Ch. 7 – Matrices and Systems of Equations

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Ch. 7 – Matrices and Systems of Equations 7.1 - Substitution

Plug it in for y in the other eqn., then solve for x! Substitution Ex: Solve the system by substitution. The first eqn. is solved for y already! Since y = x + 3, I can substitute x + 3 for y in the second equation! After substituting, solve for x! OK, so x=1. But we need an answer in the form of (x, y)… I still need a y, so plug 1 in for x and solve for y! You choose which equation to plug 1 back into. y=4, so the answer is (1, 4)! Verify by graphing! This eqn. is solved for y! Plug it in for y in the other eqn., then solve for x!

Solve the system by substitution: (1, -1) (-1, -2) (2, -1/2) (1/2, 4) (-1, 1)

Solve the system: (1, 0) (-3, 16) (3, -8) (-1, 8) No solution Infinite solutions

Ex: Solve the following system: Uh-oh, we’re going to have a quadratic! Solve the linear equation for y because the substitution will be easy! Substitute to get x2 + 4x – (2x + 1) = 7 x2 + 2x – 8 = 0 To solve, factor or use quadratic formula! (x + 4)(x – 2) = 0 x = -4 or x = 2 Plug them back in to get the y-values… Answer: (-4, -7) and (2, 5) Graph the equations to verify your answer!

Ex: Solve the following system: Solve the linear equation for y because the substitution will be easy! Substitute to get x2 + x + 4 = 3 x2 + x + 1 = 0 To solve, factor or use quadratic formula! This equation doesn’t factor, and the quadratic formula gives a non-real answer, so… No Solution! Graph each equation to verify.

Solve the system by substitution: (1, -1) and (-3, -9) (0, -3) and (2, 1) (3, 3) (-1, -5) and (3, 3) No solution

Solve the system by substitution: (0, -5) and (4, 11) (0, -5) and (1, -1) (0, 5) and (3, 7) (1, -1) and (4, 11) No solution

Find the solution to the following system by graphing: (2, 0) (2.15, -0.23) (0.18, -2.73), (1.67, -0.48) (3.41, 0.88) No solution