The systems in Exercises 25 and 26 exhibit a “lower triangular” pattern that makes them easy to solve by forward substitution. (We will encounter forward substitution again in Chapter 3.) Solve these systems. x 1 = − 1 − 1 2 x 1 + x 2 = 5 3 2 x 1 + 2 x 2 + x 3 = 7
The systems in Exercises 25 and 26 exhibit a “lower triangular” pattern that makes them easy to solve by forward substitution. (We will encounter forward substitution again in Chapter 3.) Solve these systems. x 1 = − 1 − 1 2 x 1 + x 2 = 5 3 2 x 1 + 2 x 2 + x 3 = 7
Solution Summary: The author explains how to solve the given system by back substitution.
The systems in Exercises 25 and 26 exhibit a “lower triangular” pattern that makes them easy to solve by forward substitution. (We will encounter forward substitution again in Chapter 3.) Solve these systems.
x
1
=
−
1
−
1
2
x
1
+
x
2
=
5
3
2
x
1
+
2
x
2
+
x
3
=
7
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