Trigonometry help?

Question

solve this by method of substitution
3x + y=2
x^3 - 2 + y=0
the anwers are (0,2), {sqrt(3), 2-3 sqrt(3)}, and
{-sqrt(3),2+3sqrt(3)}
Can you please show me how to do this problem, I have the answers but I don't really understand how to them. thanks in advance

Answer 1

Set both equations in terms of x, so that they both equal y.

y=2-3x

y=2-x^3

Then set both equations equal to each other, and solve for x

2-3x=2-x^3

You can drop the 2 on both sides

-3x= -x^3

...and then the negatives

3x=x^3

Then, subtract 3x from both sides...

0=x^3-3x

Factor out an x

0=x(x^2-3)

Figure out how to make this equation true....

If x = 0, then everything will equal 0, because anything mulitplied by 0 is always equal to 0.

For the part inside the ( ), you will use the same idea. Make x^2-3 equal to 0

x^2-3=0

x^2=3

x= plus or minus the sqr of 3

...because the square of a number will always be positive, so the square root should be figure for both positive and negative.

Then, to figure out the value of y....just plug and chug baby...

You'll just solve for y...which should be easy considering you just set both equations equal to y right at the beginning of the whole process.

y=2-3x

You could use either equation, but that is the easier of the two to use for this part.

y=2-3(0)
y=2, when x=0

y=2-3(sqr 3), when x=sqr 3

y=2-3(-sqr 3)
y=2+3(sqr 3), when x= -sqr 3

I hope this helps explain it. Good Luck!!!!

Answer 2

If 3x+y=2, then y=2-3x
Substitute this value where you see y in the 2nd formula to get:
x^3 - 2 + 2-3x = 0 -OR- x^3 - 3x = 0

Which simplifies to x(x^2 -3) = 0
Which becomes x (x - sqrt 3) (x + sqrt 3) = 0

That gives you x = 0 or +/- sqrt 3
Plug back into 2 -3x to get your answers.

Answer 3

Take the first equation and solve for y to get y = 2-3x, then substitute for y in the second equation to get

x^3 - 2 + (2-3x) = 0

This reduces to x^3 -3x = 0

So the solutions are x = 0 and x = +/- sqrt(3)

Then plug these values back in to get the value of y.

Answer 4

3x + y=2
x^3 - 2 + y=0
you want to know when these two systems are equal. (intersect)
Using the first equation, you can solve for y = 2-3x
Substitute this into the bottom equation, getting
x^3-2+2-3x=0
Now you get x^3-3x=0
However we can factor out an x, getting
x(x^2-3)=0
Thus x^2-3 equals 0 when x=plusminussqrt3
and x = 0 when x is equal to 0
Plug these back into an equation to get the y values.

Answer 5

I got x=2 and x= -1/2