Question
Solve the following system of linear equation graphically and shade the region between the two lines and x-axis:
3x + 2y -4 = 0
2x - 3y -7 = 0.
✓
Answer
The system of given equations is,
3x + 2y -4 = 0
2x - 3y -7 = 0
Now, 3x + 2y -4 = 0
⇒ 3x = 4 - 2y
$\Rightarrow\text{x}=\frac{4-2\text{y}}{3}$
When y = 5, we have,
$\text{x}=\frac{4-2\times5}{3}=-2$
When y = 8, we have,
$\text{x}=\frac{4-2\times8}{3}=-4$
Thus, we have the following table,
We have, 2x - 3y -7 = 0
⇒ 2x = 3y + 7
$\Rightarrow\text{x}=\frac{3\text{y}+7}{2}$
When y = 1, we have,
$\text{x}=\frac{3\times1+7}{2}=5$
When y = -1, we have,
$\text{x}=\frac{3\times(-1)+7}{2}=2$
Thus, we have the following table,
Graph of the given system of equations.
Clearly, the two lines intersect at P(2, -1).
Hence, x = 2, y = -1 is the solution of the given system of equations.
3x + 2y -4 = 0
2x - 3y -7 = 0
Now, 3x + 2y -4 = 0
⇒ 3x = 4 - 2y
$\Rightarrow\text{x}=\frac{4-2\text{y}}{3}$
When y = 5, we have,
$\text{x}=\frac{4-2\times5}{3}=-2$
When y = 8, we have,
$\text{x}=\frac{4-2\times8}{3}=-4$
Thus, we have the following table,
|
x
|
-2 |
-4
|
|
y
|
5
|
8
|
⇒ 2x = 3y + 7
$\Rightarrow\text{x}=\frac{3\text{y}+7}{2}$
When y = 1, we have,
$\text{x}=\frac{3\times1+7}{2}=5$
When y = -1, we have,
$\text{x}=\frac{3\times(-1)+7}{2}=2$
Thus, we have the following table,
|
x
|
5 |
2
|
|
y
|
1 |
-1
|
Clearly, the two lines intersect at P(2, -1).
Hence, x = 2, y = -1 is the solution of the given system of equations.
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