two college algebra questions

[ploppy123]

1st problem
x- 2y =2
{ }
y squared - x squared = 2x + 4

solve: a system of second degree equations

x=2y+2
then substituted x for the x's
y^2 - (2y + 2)^2 = 2(2y + 2) + 4
y^2-4y^2-8y-4-4y+4+4
took away the 4-4
so
y^2-4y^2-8y-4y
y^2-4y^2-12y+4

answer: (-2,-2)

second problem
graph the solution of inequaliies. find the the coordinates of all vertices and determine whether the solution set is bounded or unbounded
x^2+y^2<9
x+y<0

I changed to the y=mx+b form
then put it slope form
they are both dashed because it has no equal sign
so :
y^2/1 < x^2/1 +9/1

y/1 < -x/1 + 0/1

in the solution there were square roots in one of the vertices

answer:vertices = (3 over square root 2 - 3 square root of 2) and (-3 over square root of two and 3 square root of two) bounded

 

[JeffM]

1st problem
x- 2y =2
{ }
y squared - x squared = 2x + 4

solve: a system of second degree equations

x=2y+2
then substituted x for the x's
y^2 - (2y + 2)^2 = 2(2y + 2) + 4
y^2-4y^2-8y-4-4y+4+4
took away the 4-4
so
y^2-4y^2-8y-4y
y^2-4y^2-12y+4

answer: (-2,-2)

second problem
graph the solution of inequaliies. find the the coordinates of all vertices and determine whether the solution set is bounded or unbounded
x^2+y^2<9
x+y<0

I changed to the y=mx+b form
then put it slope form
they are both dashed because it has no equal sign
so :
y^2/1 < x^2/1 +9/1

y/1 < -x/1 + 0/1

in the solution there were square roots in one of the vertices

answer:vertices = (3 over square root 2 - 3 square root of 2) and (-3 over square root of two and 3 square root of two) bounded

Try reading Read Before Posting. It asks that you have ONE problem per thread.

Moreover what are you asking us to do? Check your answers?

Problem 1: I cannot even begin to understand your work. (-2, -2) is the correct answer, but I do not see how you get it.

\(\displaystyle x - 2y = 2 \implies x = 2y + 2.\)

\(\displaystyle So\ y^2 - x^2 = 2x + 4 \implies y^2 - (2y + 2)^2 = 2(2y + 2) + 4 \implies y^2 - (4y^2 + 8y + 4) = 4y + 4 + 4 \implies\)

\(\displaystyle y^2 - 4y^2 - 8y - 4 = 4y + 8 \implies -3y^2 - 8y - 4 = 4y + 8 \implies 3y^2 + 12y + 12 = 0 \implies y^2 + 4y + 4 = 0 \implies (y + 2)^2 = 0 \implies\)

\(\displaystyle y = - 2 \implies x = 2(-2) + 2 = - 4 + 2 = - 2.\)

 

[ploppy123]

at this point y^2-4y^2-8-4= 4y+8
what happens after?
thanks