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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