Stuck on a question - Please look over my work and help me with the last step?

Solve each non-linear system





y = 2/x


y = 4x - 2





I found the X values to be 1, and -0.5





How do the find the X value that corresponds to the original half of the system, with no possibility of error?





What I did was make it from





y = 2/x


y = 4x - 2





to





2/x = 4x - 2





to





2 = 4x^2 - 2x


0 = 4x^2 - 2x - 2





Then used the quadratic formula for find the X values.





Now I just need to know how to determine which X values match up with which halves of the system so that I can find the RIGHT Y values.





Help me!Stuck on a question - Please look over my work and help me with the last step?
2/x = 4x - 2


4x虏 - 2x = 2


x虏 - 1/4x = 1/2 + (- 1/4)虏


x虏 - 1/4x = 8/16 + 1/16


(x - 1/4)虏 = 9/16


x - 1/4 = 卤 3/4





x = 3/4 + 1/4, x = 4/4, x = 1


x = - 3/4 + 1/4, x = - 2/4, x = - 1/2





Answer: x = 1, - 1/2


-----------


y = 2/1


y = 1





y = 4(1) - 2


y = 4 - 2


y = 2





y = 2/(- 1/2)


y = - 4





Answer: y = 2, - 4Stuck on a question - Please look over my work and help me with the last step?
The correct x-value must make the same y-value come up in BOTH equations.





This occurs when x is -0.5 like you solved for.





When x=1, both equations are not the same, which they are supposed to be, so this is Not a solution.





The only solution is x = -0.5, y = - 4. ( -0.5, -4 )
You have the ';Right'; x values that solve the system of equations. It's not a case of one value solves one equation and the other solves the other. Substitute the x values into either equation to find the y- values. Each x value will generate the same y-value in both equations.





Substituting x = 1 into the first equation:


y =2/1 = 2


and into the second equation:


y = 4-2 = 2





Substituting x = -0.5 into the first equation:


y = 2/(-.5) = -4


and into the second:


y = 4(-.5) - 2 = -2 - 2 = -4





What you have are the equations of a hyperbola and a straight line. The x and y values are the points of intersection of that hyperbola and the straight line which intersect at two points. The points solve both equations simultaneously - that's one reason a system of equations like that is also called a set of ';simultaneous'; equations.
There are TWO x values and TWO y values. Substitute each values of x into the equations you get TWO answers. BOTH answers are correct.


x = 1, y = 2


x = -0.5, y = -4