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,
⇒ 2 × 0 - 3y = 6
⇒ y = -2
⇒ x = 0, y = -2
Putting y = 0 in equation (i), we get,
⇒ 2x - 3 × 0 = 6
⇒ x = 3
⇒ 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,
⇒ 0 + y = 1
⇒ y = 1
$\therefore$ x = 0, y = 1
Putting y = 0 in equation (ii), we get,
⇒ x + 0 =1
⇒ x = 1
⇒ 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,
⇒ 2 × 0 - 3y = 6
⇒ y = -2
⇒ x = 0, y = -2
Putting y = 0 in equation (i), we get,
⇒ 2x - 3 × 0 = 6
⇒ x = 3
⇒ 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,
⇒ 0 + y = 1
⇒ y = 1
$\therefore$ x = 0, y = 1
Putting y = 0 in equation (ii), we get,
⇒ x + 0 =1
⇒ x = 1
⇒ 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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