Solve the system algebraically. Then graph both equations on the same coordinate system to support your solution.
y=x^3-x
y=x
The solution set is {?}
y=x^3-x
y=x
The solution set is {?}
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Use the second equation and substitute into the first:
y = y^3 - y --> 0 = y^3 - 2y --> 0 = y(y^2 - 2) --> y = 0 or y = +-sqrt(2)
Back substitute: y = x
Solutions (0, 0) (sqrt(2), sqrt(2)) and (-sqrt(2), -sqrt(2))
y = y^3 - y --> 0 = y^3 - 2y --> 0 = y(y^2 - 2) --> y = 0 or y = +-sqrt(2)
Back substitute: y = x
Solutions (0, 0) (sqrt(2), sqrt(2)) and (-sqrt(2), -sqrt(2))