solve the system of equations y = 2x - 4 y = x^2 - 4 A. (-3,5) and (3,5) B. (0,-4) and (2,0) C. (-1,-6) and (3,2) D. (0,4) and (2,1)

We can solve the system of equations using the substitution method, meaning that we must substitute one equation into the other equation, resulting in a principal equation.

So,

We are given two equations:

[tex]y=2x-4[/tex]

and,

[tex]y=x^{2} -4[/tex]

So, making the equation equals, we have that:

[tex]2x-4=x^{2}-4\\x^{2}-2x-4+4=0\\x^{2}-2x=0\\x(x-2)=0\\[/tex]

Finding where the function tends to 0 (roots), we have:

[tex]x(x-2)=0\\x1=0\\x2=2[/tex]

Then, substituting each value of "x" in the first equation, we will find the correct options:

Substituting "x" equal to 0 into the first equation, we have:

[tex]y=2(0)-4=-4[/tex]

So, the point will be (0,-4)

Substituting "x" equal to 2 into the first equation, we have:

[tex]y=2(2)-4=4-4=0[/tex]

So, the point will be (2,0)

Therefore, the correct option is B. (0,-4) and (2,0)

Have a nice day!

kudzordzifrancis kudzordzifrancis · Answer 2 · 2019-04-17T08:41:41+02:00

kudzordzifrancis kudzordzifrancis

17-04-2019

Answer:

B. (0,-4) and (2,0)

Step-by-step explanation:

The given system of equations is:

[tex]y=2x-4[/tex]

[tex]y=x^2-4[/tex]

Equate both equations and solve for x.

[tex]x^2-4=2x-4[/tex]

[tex]x^2-4-2x+4=0[/tex]

[tex]x^2-2x=0[/tex]

Factor;

[tex]x(x-2)=0[/tex]

[tex]x=0,x-2=0[/tex]

[tex]x=0,x=2[/tex]

When x=0, y=2(0)-4=-4

Hence (0,-4) is one solution

When x=2, y=2(2)-4=0

Hence (2,0) is another solution

The correct choice is B.

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