Honors Algebra 2: Systems of 3 Linear Equations
- Class: Honors Algebra 2
- Author: Peter Atlas
- Text: Algebra and Trigonometry: Structure and Method, Brown
Calculator: You may use a calculator for help with the arithmetic, but you may not use a systems program or matrices.
Solve the following systems.
- \( 2x + 3y + 6z = 13 \\
3x + 4y + 2z = 11 \\
x - 3y + 5z = -5\)
Solution
(-1, 3, 1)
- \( 2x + y - z = -1 \\
-3x + 2y + 2z = 9 \\
x + y - z = 0\)
Solution
(-1, 2, 1)
- \( 6x + 3y + 2z = 2 \\
-7x -5y + 3z = 22 \\
x + 2y - 5 z= -24\)
Solution
(0, -2, 4)
- \( 3x + 5y - 7z = -1 \\
-2x + 7y - 3z = -2 \\
x - y + z = -1\)
Solution
(-1, -1, -1)
- \( 5x + 3y - 2z = 0 \\ 9x + 3y + 10z = 0 \\ 7x - 9y + 3z = 0\)
Solution
This is a homogeneous system with the single solution (0, 0, 0)
- \( -2x + 4z = 9y + 38 \\ z = 2x - 3y -19 \\ 5x + 3y - 4z -1 = 0\)
Solution
(5, -4, 3)
- \( 3x - 9y + 7z = -21 \\ -4x + 5y - 2z = 5 \\ -11x + y + 12z = -53\)
Solution
(-2, -3, -6)
- \( -2x - 8y - 7z = 87 \\ x + 9y - 6z = -2 \\ -10x + 4y - 5z = 5\)
Solution
(1, -5, -7)
- \( -11x + 3y + 12z = 61 \\ -12x - 11y + 8z = 195 \\ -10x - 6y - 2z = 134\)
Solution
(-8, -9, 0)
- \( 3x + 4y + 9z = -51 \\ -y - 5z = 24 \\ -11x - 7y - 12z = 55\)
Solution
(4, -9, -3)
- \( 10x + 3y - z = -55 \\ 3x - 6y - 6z = 36 \\ -6x + 2y - 9z = 106\)
Solution
(-6, -1, -8)
- \( 4y + 11z = 85 \\ -3x + 6y - 3z = 3 \\ -8x - 9y - 12z = -70\)
Solution
(-4, 2, 7)
- \( 4x + y - 2z = 0 \\ 2x - 3y + 3z = 9 \\ -6x - 2y + z = 0\)
Solution
\( \displaystyle \left( \frac{3}{4}, -2, \frac{1}{2} \right) \)
- \( x - y = 2 \\
3x + z = 11 \\
y - 2z = -3\)
Solution
(3, 1, 2)
- In a coin bank, there are three times as many dimes as there are nickles and quarters combined. The total value of the 24 coins (all dimes, nickles, and quarters) is $2.90. How many of each kind of coin are there?
Solution
18 dimes, 4 quarters, 2 nickles