Back-Substitution and Triangular Form
Back-substitution and triangular form is a Grade 7 algebra skill from Yoshiwara Intermediate Algebra used to solve systems of linear equations. After using elimination to reach upper triangular form (where each equation has one fewer variable), students substitute known values upward to find all unknowns.
Key Concepts
Property A special case is called back substitution . It works when one of the equations involves exactly one variable, and a second equation involves that same variable and just one other variable. A $3 \times 3$ system with these properties is said to be in triangular form .
Examples Solve the system: $x+y+z=8$, $y z=1$, and $2z=4$. From the third equation, $z=2$. Substitute into the second: $y 2=1$, so $y=3$. Substitute both into the first: $x+3+2=8$, so $x=3$. The solution is $(3, 3, 2)$.
Use back substitution on the system $2x y+z=5$, $3y+z=11$, and $z=2$. Substitute $z=2$ into the second equation to find $3y+2=11$, so $y=3$. Substitute both values into the first: $2x 3+2=5$, so $x=3$. The solution is $(3, 3, 2)$.
Common Questions
What is triangular form in a system of equations?
A system is in triangular form when, after elimination, you have one equation in one variable, another in two, and so on. The system looks like a triangle of variables.
How does back-substitution work?
Solve the simplest equation (one variable) first, then substitute that value into the next equation to find another variable, continuing upward until all variables are known.
When do you use back-substitution?
Use back-substitution after reducing a system to triangular form using Gaussian elimination, row operations, or structured elimination.
What is an example of back-substitution?
If z = 3, then 2y + z = 7 gives 2y = 4, y = 2. Then x + y + z = 6 gives x = 1.