Question
Solve graphically the following system of linear equation. Also find the coordinates of the points where the lines meet axis of y.
2x - 5y + 4 = 0,
2x + y - 8 = 0.
✓
Answer
We have,
2x - 5y + 4 = 0
2x + y - 8 = 0
Now, 2x - 5y + 4 = 0
⇒ 2x = 5y - 4
$\Rightarrow\text{x}=\frac{5\text{y}-4}{2}$
When y = 2, we have,
$\text{x}=\frac{5\times2-4}{2}=3$
When y = 4, we have,
$\text{x}=\frac{5\times4-4}{2}=8$
Thus, we have the following table giving points on the line 2x - 5y + 4 = 0
Now, 2x + y - 8 = 0
⇒ 2x = 8 - y
$\Rightarrow\text{x}=\frac{8-\text{y}}{2}$
When y = 4, we have,
$\text{x}=\frac{8-4}{2}=2$
When y = 2, we have,
$\text{x}=\frac{8-2}{2}=3$
Thus, we have the following table points on the line 2x + y - 8 = 0
Graph of the given equations,
Clearly, two intersect at P(3, 2).
Hence, x = 3, y = 2 is the solution of the given system of equations.
We also observe that lines represented by 2x - 5y + 4 and 2x + y - 8 = 0 meet y-axis at $\text{A}\Big(0,\frac{4}{5}\Big)$ and B(0, 8) respectively.
2x - 5y + 4 = 0
2x + y - 8 = 0
Now, 2x - 5y + 4 = 0
⇒ 2x = 5y - 4
$\Rightarrow\text{x}=\frac{5\text{y}-4}{2}$
When y = 2, we have,
$\text{x}=\frac{5\times2-4}{2}=3$
When y = 4, we have,
$\text{x}=\frac{5\times4-4}{2}=8$
Thus, we have the following table giving points on the line 2x - 5y + 4 = 0
|
x
|
3
|
8
|
|
y
|
2
|
4
|
⇒ 2x = 8 - y
$\Rightarrow\text{x}=\frac{8-\text{y}}{2}$
When y = 4, we have,
$\text{x}=\frac{8-4}{2}=2$
When y = 2, we have,
$\text{x}=\frac{8-2}{2}=3$
Thus, we have the following table points on the line 2x + y - 8 = 0
|
x
|
2
|
3
|
|
y
|
4
|
2
|
Clearly, two intersect at P(3, 2).
Hence, x = 3, y = 2 is the solution of the given system of equations.
We also observe that lines represented by 2x - 5y + 4 and 2x + y - 8 = 0 meet y-axis at $\text{A}\Big(0,\frac{4}{5}\Big)$ and B(0, 8) respectively.
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