Question
Solve the following systems of equations graphically:
x - 2y = 5
2x + 3y = 10
✓
Answer
The given equations are:
x - 2y = 5 ......(i)
2x + 3y = 10 .....(ii)
Puting x = 0 in equation (i), we get,
⇒ 0 - 2y = 5
$\Rightarrow\text{y}=\frac{-5}{2}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{-5}{2}$
Puting y = 0 in equation (i), we get,
⇒ x + 2 × 0 = 5
⇒ x = 5
⇒ x = 5, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points $\text{A}\Big(0,\frac{-5}{2}\Big)$ and B(5, 0) from table.
Graph the equation (ii),
⇒ 2x + 3y = 10 ......(ii)
Putting x = 0 in equation (ii), we get,
⇒ 2 × 0 + 3y = 10
$\Rightarrow\text{y}=\frac{10}{3}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{10}{3}$
Putting y = 0 in equation (ii), we get,
⇒ 2x + 3 × 0 = 10
⇒ x = 5
x = 5, y = 0
use the following table to draw the graph.
Draw the graph by plotting the two points $\text{C}\Big(0,\frac{10}{3}\Big)$ and B(5, 0) from table.
The two lines intersects at points B(5, 0).
Hence x = 5, y = 0 is the solution.
x - 2y = 5 ......(i)
2x + 3y = 10 .....(ii)
Puting x = 0 in equation (i), we get,
⇒ 0 - 2y = 5
$\Rightarrow\text{y}=\frac{-5}{2}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{-5}{2}$
Puting y = 0 in equation (i), we get,
⇒ x + 2 × 0 = 5
⇒ x = 5
⇒ x = 5, y = 0
Use the following table to draw the graph.
|
x
|
0
|
5
|
|
y
|
$\frac{-5}{2}$
|
0
|
Graph the equation (ii),
⇒ 2x + 3y = 10 ......(ii)
Putting x = 0 in equation (ii), we get,
⇒ 2 × 0 + 3y = 10
$\Rightarrow\text{y}=\frac{10}{3}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{10}{3}$
Putting y = 0 in equation (ii), we get,
⇒ 2x + 3 × 0 = 10
⇒ x = 5
x = 5, y = 0
use the following table to draw the graph.
|
x
|
0
|
5
|
|
y
|
$\frac{10}{3}$
|
0
|
The two lines intersects at points B(5, 0).
Hence x = 5, y = 0 is the solution.
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