Solving three-variable systems
Solve three-variable linear systems in Grade 10 algebra using elimination or substitution to reduce to two equations, finding the ordered triple (x, y, z) solution.
Key Concepts
To solve a system of three equations, first isolate one variable in one equation. Then, substitute this expression into the other two equations to create a new system with only two variables and two equations. Solve this smaller system for those two variables. Finally, substitute those values back into an original equation to find the value of the third variable.
Given $x 2y + z = 7$, solve for $x$ to get $x = 2y z + 7$. Now substitute $(2y z + 7)$ for $x$ in other equations. After substitution, you might get a 2 variable system like $y + 5z = 10$ and $2y z = 4$. Solve this smaller system first. If you find $y=3$ and $z=2$ from the smaller system, plug them back into an original equation to find the value of $x$.
It's like a detective mission! Use one clue (equation) to get a description of a suspect (variable). Then, plug that description into your other two clues. This creates a simpler two suspect mystery that’s much easier to crack! Once you solve that, you can find out who the last suspect is.
Common Questions
How do you solve a three-variable linear system?
Use elimination to reduce three equations to two by eliminating one variable, then reduce to one equation with one variable. Back-substitute to find all three values.
What is an ordered triple and how do you verify it?
An ordered triple (x,y,z) is the solution to a three-variable system. Verify by substituting all three values into each of the original three equations to check that all equations are satisfied.
What indicates a three-variable system has no unique solution?
If elimination produces a contradictory equation (e.g., 0=5), there is no solution. If it produces an identity (0=0), the system has infinitely many solutions (a line or plane of solutions).