How do you solve this system of nonlinear equations? x² + y² = 25 2x + 4y = 10 When solving the system of those two equations, you are finding the point(s) of intersection of a circle and a straight line. x² + y² = 25 {a circle} 2x + 4y = 10 {a straight line} Isolate a variable on the linear equation: 2x + 4y = 10 {the second equation} 2x = 4y  10 {subtracted 4y from each side} x = 2y  5 {divided each side by 2} Substitute the value of that variable into the nonlinear equation: x² + y² = 25 {the first equation} (2y  5)² + y² = 25 {substituted (2y  5) , in for x, into top equation} (2y  5)(2y  5) + y² = 25 {when squaring a binomial, multiply it by itself} 4y² + 20y + 25 + y² = 25 {used foil method} 5y² + 20y + 25 = 25 {combined like terms} 5y² + 20y = 0 {subtracted 25 from each side} 5y(y + 4) = 0 {factored out the common factor, 5y} 5y = 0 or y + 4 = 0 {set each factor equal to 0} y = 0 or y = 4 {solved each equation} x = 2y  5 {the rearranged second equation} If y = 0: x = 2(0)  5 {substituted 0 for y} x = 0  5 {multiplied} x = 5 {subtracted} (5,0) is a solution {a point of intersection of the line and circle} x = 2y  5 {the rearranged second equation} If y = 4: x = 2(4)  5 {substituted 4 for y} x = 8  5 {multiplied} x = 3 {subtracted} (3,4) is a solution {a point of intersection of the line and circle} x = 5 when y = 0 and x = 3 when y = 4 (5,0) and (3,4) are the points of intersection of the line and circle  Algebra House Comments are closed.

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