Systems of Nonlinear Equations Substitution

Substitution Just like linear systems, solving by substitution means: Rewrite one equation and plug it into the other. You will have to use FOIL on most problems. Remember, if you have a binomial you cannot just distribute the exponent! (x – 2)2 = x2 – 4x + 4 NOT (x – 2)2 ≠ x2 + 4 We will be working with systems that can have multiple solutions, so keep your work organized.

Substitution Let’s give it a try: x2 + y2 = 5 x2 + (x + 1)2 = 5 x2 + x2 + 2x + 1 = 5 2x2 + 2x + 1 = 5 2x2 + 2x – 4 = 0 2(x2 + x – 2) = 0 2(x – 1)(x + 2) = 0 x = 1 x = -2 y = 1 + 1 y = -2 + 1 y = 2 y = -1 (1, 2) (-2, -1) Pick the variable you want to solve for. Rewrite the equation. Plug it into the other equation. FOIL Combine like terms. Factor Solve for x. Plug x’s back in to find y’s.

Substitution Try it again. x2 + y2 = 13 x + y = 5 y = -x + 5 x2 + (-x + 5)2 = 13 x2 + x2 – 10x + 25 = 13 2x2 – 10x + 12 = 0 2(x2 – 5x + 6) = 0 2(x – 3)(x – 2) = 0 x = 3 x = 2 y = -3 + 5 y = -2 + 5 y = 2 y = 3 (3, 2) (2, 3)

Substitution Things to watch out for: Watch your signs! If there is a negative, don’t forget to carry it through the problem. Make sure you have the right factors. If you factor incorrectly you will get the wrong answers. Check your answers by plugging them back into both equations. Keep your work organized. If you can’t tell what you are doing, you will get lost!

Substitution Try one for me! x2 + y2 – 2x – 4y = 0 -2x + y = -3 x2 + (2x – 3)2 – 2x – 4(2x – 3) = 0 x2 + 4x2 – 12x + 9 – 2x – 8x + 12 = 0 5x2 – 22x +21 = 0 x2 – 22x +105 = 0 (x -15)(x – 7) = 0 (x – 3)(5x – 7) = 0 x = 3 x = 7/5 y = 2(3) – 3 y = 2(7/5) – 3 y = 3 y = -1/5 Try one for me! x2 + y2 – 2x – 4y = 0 -2x + y = -3 (3 ,3) ( 𝟕 𝟓 , − 𝟏 𝟓 )