Practice 2 Solutions

Solve the system of equations.

$- 2 x + 3 y + 5 z = - 7 - 6 x - 2 y - z = - 15 - 4 x + 4 y + 5 z = - 15$

$\begin{matrix} - 3 (- 2 x + 3 y + 5 z = - 7) & \Rightarrow \end{matrix} \begin{array}{l} 6 x - 9 y - 15 z = 21 \\ + - 6 x - 2 y - z = - 15 \\ - 11 y - 16 z = 6 \end{array}$

$\begin{matrix} - 2 (- 2 x + 3 y + 5 z = - 7) & \Rightarrow \end{matrix} \begin{array}{l} 4 x - 6 y - 10 z = 14 \\ + - 4 x + 4 y + 5 z = - 15 \\ - 2 y - 5 z = - 1 \end{array}$

$\begin{matrix} - 2 (- 11 y - 16 z = 6) & \Rightarrow \\ 11 (- 2 y - 5 z = - 1) & \Rightarrow \end{matrix} \begin{array}{l} 22 y + 32 z = - 12 \\ + - 22 y - 55 z = - 11 \\ - 23 z = - 23 \\ z = 1 \end{array}$

$\begin{array}{rcl} - 2 y - 5 (1) & = & - 1 \begin{matrix} - 2 x + 3 (- 2) + 5 (1) = - 7 \end{matrix} \\ - 2 y - 5 & = & - 1 \begin{matrix} - 2 x - 6 + 5 = - 7 \end{matrix} \\ - 2 y & = & 4 \begin{matrix} - 2 x = - 6 \end{matrix} \\ y & = & - 2 \begin{matrix} x = 3 \end{matrix} \end{array}$

(3, −2, 1)

$2 x - 5 y + 2 z = - 5 - 4 x + 10 y - 4 z = 10 5 x + 6 y - z = 18$

$\begin{matrix} 2 (5 x + 6 y - z = 18) & \Rightarrow \end{matrix} \begin{array}{l} 10 x + 12 y - 2 z = 36 \\ + 2 x - 5 y + 2 z = - 5 \\ 12 x + 7 y = 31 \end{array}$

$\begin{matrix} - 4 (5 x + 6 y - z = 18) & \Rightarrow \end{matrix} \begin{array}{l} - 20 x - 24 y + 4 z = - 72 \\ + - 4 x + 10 y - 4 z = 10 \\ - 24 x - 14 y = - 62 \end{array}$

$\begin{matrix} 2 (12 x + 7 y = 31) & \Rightarrow \end{matrix} \begin{array}{l} 24 x + 14 y = 62 \\ + - 24 x - 14 y = - 62 \\ 0 = 0 \end{array}$

$\begin{matrix} 2 (2 x - 5 y + 2 z = - 5) & \Rightarrow \end{matrix} \begin{array}{l} 4 x - 10 y + 4 z = - 10 \\ + - 4 x + 10 y - 4 z = 10 \\ 0 = 0 \end{array}$

This system has infinitely many solutions.

$6 x + 3 y - 5 z = 5 - 2 x - 3 y - z = - 1 2 x + y - 3 z = - 1$

$\begin{array}{l} 6 x + 3 y - 5 z = 5 \\ + - 2 x - 3 y - z = - 1 \\ 4 x - 6 z = 4 \end{array}$

$\begin{array}{ccl} 3 (2 x + y - 3 z = - 1) & \Rightarrow & 6 x + 3 y - 9 z = - 3 \\ + - 2 x - 3 y - z = - 1 \\ 4 x - 10 z = - 4 \end{array}$

$\begin{array}{ccl} - 1 (4 x - 6 z = 4) & \Rightarrow & - 4 x + 6 z = - 4 \\ + 4 x - 10 z = - 4 \\ - 4 z = - 8 \\ z = 2 \end{array}$

$\begin{array}{rcl} 4 x - 6 (2) & = & 4 \begin{matrix} 2 (4) + y - 3 (2) = - 1 \end{matrix} \\ 4 x & = & 16 \begin{matrix} y + 2 = - 1 \end{matrix} \\ x & = & 4 \begin{matrix} y = - 3 \end{matrix} \end{array}$

(4, −3, 2)

$x - 2 y + 4 z = - 4 3 x + 4 y - 5 z = 25 5 x - 3 y + 2 z = 12$

$\begin{array}{ccl} - 3 (x - 2 y + 4 z = - 4) & \Rightarrow & - 3 x + 6 y - 12 z = 12 \\ + 3 x + 4 y - 5 z = 25 \\ 10 y - 17 z = 37 \end{array}$

$\begin{array}{ccl} - 5 (x - 2 y + 4 z = - 4) & \Rightarrow & - 5 x + 10 y - 20 z = 20 \\ + 5 x - 3 y + 2 z = 12 \\ 7 y - 18 z = 32 \end{array}$

$\begin{array}{ccl} - 7 (10 y - 17 z = 37) & \Rightarrow & - 70 y + 119 z = - 259 \\ 10 (7 y - 18 z = 32) & \Rightarrow & + 70 y - 180 z = 320 \\ - 61 z = 61 \\ z = - 1 \end{array}$

$\begin{array}{rcl} 10 y - 17 (- 1) & = & 37 \begin{matrix} x - 2 (2) + 4 (- 1) = - 4 \end{matrix} \\ 10 y + 17 & = & 37 \begin{matrix} x - 8 = - 4 \end{matrix} \\ 10 y & = & 20 \begin{matrix} x = 4 \end{matrix} \\ y & = & 2 \begin{matrix} \end{matrix} \end{array}$

(4, 2, −1)

Note

Problems 5–7
You may use any variables you prefer. Remember to define your variables before writing equations.

The perimeter of a right triangle is 30 cm. The length of the hypotenuse is the sum of twice the length of the shorter leg plus three. The length of the longer leg is one less than the hypotenuse. Find the measurement of each side.

a = shorter leg, b = longer leg, c = hypotenuse

$a + b + c = 30 c = 2 a + 3 b = c - 1$

$\begin{array}{rcl} a + (c - 1) + (2 a + 3) & = & 30 \\ a + (2 a + 3) - 1 + 2 a + 3 & = & 30 \\ 5 a + 5 & = & 30 \\ 5 a & = & 25 \\ a & = & 5 \end{array}$

$b = 13 - 1 b = 12$

$\begin{array}{rcl} c & = & 2 (5) + 3 \\ c & = & 13 \end{array}$

The sides of the triangle are 5 cm, 12 cm, 13 cm.

Note

Recall that the longest side of a right triangle is the hypotenuse.

Alternate version to solve using elimination

$c = 2 a + 3 \Rightarrow - 2 a + c = 3 b = c - 1 \Rightarrow b - c = - 1$

$a + b + c = 30 + b - c = - 1 a + 2 b = 29$

$\begin{matrix} (- 2 a + c = 3) (- 1) & \Rightarrow \end{matrix} \begin{array}{l} a + b + c = 30 \\ + 2 a - c = - 3 \\ 3 a + b = 27 \end{array}$

$\begin{matrix} (3 a + b = 27) (- 2) & \Rightarrow \end{matrix} \begin{array}{l} - 6 a - 2 b = - 54 \\ a + 2 b = 29 \\ - 5 a = - 25 \\ a = 5 \end{array}$

$\begin{array}{rcl} c & = & 2 (5) + 3 \begin{matrix} b = (13) - 1 \end{matrix} \\ c & = & 13 \begin{matrix} b = 12 \end{matrix} \end{array}$

A local library needs to buy 44 children’s, juvenile, and young adult books for its inventory. Each children’s book costs the library $2, the juvenile books cost $5 each, while the young adult books cost $6 each. The library has a budget of $184. Twice the amount of young adult novels minus one is equal to the total of children’s and juvenile books. Determine the number of each type of book.

c = children’s books, j = juvenile, y = young adult

$\begin{array}{rcl} c + j + y & = & 44 \\ 2 c + 5 j + 6 y & = & 184 \\ 2 y - 1 & = & c + j \end{array}$

$\begin{array}{ccl} 2 y - 1 = c + j & \Rightarrow & - c - j + 2 y = 1 \\ + c + j + y = 44 \\ 3 y = 45 \\ y = 15 \end{array}$

$c + j + 15 = 44 \Rightarrow c + j = 29 2 c + 5 j + 6 (15) = 184 \Rightarrow 2 c + 5 j = 94$

$\begin{array}{ccl} - 2 (c + j = 29) & \Rightarrow & - 2 c - 2 j = - 58 \\ + 2 c + 5 j = 94 \\ 3 j = 36 \\ j = 12 \end{array}$

$\begin{array}{rcl} c + 12 & = & 29 \\ c & = & 17 \end{array}$

Note

Remember to define your variables before writing your equations and solving.

The library bought 17 children’s books, 12 juvenile books, and 15 young adult books.