How do I solve the equations x squared - y squared=1
2x squared - y squared= x+3 using substitution, elimination and graphing method?
What, exactly, is your difficulty? Do you not know what any of those words mean?
To solve a pair of equation "by substitution" means to solve one equation for something and substitute that into the othe equation. Solve the equation $\displaystyle x^2- y^2= 1$ for $\displaystyle y^2$ as a function of x (or $\displaystyle x^2$) and then substitute that into $\displaystyle 2x^2- y^2= x+ 3$.
To solve a pair of equation "by elimination" means to algebraically combine the two equation to eliminate one of the unknowns. Here, I suggest that you subtract $\displaystyle x^2- y^2= 1$ from $\displaystyle 2x^2- y^2= x+ 3$ eliminating y.
Finally, to solve a pair of equations "by graphing" means to do exactly that- graph both equations and see where the graphs intersect.
With each of those methods, I get 3 distinct solutions.
I realize the steps however when I do the arithmetic is doesnt seem to come out right. I think Im doing something wrong but I dont know what. I need to see it worked out to figure out where I went wrong.
A better approach would be for you to post what you have, then someone here can point out where you are going wrong. This will do two things...demonstrate that you are making a genuine effort and help you find exactly where the error in your approach is, and then you can avoid this error in the future.
when I solved for x in the equation x^2 - y^2=1 I got x=y+1, is that correct? If so when I plugged it into the equation 2x^2-y^2=x+3 and got the variables set to zero I got y^2+3y-2=0 (this is in regards to the substitution method.)
No, I suggest you read carefully the suggestions provided by HallsofIvy, and follow the steps he has provided.
For example, he suggests solving $\displaystyle x^2-y^2=1$ for $\displaystyle y^2$...what do you get?
You don't need to take the square root of each side (and you did so incorrectly), because the second equation contains $\displaystyle y^2$. Try again to solve for $\displaystyle y^2$ as you did not get it quite right. Add $\displaystyle y^2-1$ to both sides.
No, that is NOT correct. You can get x^2= y^2+ 1 but the square root of y^2+ 1 is NOT y+ 1 For example, if y= 2 y^2+ 1= 5. The square root of 5 is NOT equal to 2+ 1= 3!
When I plugged it into the equation 2x^2-y^2=x+3 and got the variables set to zero I got y^2+3y-2=0 (this is in regards to the substitution method.)