Question
Solve the following system of equations graphically:
Shade the region between the lines and the y-axis.
Shade the region between the lines and the y-axis.
4x - y = 4,
3x + 2y = 14.
✓
Answer
The given system of equations is
4x - y = 4
3x + 2y = 14
Now, 4x - y = 4
⇒ 4x - 4 = y
⇒ y = 4x - 4
When x = 0, we have
y = 4 × 0 - 4 = -4
When x = -1, we have
y = 4 × (-1) - 4 = -8
Thus, we have the following table.
We have,
3x + 2y = 14
⇒ 2y = 14 - 3x
$\Rightarrow\text{y}=\frac{14-3\text{x}}{2}$
When x = 0, we have
$\text{y}=\frac{14-3\times0}{2}=7$
When x = 4, we have
$\text{y}=\frac{14-3\times4}{2}=1$
Thus, we have the following table.
Graph of the given system of equations
Clearly, the two lines intersect at A(2, 4). Hence, x = 2, y = 4 is the solution of the given system of equations.
We also observe $\triangle\text{ABC}$ is the required shaded region.
4x - y = 4
3x + 2y = 14
Now, 4x - y = 4
⇒ 4x - 4 = y
⇒ y = 4x - 4
When x = 0, we have
y = 4 × 0 - 4 = -4
When x = -1, we have
y = 4 × (-1) - 4 = -8
Thus, we have the following table.
|
x
|
0
|
-1
|
|
y
|
-4
|
-8
|
3x + 2y = 14
⇒ 2y = 14 - 3x
$\Rightarrow\text{y}=\frac{14-3\text{x}}{2}$
When x = 0, we have
$\text{y}=\frac{14-3\times0}{2}=7$
When x = 4, we have
$\text{y}=\frac{14-3\times4}{2}=1$
Thus, we have the following table.
|
x
|
0
|
4
|
|
y
|
7
|
1
|
Clearly, the two lines intersect at A(2, 4). Hence, x = 2, y = 4 is the solution of the given system of equations.
We also observe $\triangle\text{ABC}$ is the required shaded region.
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