Three Unknowns
Part of the Graphs & Systems section of Coddy's Algebra journey. Lesson 58 of 64.
A system can have three unknowns and three equations. You solve it with the same two methods. First eliminate one variable to get a system with two unknowns. Then solve that smaller system. In x + y + z = 6; x - y = 1; y + z = 4, the third equation gives z = 4 - y. Substitute this into the first equation: x + y + (4 - y) = 6. The y terms cancel, so x + 4 = 6 and x = 2. The second equation then gives 2 - y = 1, so y = 1. Finally z = 4 - 1, so z = 3.
Substitution and elimination also work for three unknowns. Eliminate one unknown to obtain equations in two unknowns, solve those equations, and substitute back to find the remaining value.
Challenge
MediumSolve x + y + z = 11; x - y = 3; z = 2y for x, y and z.
Try it yourself
Solve for x,y,z
Test yourselfThis lesson includes a short quiz. Start the lesson to answer it and track your progress.
All lessons in Graphs & Systems
1The Coordinate Plane
Two Numbers, One PointThe Four QuadrantsPoints on the AxesMirror PointsRecap: The Plane4Slope-Intercept Form
y = mx + bThe Starting HeightWrite the EquationSolve for yDraw From Any FormRecap: y = mx + b7Parallel & Perpendicular
Same Slope, Never MeetThe Perpendicular SlopeA Right-Angle LineRecap: Two Relations2From Rule to Line
A Rule Makes PointsThe Points Line UpOn the Line or Not?Between the DotsRecap: Rule to Line5Point-Slope Form
A Point and a SlopeFrom Two PointsFlat or Vertical ThroughDraw From a PointRecap: Point-Slope11Solving Systems Exactly
SubstitutionIsolate FirstEliminationScale, Then AddThree UnknownsTwo Numbers, Two FactsRecap: Two Methods