Give a geometric description of the following system of equations.a. 2x−4y=12 −3x+6y=−15.b. 2x−4y=12 −5x+3y=10.a. 2x−4y=12 −3x+6y=−18.

The geometric description of a system of equations in 2 variables mean the system of equations will represent the number of lines equal to the number of equations in the system given.

i.e.

Number of planes = Number of variables

Number of lines = Number of equations in the system.

Here, we are given 2 variables and 2 equation in each system.

So, they can be represented in the xy-coordinates plane.

And the number of solutions to the system depends on the following condition.

Let the system of equations be:

[tex]A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0[/tex]

1. One solution:

There will be one solution to the system of equations, If we have:

[tex]\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}[/tex]

2. Infinitely Many Solutions: (Identical lines in the system)

[tex]\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}= \dfrac{C_1}{C_2}[/tex]

3. No Solution:(Parallel lines)

[tex]\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}[/tex]

Now, let us discuss the system of equations one by one:

a. [tex]2x-4y=12[/tex] OR [tex]2x-4y-12=0[/tex]

[tex]-3x+6y=-15[/tex] OR [tex]-3x+6y+15=0[/tex]

[tex]A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -3, B_2 = 6, C_2= 15[/tex]

Here, the ratio:

[tex]\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2} = -\dfrac{2}{3}\\\dfrac{C_1}{C_2} = -\dfrac{4}{5}[/tex]

[tex]\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}[/tex]

Therefore, no solution i.e. parallel lines.

b. [tex]2x-4y=12[/tex] OR [tex]2x-4y-12=0[/tex]

[tex]-5x+3y=10[/tex] OR [tex]-5x+3y-10=0[/tex]

[tex]A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -5, B_2 = 3, C_2 = -10[/tex]

[tex]\dfrac{A_1}{A_2}= -\dfrac{2}{5}\\\dfrac{B_1}{B_2} = -\dfrac{4}{3}\\\dfrac{C_1}{C_2} = -\dfrac{6}{5}[/tex]

[tex]\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}[/tex]

So, one solution.

Kindly refer to the images attached for the graphical representation of the given system of equations.