$\Rightarrow y = -2$
$\Rightarrow x = 0, y = -2$
Putting y = 0 in equation (i), we get,
$\Rightarrow 2x - 3 \times 0 = 6$
$\Rightarrow x = 3$
$\Rightarrow x = 3, y = 0$
Use the following table to draw the graph.
|
$x$
|
$0$
|
$3$
|
|
$y$
|
$-2$
|
$0$
|
Graph of the equation.
$x + y = 1 .......(ii)$
Putting $x = 0$ in equation $(ii)$, we get,
$\Rightarrow 0 + y = 1$
$\Rightarrow y = 1$
$\therefore x = 0, y = 1$
Putting $y = 0$ in equation $(ii)$, we get,
$\Rightarrow x + 0 =1$
$\Rightarrow x = 1$
$\Rightarrow x = 1, y = 0$
Use the following table to draw the graph.
|
$x$
|
$0$
|
$1$
|
|
$y$
|
$1$
|
$0$
|
Hence the equations have unique solution.
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