Question

Solve the following system of linear equations graphically. 3x + y - 12 = 0 and x - 3y + 6 = 0. Shade the region bounded by the lines and Y-axis. Also , find the area of the shaded region.

Answer 1

EXPLANATION.

System of linear equation graphically.

⇒ 3x + y - 12 = 0. - - - - - (1).

⇒ x - 3y + 6 = 0. - - - - - (2).

As we know that,

From equation (1), we get.

⇒ 3x + y - 12 = 0. - - - - - (1).

Put the value of x = 0 in the equation, we get.

⇒ 3(0) + y - 12 = 0.

⇒ y - 12 = 0.

⇒ y = 12.

Their Co-ordinates = (0,12).

Put the value of y = 0 in the equation, we get.

⇒ 3x + (0) - 12 = 0.

⇒ 3x - 12 = 0.

⇒ 3x = 12.

⇒ x = 12/3.

⇒ x = 4.

Their Co-ordinates = (4,0).

From equation (2), we get.

⇒ x - 3y + 6 = 0. - - - - - (2).

Put the value of x = 0 in the equation, we get.

⇒ (0) - 3y + 6 = 0.

⇒ - 3y + 6 = 0.

⇒ 3y = 6.

⇒ y = 2.

Their Co-ordinates = (0,2).

Put the value of y = 0 in the equation, we get.

⇒ x - 3(0) + 6 = 0.

⇒ x + 6 = 0.

⇒ x = - 6.

Their Co-ordinates = (-6,0).

Both curves intersects at a point = (3,3).

As we know that,

Area of triangle = 1/2 x Base x Height.

Height = 3.

Base = 2 - 12 = - 10.

⇒ Area of triangle = | 1/2 x - 10 x 3 |.

⇒ Area of triangle = |- 15|.

⇒ Area of triangle = 15 sq. units.

Answer 2

Answer:

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Step-by-step explanation:

3x+y_12=0

3x=12–y=eq 1

MULTIPLY IT BY 3 WE GET

3X= 9Y– 18 EQ 2

FROM 1,2 WE HAVE

12_Y =9Y–18

9Y+Y=12+18

10Y=30

Y=3

NOW PUT Y=3 IN EQ 1

3X=12–Y

3X=12–3

3X=9

X=3