Question

Solve the following system of linear equations graphically. Shade the region bounded

by these lines and x - axis. Find the area of the shaded region. x + y = 3 and 3x - 2y = 4​

Answer 1

Answer:

The given equation is

2x+3y=−5 and 3x−2y=12

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for2x+3y=−5 or y=

3

−(2x+5)

Now, table for 3x−2y=12 or y=

2

3x−12

Here, the lines intersecting at point C (2, -3)

The coordinates of the vertices of △CBD are C (2, -3), B(−

2

5

, 0) and D(4, 0).

Area = $$\frac{1}

{2} \times base \times height$$

=

2

1

×

2

13

×3(∴base=(4−(

2

5

))=

2

13

)

=

4

39

sq.units

Answer 2

Answer:

We draw graphs of the equations x+y=3 and 3x−2y=4. Then locate their point of intersection.

Recall, that it is enough to know two points to draw a straight line. Consider x+y=3. Taking x=0, we get y=3. Taking y=0, we get x=3. Thus (0,3) and (3,0) are on the straight line x+y=3.

Take a graph paper. Fix your coordinate system x⇔y axis on the graph paper. Locate A=(3,0) and B=(0,3) with respect to this coordinate system.

Join A and B and extend it to a straight line.

Consider the equation 3x−2y=4. If we take x=0, we obtain y=−2. Similarly, x=4 gives y=4. Thus C=(0,−2) and D=(4,4) are points on the straight line 3x−2y=4. Locate these points on the graph paper. Join them and extend to a straight line.

Now we have two straight line. They intersect at a point E. Looking at the graph, you see that E has coordinates (2,1). Thus x=2 and y=1 is the solution. We may verify it: 2+1=3 and 3(2)−2(1)=6−2=4.