Question

Draw the graph of the following equations :

x + y = 3, X - y = 1
i) Find the solution of the equation from the graph.

ii) Shade the triangular region formed by the lines and the y axis,



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Answer 1

Step-by-step explanation:

x−y+1=0

x−y=−1 ...(i)

3x+2y−12=0

3x+2y=12 ...(ii)

x−y=−1

y=1+x

x 0 −1

y 1 0

Plot (0,1) and (1,2) on graph and join them to get eqn (i)

For 3x+2y=12

⇒2y=12−3x

⇒r=

2

12−3x

x 0 4

y 6 0

Plot point (0,6) and (4,0) on graph and join them to get eqn (ii)

△ABC is the required triangle with coordinates of A as (2,3),B(−1,0) and C(4,0).

solution

Answer 2

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

Given pair of equations are

$\rm :\longmapsto\:x + y = 3 - - - (1)$

and

$\rm :\longmapsto\:x - y = 1 - - - (2)$

Now, Consider Equation (1)

$\rm :\longmapsto\:x + y = 3$

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

$\rm :\longmapsto\:0+ y = 3$

$\rm :\longmapsto\:y = 3$

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

$\rm :\longmapsto\:x + 0 = 3$

$\rm :\longmapsto\:x = 3$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 3 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 3) & (3 , 0)

➢ See the attachment graph.

Now, Consider Equation (2)

$\rm :\longmapsto\:x - y = 1$

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

$\rm :\longmapsto\:0 - y = 1$

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

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

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

$\rm :\longmapsto\:x - 0 = 1$

$\rm :\longmapsto\:x = 1$

Hᴇɴᴄᴇ,

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

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

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

➢ See the attachment graph.

From graph we concluded that

Solution is

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = 2} \\ \\ &\sf{y = 1} \end{cases}\end{gathered}\end{gathered}}$

And

Th triangular region bounded by the lines with x - axis is ABC having coordinates

Coordinates of A (3, 0)

Coordinates of B (1, 0)

Coordinates of C (2, 1)