Question

Solve the following pairs of simultaneous equation graphically 1)3x+2y=16
4y+3x=24

Answer 1

EXPLANATION.

Graph of the equations,

(1) = 3x + 2y = 16.

(2) = 4y + 3x = 24.

As we know that,

From equation (1), we get.

⇒ 3x + 2y = 16.

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

⇒ 3(0) + 2y = 16.

⇒ 2y = 16.

⇒ y = 16/2.

⇒ y = 8.

Their Co-ordinates = (0,8).

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

⇒ 3x + 2(0) = 16.

⇒ 3x = 16.

⇒ x = 16/3.

⇒ x = 5.33.

Their Co-ordinates = (5.33,0).

From equation (2), we get.

⇒ 4y + 3x = 24.

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

⇒ 4y + 3(0) = 24.

⇒ 4y = 24.

⇒ y = 24/4.

⇒ y = 6.

Their Co-ordinates = (0,6).

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

⇒ 4(0) + 3x = 24.

⇒ 0 + 3x = 24.

⇒ 3x = 24.

⇒ x = 24/3.

⇒ x = 8.

Their Co-ordinates = (8,0).

Both curve intersects at a point = (2.667,4)

Answer 2

Answer:

The solution of the system of equations is (-1,-1)

The point (3,2) lies on the line 3x-4y=13x−4y=1

Explanation:

Given that the system of equations is 2x-3y=12x−3y=1 and 3x-4y=13x−4y=1

First, we shall plot the equation 2x-3y=12x−3y=1 in the graph.

$When \: x=0x=0 ⇒ -3y=1\implies$

$y=-0.3333−3y=1⟹y=−0.3333$

$When \: y=0y=0 ⇒ 2x=1\implies x=0.52x=1⟹x=0.5$

Hence, plotting the coordinates (0,-0.3333) and (0.5,0) and joining the line, we get, the line for the equation 2x-3y=12x−3y=1

Similarly, we shall plot the equation 3x-4y=13x−4y=1 in the graph.

$When \: x=0x=0 ⇒ -4y=1\implies$

$y=-0.25−4y=1⟹y=−0.25$

$When \: y=0y=0 ⇒ 3x=1\implies$

$x=0.33333x=1⟹x=0.3333$

Hence, plotting the coordinates (0,-0.3333) and (0.5,0) and joining the line, we get, the line for the equation 3x-4y=13x−4y=1

The solution to the system of equations is the point of intersection of the two lines.

Hence, the solution is (-1,-1)

To determine the point (3,2) lie in the equation 2x-3y=12x−3y=1

Substituting (3,2) in the equation

$2x-3y=12x−3y=1 , we \: get,$

$2(3)-3(2)=12(3)−3(2)=1$

$6-6=16−6=1$

$0\neq 10$

$=1$

Hence, the point (3,2) does not lie in the equation 2x-3y=12x−3y=1

To determine the point (3,2) lie in the equation 3x-4y=13x−4y=1

Substituting (3,2) in the equation 3x-4y=13x−4y=1 , we get,

$3(3)-4(2)=13(3)−4(2)=1$

$9-8=19−8=1$

$1=11=1$

Hence, the point (3,2) lie in the equation 3x-4y=13x−4y=1