Question

I need help on this I would appreciate if you could Help me thank you

Answer 1

Answer:

(x1, y1)=(-1, -2)

(x2, y2)=(3, 6)

Step-by-step explanation:

y=2x

y=x^2-3

Since both of the expressions 2x and x^2-3 are equal to y, set them equal to each other forming an equation in x

2x=x^2-3

solve the equation for x

x=-1

x=3

Substitute the given value of x into the equation y=2x

y=2 x (-1)

x=3      y=2 x 3

Solve the equation for y

y=-2

y=2 x 3      y=6

The possible solutions of the system are the ordered pairs (x, y)

(x1, y1)=(-1, -2)

(x2, y2)=(3, 6)

Check if the given ordered pairs are the solutions of the system of equations

-2=2x(-1)

-2=(-1)^2-3

6=2x3

6=3^2-3

Simplify the equalities

-2=-2

-2=-2

6=2x3

6=3^2-3

-2=-2

-2=-2

6=6

6=6

Since all of the equalites are true, the ordered pairs are the solution of the system

(x1, y1)=(-1, -2)

(x2, y2)=(3, 6)

Answer 2

The points of intersection are at (3, 6) and (-1, -2).

Since both of these equations have y isolated, we can set them equal to each other:

2x=x²-3

We want all of the variables on one side, so subtract 2x:
2x-2x = x²-3-2x
0=x²-3-2x

Write the quadratic in standard form:
0=x²-2x-3

This is easily factorable, as there are factors of -3 that will sum to -2. -3(1)=-3 and -3+1=-2:
0=(x-3)(x+1)

Using the zero product property we know that either x-3=0 or x+1=0; therefore x=3 or x=-1.

Substituting this into the first equation (it is simpler):
y=2(3) = 6
y=2(-1)=-2

Therefore the coordinates are (3, 6) and (-1, -2).