Solving nonlinear eqns for real solns: y=x+1, x^2-y^2=1

[lp_princess05]

{ y=x+1}
{ x^2-y^2=1}

So far all I know is that you have to solve for x or y and substitute it into the other problem.

 

[stapel]

So far all I know is that you have to solve for x or y and substitute it into the other problem.

So do that: The first equation is already solved for y. Plug this in for y in the second equation, and solve the resulting quadratic for the value(s) of x. Plug the x-value(s) back into the first equation, and simplify to find the corresponding y-value(s).

Eliz.

 

[lp_princess05]

Re: Solving nonlinear equations for real numbers

The problem is that I get stuck because when I work it out I get x^2-x^2+1=1 and that cancels them out. Then it doesn't work out.

 

[pka]

Re: Solving nonlinear equations for real numbers

\(\displaystyle x^2 - \left( {x + 1} \right)^2 = - 2x - 1\)

 

[lp_princess05]

Re: Solving nonlinear equations for real numbers

But then what would happen when you multiply out the (x+1)^2? Wouldn't you get x^2-X^2? Then that would cancel those two out, then what would you do?

 

[Mrspi]

Re: Solving nonlinear equations for real numbers

{ y=x+1}
{ x^2-y^2=1}

So far all I know is that you have to solve for x or y and substitute it into the other problem.

So...substitute (x + 1) for y in the second equation:

x[sup:1jwsoms5]2[/sup:1jwsoms5] - (x + 1)[sup:1jwsoms5]2[/sup:1jwsoms5] = 1

Now, simplify the left side. You need to start by multiplying out (x + 1)[sup:1jwsoms5]2[/sup:1jwsoms5]. That's (x +1)(x + 1)....and it is NOT x[sup:1jwsoms5]2[/sup:1jwsoms5] + 1

Try that again...and see what you get.

 

[lp_princess05]

Re: Solving nonlinear equations for real numbers

So when I do that I get x^2-x^2+2x+1=1. But then how would I solve that?

 

[stapel]

Re: Solving nonlinear equations for real numbers

So when I do that I get x^2-x^2+2x+1=1.

How did you arrive at this result? Please reply with all of your steps.

But then how would I solve that?

What is x[sup:uq88j0la]2[/sup:uq88j0la] - x[sup:uq88j0la]2[/sup:uq88j0la]?

Eliz.

 

[lp_princess05]

Re: Solving nonlinear equations for real numbers

So if it is 2x+1=1, then y would equal 1. So how would you graph that?

 

[masters]

Re: Solving nonlinear equations for real numbers

x[sup:1c43ilvv]2[/sup:1c43ilvv] - (x + 1)[sup:1c43ilvv]2[/sup:1c43ilvv] = 1
x[sup:1c43ilvv]2[/sup:1c43ilvv] - (x[sup:1c43ilvv]2[/sup:1c43ilvv] + 2x + 1) = 1
x[sup:1c43ilvv]2[/sup:1c43ilvv] - x[sup:1c43ilvv]2[/sup:1c43ilvv] -2x - 1 = 1
-2x - 1 = 1
-2x = 2
x = -1

Since y = x + 1
y = -1 + 1
y = 0

Solution(-1, 0)

The graph is a line that intersects a hyperbola in one point. The hyperbola's transverse axis is the x-axis and the line intersects the left branch at the vertex (-1, 0)