Question

4. Draw the graphs of following Linear equation on same paper. 2x +3y = 12. x-y =1
​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider first linear equation

$\rm \: 2x + 3y = 12$

Substituting 'x = 0' in the given equation, we get

$\rm \: 2 \times 0 + 3y = 12$

$\rm \: 3y = 12$

$\rm\implies \:y = 4$

Substituting 'y = 0' in the given equation, we get

$\rm \: 2x + 3 \times 0 = 12$

$\rm \: 2x + 0 = 12$

$\rm \: 2x = 12$

$\rm\implies \:x = 6$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 6 & \sf 0 \end{array}} \\ \end{gathered}$

Consider second equation of line

$\rm \: x - y = 1$

Substituting 'x = 0' in the given equation, we get

$\rm \: 0 - y = 1$

$\rm \: - y = 1$

$\rm\implies \:y = - 1$

Substituting 'y = 0' in the given equation, we get

$\rm \: x - 0 = 1$

$\rm\implies \:x = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 0 \\ \\ \sf 0 & \sf - 1 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 4), (6 , 0), (0, - 1) & (1 , 0)

➢ See the attachment graph.

Answer 2

Step-by-step explanation:

Solution

$\[ \begin{array}{l} \rm 2 x+3 y=12 \\ \\ \rm x=\dfrac{12-3 y}{2} \\ \\ \rm \text{When \( \rm y=0 \), then \( \rm x=6 \)} \\ \\ \rm \text{ When \( \rm y=2 \), then \( \rm x=3 \)} \end{array} \]$

$\color{green}\pmb{\large\begin{array}{|l|l|l | } \hline \rm x & 6&3 \\ \hline \rm y & 0&2 \\ \hline \end{array}}$

We have,

$\[ \begin{array}{l}\rm x-y=1 \\ \\ \rm x=1+y \\ \\ \text{When \(\rm y=0 \), then \(\rm x=1 \)} \\ \\ \text{When \(\rm y=-1 \), then \(\rm x=0 \)}\end{array} \]$

$\color{blue}\pmb{\large\begin{array}{|l|l|l | } \hline \rm x & 1&0 \\ \hline \rm y & 0& - 1\\ \hline \end{array}}$