In Exercises triangfirst - trianglast, put each system of linear equations into triangular form and solve the system if possible. Classify each system as consistent independent, consistent dependent, or inconsistent.

NOTE: Because triangular form is not unique, we give only one possible answer to that part of the question. Yours may be different and still be correct.

\(\left \{ \begin{array}{rcr} -5x + y & = & 17 \\ x + y & = & 5 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x + y & = & 5 \\ y & = & 7 \end{array} \right .\)

Consistent independent
Solution \((-2, 7)\)

\(\left \{ \begin{array}{rcr} x + y + z & = & 3 \\ 2x - y + z & = & 0 \\ -3x + 5y + 7z & = & 7 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x - \frac {5}{3}y - \frac {7}{3}z & = & -\frac {7}{3} \\ y + \frac {5}{4}z & = & 2 \\ z & = & 0 \end{array} \right .\)

Consistent independent
Solution \((1, 2, 0)\)

\(\left \{ \begin{array}{rcr} 4x - y + z & = & 5 \\ 2y + 6z & = & 30 \\ x + z & = & 5 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x - \frac {1}{4}y + \frac {1}{4}z & = & \frac {5}{4} \\ y + 3z & = & 15 \\ 0 & = & 0 \end{array} \right .\)

Consistent dependent
Solution \((-t + 5, -3t + 15, t)\)
for all real numbers \(t\)

\(\left \{ \begin{array}{rcr} 4x - y + z & = & 5 \\ 2y + 6z & = & 30 \\ x + z & = & 6 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x - \frac {1}{4}y + \frac {1}{4}z & = & \frac {5}{4} \\ y + 3z & = & 15 \\ 0 & = & 1 \end{array} \right .\)

Inconsistent
No solution

\(\left \{ \begin{array}{rcr} x + y + z & = & -17 \\ y - 3z & = & 0 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x + y + z & = & -17 \\ y - 3z & = & 0 \end{array} \right .\)

Consistent dependent
Solution \((-4t - 17, 3t, t)\)
for all real numbers \(t\)

\(\left \{ \begin{array}{rcr} x-2y+3z & = & 7 \\ -3x+y+2z & = & -5 \\ 2x+2y+z & = & 3 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x-2y+3z & = & 7 \\ y - \frac {11}{5}z & = & -\frac {16}{5} \\ z & = & 1 \\ \end{array} \right .\)

Consistent independent
Solution \((2,-1,1)\)

\(\left \{ \begin{array}{rcr} 3x-2y+z & = & -5 \\ x+3y-z & = & 12 \\ x+y+2z & = & 0 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x+y+2z & = & 0 \\ y - \frac {3}{2}z & = & 6 \\ z & = & -2 \\ \end{array} \right .\)

Consistent independent
Solution \((1,3,-2)\)

\(\left \{ \begin{array}{rcr} 2x-y+z& = & -1 \\ 4x+3y+5z & = & 1 \\ 5y+3z & = & 4 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x - \frac {1}{2} y + \frac {1}{2} z & = & -\frac {1}{2} \\ y + \frac {3}{5} z & = & \frac {3}{5} \\ 0 & = & 1 \\ \end{array} \right .\)

Inconsistent
no solution

\(\left \{ \begin{array}{rcr} x-y+z & = & -4 \\ -3x+2y+4z & = & -5 \\ x-5y+2z & = & -18 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x-y+z & = & -4 \\ y - 7z & = & 17 \\ z & = & -2 \\ \end{array} \right .\)

Consistent independent
Solution \((1,3,-2)\)

\(\left \{ \begin{array}{rcr} 2x-4y+z & = & -7 \\ x-2y+2z & = & -2 \\ -x+4y-2z & = & 3 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x-2y+2z & = & -2 \\ y & = & \frac {1}{2} \\ z & = & 1 \\ \end{array} \right .\)

Consistent independent
Solution \(\left (-3,\frac {1}{2},1\right )\)

\(\left \{ \begin{array}{rcr} 2x-y+z & = & 1 \\ 2x+2y-z & = & 1 \\ 3x+6y+4z & = & 9 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x-\frac {1}{2} y+\frac {1}{2} z & = & \frac {1}{2} \\ y - \frac {2}{3} z & = & 0 \\ z & = & 1 \\ \end{array} \right .\)

Consistent independent
Solution \(\left (\frac {1}{3},\frac {2}{3},1\right )\)

\(\left \{ \begin{array}{rcr} x-3y-4z & = & 3 \\ 3x+4y-z & = & 13 \\ 2x-19y-19z & = & 2 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x-3y-4z & = & 3 \\ y + \frac {11}{13} z & = & \frac {4}{13} \\ 0 & = & 0 \\ \end{array} \right .\)

Consistent dependent
Solution \(\left (\frac {19}{13} t + \frac {51}{13},-\frac {11}{13} t+\frac {4}{13},t\right )\)
for all real numbers \(t\)

\(\left \{ \begin{array}{rcr} x+y+z & = & 4 \\ 2x-4y-z& = & -1 \\ x-y & = & 2 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x+y+z & = & 4 \\ y + \frac {1}{2} z & = & \frac {3}{2} \\ 0 & = & 1 \\ \end{array} \right .\)

Inconsistent
no solution

\(\left \{ \begin{array}{rcr} x-y+z & = & 8 \\ 3x+3y-9z & = & -6 \\ 7x-2y+5z & = & 39 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x- y + z & = & 8 \\ y -2z & = & -5 \\ z & = & 1 \\ \end{array} \right .\)

Consistent independent
Solution \(\left (4,-3,1\right )\)

\(\left \{ \begin{array}{rcr} 2x-3y+z & = & -1 \\ 4x-4y+4z & = & -13 \\ 6x-5y+7z & = & -25 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x- \frac {3}{2} y + \frac {1}{2} z & = & -\frac {1}{2} \\ y + z & = & -\frac {11}{2} \\ 0 & = & 0 \\ \end{array} \right .\)

Consistent dependent
Solution \(\left (-2t - \frac {35}{4},-t - \frac {11}{2},t\right )\)
for all real numbers \(t\)

\(\left \{ \begin{array}{rcr} 2x_1 + x_2 - 12x_3 - x_4 & = & 16 \\ -x_1 + x_2 + 12x_3 - 4x_4 & = & -5 \\ 3x_1 + 2x_2 - 16x_3 - 3x_4 & = & 25 \\ x_1 + 2x_2 - 5x_4 & = & 11 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x_1 + \frac {2}{3}x_2 - \frac {16}{3}x_3 - x_4 & = & \frac {25}{3} \\ x_2 + 4x_3 - 3x_4 & = & 2 \\ 0 & = & 0 \\ 0 & = & 0 \end{array} \right .\)

Consistent dependent
Solution \((8s - t + 7, -4s + 3t + 2, s, t)\)
for all real numbers \(s\) and \(t\)

\(\left \{ \begin{array}{rcr} x_1 - x_3 & = & -2 \\ 2x_2 - x_4 & = & 0 \\ x_1 - 2x_2 + x_3 & = & 0 \\ -x_3 + x_4 & = & 1 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x_1 - x_3 & = & -2 \\ x_2 - \frac {1}{2}x_4 & = & 0 \\ x_3 - \frac {1}{2} x_4 & = & 1 \\ x_4 & = & 4 \end{array} \right .\)

Consistent independent
Solution \((1, 2, 3, 4)\)

\(\left \{ \begin{array}{rcr} x_1 - x_2 - 5x_3 + 3x_4 & = & -1 \\ x_1 + x_2 + 5x_3 - 3x_4 & = & 0 \\ x_2 + 5x_3 - 3x_4 & = & 1 \\ x_1 - 2x_2 - 10x_3 + 6x_4 & = & -1 \end{array} \right .\)

\(\left \{ \begin{array}{rcr} x_1 - x_2 - 5x_3 + 3x_4 & = & -1 \\ x_2 + 5x_3 - 3x_4 & = & \frac {1}{2} \\ 0 & = & 1 \\ 0 & = & 0 \end{array} \right .\)

Inconsistent
No solution

Find two other forms of the parametric solution to Exercise dependentsystemmuliple above by reorganizing the equations so that \(x\) or \(y\) can be the free variable.

If \(x\) is the free variable then the solution is \((t, 3t, -t + 5)\) and if \(y\) is the free variable then the solution is \(\left (\frac {1}{3}t, t, -\frac {1}{3}t + 5\right )\).
At The Crispy Critter’s Head Shop and Patchouli Emporium along with their dried up weeds, sunflower seeds and astrological postcards they sell an herbal tea blend. By weight, Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile?

\(\frac {4}{3}- \frac {1}{2}t\) pounds of Type I, \(\frac {2}{3} - \frac {1}{2}t\) pounds of Type II and \(t\) pounds of Type III where \(0 \leq t \leq \frac {4}{3}\).
Discuss with your classmates how you would approach Exercise herbalteablend above if they needed to use up a pound of Type I tea to make room on the shelf for a new canister.
If you were to try to make 100 mL of a \(60\%\) acid solution using stock solutions at \(20\%\) and \(40\%\), respectively, what would the triangular form of the resulting system look like? Explain.