### About 3 Variable System of Equations:

To solve a linear system in three variables, we can use an extension of the elimination method. We begin by eliminating one of the variables from any two equations. We then eliminate the same variable in any other two equations. This creates a linear system in two variables. We solve this system and use substitution to find the third unknown.

Test Objectives

- Demonstrate the ability to eliminate one of the variables from a three variable linear system
- Demonstrate the ability to solve a linear system in two variables
- Demonstrate the ability to solve a linear system in three variables

#1:

Instructions: Solve each linear system.

a) $$-5y - 7z = -9$$ $$x - y + 4z = 5$$ $$-x + 3y + 3z = 7$$

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#2:

Instructions: Solve each linear system.

a) $$-4x + 2y + z = 11$$ $$-3x + 5y + 6z = -4$$ $$-2x + 4y - z = 25$$

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#3:

Instructions: Solve each linear system.

a) $$5x - y + z = 22$$ $$-4x - 4y + 4z = 20$$ $$-4x - 3y + 3z = -13$$

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#4:

Instructions: Solve each linear system.

a) $$5x + y + 4z = -7$$ $$x + 7y - 6z = -15$$ $$-7x - 6y - z = 19$$

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#5:

Instructions: Solve each linear system.

a) $$6x + 6y - 10z = 17$$ $$6x + 21y + 3z = 28$$ $$30x - 12y - 12z = 17$$

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Written Solutions:

#1:

Solutions:

a) {(-4,-1,2)} : x = -4, y = -1, z = 2

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#2:

Solutions:

a) {(-2,4,-5)} : x = -2, y = 4, z = -5

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#3:

Solutions:

a) No solution : ∅

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#4:

Solutions:

a) {(-∞,∞)} : infinite number of solutions

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#5:

Solutions:

a) $$\left\{\left(\frac{5}{6},\frac{7}{6},-\frac{1}{2}\right) \right\} : x = \frac{5}{6},\hspace{.25em} y = \frac{7}{6},\hspace{.25em} z = -\frac{1}{2}$$ $$\left\{\left(\frac{5}{6},\frac{7}{6},-\frac{1}{2}\right) \right\} :$$ $$x = \frac{5}{6},\hspace{.25em} y = \frac{7}{6},\hspace{.25em} z = -\frac{1}{2}$$