Nonlinear Systems   To solve a system of two equations and two unknowns when the equations are not linear, we use the methods of substitution or elimination and hope that the resulting equation becomes a hidden quadratic or other solvable equation.      Example:   Solve         x2 - y  =  0         x2 - 2x + y  =  6   Solution We use substitution: solving for y in the first equation we get:         y  =  x2 Putting it into the second equation, we have:         x2 - 2x + x2 = 6 so that         2x2 -2x - 6 = 0 or         x2 - x - 3 = 0 The quadratic formula gives us: so          x   =  2.3     or     x   =   -1.3 since         y  =   x2 we get          y = 5.29     or     y = 1.69 We arrive at the two points:         (2.3,5.29)     and     (-1.3,1.69)     Example 2Solve         x2 + y2  =  29         x2 - y2  =  3   Solution We use elimination:  adding the two equations, we have         2x2  =  32         x2  =  16         x   =   4         or         x   =   -4 Putting          x   =   4  into equation 2, we have:         16 - y2 = 3         -y2 = -13         y = or         y = - Putting          x = -4  into equation 2, we have:         16  -  y2   =   3         -y2   =   -13         y   = or         y   =   - Therefore we obtain the four points:         (4, ),     (4,- ),     (-4, ),     (-4,- ) Exercises Solve the following:  x2 - y2  =  4 2x2 + y2  =  16 y - 2x2  =  0 x2 +5x - y  =  -6 x2 - y2  =  21 x2  + xy - y2  =  31 Back to the Intermediate Algebra (Math 154) Home Page e-mail Questions and Suggestions