Lesson 76: Using Both Substitution and Elimination, Negative Vectors

We have found that we can use either the substitution method of the elimination method to solve a system of equations in two unknowns.

When we have to solve a system of three equations in three unknowns, it is sometimes helpful if we begin by using substitution and then finish by using elimination.

Example: Use substitution and elimination as necessary to solve this system of equations. x = 2y x + y + z = 9 x – 3y – 2z = -8

Answer: x = 4 y = 2 z = 3 We find that the solution to this system of three equations in three unknowns is the ordered triple (4, 2, 3)

Example: Us substitution and elimination as necessary to solve this system of equations. 2x + 2y – z = 12 3x – y + 2z = 21 x – 3z = 0

Answer: x = 6 y = 1 z = 2 (6, 1, 2)

In lesson 72 we noted that there is only one way to use rectangular coordinates to designate the location of a point, but more than one form of polar coordinates is possible because either positive angles or negative angles may be used.

To make matters even more confusing, we note that it is also possible to use negative magnitudes to locate a point.

Open books to page 318. Look at the diagram of -10<210 degrees.

Example: Add -4<-20° + 5<135°

Answer: -7.29R + 4.90U

HW: Lesson 76 #1-30