Question
Solve the following systems of equations graphically:
2x - 3y + 13 = 0
3x - 2y + 12 = 0
✓
Answer
We have,
2x - 3y + 13 = 0
3x - 2y + 12 = 0
Now, 2x - 3y + 13 = 0
⇒ 2x = 3y - 13
$\Rightarrow\text{x}=\frac{3\text{y}-13}{2}$
When y = 1, we have,
$\text{x}=\frac{3\times1-13}{2}=-5$
When y = 3, we have,
$\text{x}=\frac{3\times3-13}{2}=-2$
Thus, we have the following table giving points on the line 2x - 3y + 13 = 0
Now, 3x - 2y + 12 = 0
⇒ 3x = 2y - 12
$\Rightarrow\text{x}=\frac{2\text{y}-12}{3}$
When y = 0, we have,
$\Rightarrow\text{x}=\frac{2\times0-12}{3}=-4$
When y = 3, we have,
$\text{x}=\frac{2\times3-12}{3}=-2$
Thus, we have the following table giving points on the line 3y - 2y + 12 = 0
Graph of the given equations are,
Clearly, two lines intersect at (-2, 3)
Hence, x = -2, y = 3 is the solution of the given system of equations.
2x - 3y + 13 = 0
3x - 2y + 12 = 0
Now, 2x - 3y + 13 = 0
⇒ 2x = 3y - 13
$\Rightarrow\text{x}=\frac{3\text{y}-13}{2}$
When y = 1, we have,
$\text{x}=\frac{3\times1-13}{2}=-5$
When y = 3, we have,
$\text{x}=\frac{3\times3-13}{2}=-2$
Thus, we have the following table giving points on the line 2x - 3y + 13 = 0
|
x
|
-5
|
-2
|
|
y
|
1
|
3
|
⇒ 3x = 2y - 12
$\Rightarrow\text{x}=\frac{2\text{y}-12}{3}$
When y = 0, we have,
$\Rightarrow\text{x}=\frac{2\times0-12}{3}=-4$
When y = 3, we have,
$\text{x}=\frac{2\times3-12}{3}=-2$
Thus, we have the following table giving points on the line 3y - 2y + 12 = 0
|
x
|
-4
|
-2
|
|
y
|
0
|
3
|
Clearly, two lines intersect at (-2, 3)
Hence, x = -2, y = 3 is the solution of the given system of equations.
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