4.2 Notes - Step 2: Choose a pair of equations and add them...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
4.2 Solving Systems of Equations in Three Variables Goal:  Solve three equations with three unknowns The solution of a linear equation in three variables is an  ordered triple (x, y, z) that makes the equation true. HW problem #1: Which equations have (-1, 3, 1) as a solution? a) x + y + z = 3 b) –x + y + z = 5 c) –x + y + 2z = 0 d) x + 2y – 3z = 2 We can use elimination or substitution to solve a system of three equations, just like we  did with a system of two equations.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Solving a System of Three Equations using Elimination: Step 1: Write each equation in standard form Ax + By+ Cz = D
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Step 2: Choose a pair of equations and add them to eliminate a variable. Step 3: Choose another pair of equations and add to eliminate the same variable as in step 2. Step 4: Solve the resulting system of two equations for both variables. Step 5: Substitute the values of the two known variables into any of the original equations and solve for the third variable. HW problem #5: x y + z = -4 3x + 2y z = 5-2x + 3y z = 15 Homework: p. 233 #1-16...
View Full Document

Page1 / 2

4.2 Notes - Step 2: Choose a pair of equations and add them...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online