Practice 1 Solutions

Note

Remember to check your solution in every equation to confirm your solution is correct. Perseverance is critical when solving systems of equations with three variables.
The systems can be solved in more than one way, but the answers will be the same.
5–6) You may use any variables you prefer. Remember to define the variables before writing equations.

Solve the system of equations.

$3 x + 6 y - 4 z = 17 - x + 5 y + 4 z = 11 x + y - 5 z = 0$

$- x + 5 y + 4 z = 11 + x + y - 5 z = 0 6 y - z = 11$

$\begin{matrix} 3 (- x + 5 y + 4 z = 11) & \Rightarrow \end{matrix} \begin{array}{l} - 3 x + 15 y + 12 z = 33 \\ + 3 x + 6 y - 4 z = 17 \\ 21 y + 8 z = 50 \end{array}$

$\begin{matrix} 8 (6 y - z = 11) & \Rightarrow \end{matrix} \begin{array}{l} 48 y - 8 z = 88 \\ + 21 y + 8 z = 50 \\ 69 y = 138 \\ y = 2 \end{array}$

$\begin{array}{rcl} 6 (2) - z & = & 11 \begin{matrix} x + (2) - 5 (1) = 0 \end{matrix} \\ 12 - z & = & 11 \begin{matrix} x + 2 - 5 = 0 \end{matrix} \\ - z & = & - 1 \begin{matrix} x = 3 \end{matrix} \\ z & = & 1 \begin{matrix} \end{matrix} \end{array}$

(3, 2, 1)

$- 3 x - y - 2 z = - 13 2 x + 2 y + z = 16 x + 3 y + 3 z = 13$

$\begin{matrix} 2 (2 x + 2 y + z = 16) & \Rightarrow \end{matrix} \begin{array}{l} 4 x + 4 y + 2 z = 32 \\ + - 3 x - y - 2 z = - 13 \\ x + 3 y = 19 \end{array}$

$\begin{matrix} - 3 (2 x + 2 y + z = 16) & \Rightarrow \end{matrix} \begin{array}{l} - 6 x - 6 y - 3 z = - 48 \\ + x + 3 y + 3 z = 13 \\ - 5 x - 3 y = - 35 \end{array}$

$\begin{array}{l} x + 3 y = 19 \\ + - 5 x - 3 y = - 35 \\ - 4 x = - 16 \\ x = 4 \end{array}$

$\begin{array}{rcl} (4) + 3 y & = & 19 \begin{matrix} (4) + 3 (5) + 3 z = 13 \end{matrix} \\ 3 y & = & 15 \begin{matrix} 19 + 3 z = 13 \end{matrix} \\ y & = & 5 \begin{matrix} 3 z = - 6 \\ z = - 2 \end{matrix} \end{array}$

(4, 5, –2)

$x + 2 y + 3 z = 32 x - 3 y + z = 7 - 2 x + 6 y - 2 z = 4$

$\begin{matrix} 2 (x - 3 y + z = 7) & \Rightarrow \end{matrix} \begin{array}{l} 2 x - 6 y + 2 z = 14 \\ + - 2 x + 6 y - 2 z = 4 \\ 0 = 18 \end{array}$

no solution

$x - 8 y + z = 6 2 x + 7 y - z = 11 2 x - 10 y - 3 z = - 22$

$x - 8 y + z = 6 + 2 x + 7 y - z = 11 3 x - y = 17$

$\begin{matrix} 3 (x - 8 y + z = 6) & \Rightarrow \end{matrix} \begin{array}{l} 3 x - 24 y + 3 z = 18 \\ + 2 x - 10 y - 3 z = - 22 \\ 5 x - 34 y = - 4 \end{array}$

$\begin{matrix} - 34 (3 x - y = 17) & \Rightarrow \end{matrix} \begin{array}{l} - 102 x + 34 y = - 578 \\ + 5 x - 34 y = - 4 \\ - 97 x = - 582 \\ x = 6 \end{array}$

$\begin{array}{rcl} 3 (6) - y & = & 17 \begin{matrix} 6 - 8 (1) + z = 6 \end{matrix} \\ 18 - y & = & 17 \begin{matrix} - 2 + z = 6 \end{matrix} \\ y & = & 1 \begin{matrix} z = 8 \end{matrix} \end{array}$

(6, 1, 8)

Chari has a total of 17 coins in her pocket that are pennies, nickels, and dimes. The value is $0.87. There are two more dimes than nickels. How many of each type of coin does Chari have?

Note

If needed, refer back to Algebra 1: Principles of Secondary Mathematics Lesson 18 to set up a system of equations with coins.

$p + n + d = 17 (0.01 p + 0.05 n + 0.10 d = 0.87) (100) \Rightarrow p + 5 n + 10 d = 87 d = n + 2$

$p + n + (n + 2) = 17 \Rightarrow p + 2 n + 2 = 17 \Rightarrow p + 2 n = 15 p + 5 n + 10 (n + 2) = 87 \Rightarrow p + 15 n + 20 = 87 \Rightarrow p = 15 n = 67$

$\begin{matrix} - (p + 2 n = 15) & \Rightarrow \end{matrix} \begin{array}{l} - p - 2 n = - 15 \\ + p + 15 n = 67 \\ 13 n = 52 \\ n = 4 \end{array}$

$d = (4) + 2 = 6 \begin{matrix} p + 4 + 6 = 17 \\ p = 7 \end{matrix}$

Chari has 7 pennies, 4 nickels, and 6 dimes.

Note

Alternate version to solve using elimination

$d = n + 2 \Rightarrow - n + d = 2 \begin{array}{rcl} p + n + d & = & 17 \end{array} + \begin{matrix} - n + d & = & 2 \end{matrix} \begin{matrix} \begin{array}{r} \begin{array}{r} p + 2 d \end{array} \end{array} & = & 19 \end{matrix}$

$\begin{matrix} (- n + d = 2) (5) & \Rightarrow \end{matrix} \begin{array}{l} p + 5 n + 10 d = 87 \\ + - 5 n + 5 d = 10 \\ p + 15 d = 97 \end{array}$

$\begin{matrix} (p + 2 d = 19) (- 1) & \Rightarrow \end{matrix} \begin{array}{l} p + 15 d = 97 \\ + - p - 2 d = - 19 \\ 13 d = 78 \end{array} \begin{matrix} d = 6 \end{matrix}$

$\begin{array}{rcl} (6) & = & n + 2 \begin{matrix} p + (4) + (6) = 17 \end{matrix} \\ n & = & 4 \begin{matrix} p = 7 \end{matrix} \end{array}$

The average of three numbers is 2. The second is three more than the third. The sum of the first and third is −1. Find the three numbers.

x = first number, y = second number, z = third number

$\frac{x + y + z}{3} = 2 y = 3 + z x + z = - 1$

$\frac{x + y + z}{3} = 2 \Rightarrow x + y + z = 6$

$x + (3 + z) + z = 6 \Rightarrow x + 2 z = 3$

$\begin{matrix} - 1 (x + z = - 1) & \Rightarrow \end{matrix} \begin{array}{l} - x - z = 1 \\ + x + 2 z = 3 \\ z = 4 \end{array}$

$\begin{array}{rcl} x + 2 (4) & = & 3 \begin{matrix} y = 3 + (4) \end{matrix} \\ x & = & - 5 \begin{matrix} y = 7 \end{matrix} \end{array}$

(−5, 7, 4)