9.1E: Exercises
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Exercises
(Review Exercises) In Exercises 1 - 8, take a trip down memory lane and solve the given system using substitution and/or elimination. Classify each system as consistent independent, consistent dependent, or inconsistent. Check your answers both algebraically and graphically.
{ 𝑥 + 2 𝑦 = 5 𝑥 = 6 { 2 𝑦 − 3 𝑥 = 1 𝑦 = − 3 ⎧ { { ⎨ { { ⎩ 𝑥 + 2 𝑦 4 = − 5 3 𝑥 − 𝑦 2 = 1 ⎧ { { ⎨ { { ⎩ 2 3 𝑥 − 1 5 𝑦 = 3 1 2 𝑥 + 3 4 𝑦 = 1 { 1 2 𝑥 − 1 3 𝑦 = − 1 2 𝑦 − 3 𝑥 = 6 { 𝑥 + 4 𝑦 = 6 1 1 2 𝑥 + 1 3 𝑦 = 1 2 ⎧ { { ⎨ { { ⎩ 3 𝑦 − 3 2 𝑥 = − 1 5 2 1 2 𝑥 − 𝑦 = 3 2 ⎧ { { ⎨ { { ⎩ 5 6 𝑥 + 5 3 𝑦 = − 7 3 − 1 0 3 𝑥 − 2 0 3 𝑦 = 1 0
In Exercises 9 - 26, 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.
{ − 5 𝑥 + 𝑦 = 1 7 𝑥 + 𝑦 = 5 ⎧ { { ⎨ { { ⎩ 𝑥 + 𝑦 + 𝑧 = 3 2 𝑥 − 𝑦 + 𝑧 = 0 − 3 𝑥 + 5 𝑦 + 7 𝑧 = 7 ⎧ { { ⎨ { { ⎩ 4 𝑥 − 𝑦 + 𝑧 = 5 2 𝑦 + 6 𝑧 = 3 0 𝑥 + 𝑧 = 5 ⎧ { { ⎨ { { ⎩ 4 𝑥 − 𝑦 + 𝑧 = 5 2 𝑦 + 6 𝑧 = 3 0 𝑥 + 𝑧 = 6 { 𝑥 + 𝑦 + 𝑧 = − 1 7 𝑦 − 3 𝑧 = 0 ⎧ { { ⎨ { { ⎩ 𝑥 − 2 𝑦 + 3 𝑧 = 7 − 3 𝑥 + 𝑦 + 2 𝑧 = − 5 2 𝑥 + 2 𝑦 + 𝑧 = 3 ⎧ { { ⎨ { { ⎩ 3 𝑥 − 2 𝑦 + 𝑧 = − 5 𝑥 + 3 𝑦 − 𝑧 = 1 2 𝑥 + 𝑦 + 2 𝑧 = 0 ⎧ { { ⎨ { { ⎩ 2 𝑥 − 𝑦 + 𝑧 = − 1 4 𝑥 + 3 𝑦 + 5 𝑧 = 1 5 𝑦 + 3 𝑧 = 4 ⎧ { { ⎨ { { ⎩ 𝑥 − 𝑦 + 𝑧 = − 4 − 3 𝑥 + 2 𝑦 + 4 𝑧 = − 5 𝑥 − 5 𝑦 + 2 𝑧 = − 1 8 ⎧ { { ⎨ { { ⎩ 2 𝑥 − 4 𝑦 + 𝑧 = − 7 𝑥 − 2 𝑦 + 2 𝑧 = − 2 − 𝑥 + 4 𝑦 − 2 𝑧 = 3 ⎧ { { ⎨ { { ⎩ 2 𝑥 − 𝑦 + 𝑧 = 1 2 𝑥 + 2 𝑦 − 𝑧 = 1 3 𝑥 + 6 𝑦 + 4 𝑧 = 9 ⎧ { { ⎨ { { ⎩ 𝑥 − 3 𝑦 − 4 𝑧 = 3 3 𝑥 + 4 𝑦 − 𝑧 = 1 3 2 𝑥 − 1 9 𝑦 − 1 9 𝑧 = 2 ⎧ { { ⎨ { { ⎩ 𝑥 + 𝑦 + 𝑧 = 4 2 𝑥 − 4 𝑦 − 𝑧 = − 1 𝑥 − 𝑦 = 2 ⎧ { { ⎨ { { ⎩ 𝑥 − 𝑦 + 𝑧 = 8 3 𝑥 + 3 𝑦 − 9 𝑧 = − 6 7 𝑥 − 2 𝑦 + 5 𝑧 = 3 9 ⎧ { { ⎨ { { ⎩ 2 𝑥 − 3 𝑦 + 𝑧 = − 1 4 𝑥 − 4 𝑦 + 4 𝑧 = − 1 3 6 𝑥 − 5 𝑦 + 7 𝑧 = − 2 5 ⎧ { { { ⎨ { { { ⎩ 2 𝑥 1 + 𝑥 2 − 1 2 𝑥 3 − 𝑥 4 = 1 6 − 𝑥 1 + 𝑥 2 + 1 2 𝑥 3 − 4 𝑥 4 = − 5 3 𝑥 1 + 2 𝑥 2 − 1 6 𝑥 3 − 3 𝑥 4 = 2 5 𝑥 1 + 2 𝑥 2 − 5 𝑥 4 = 1 1 ⎧ { { { ⎨ { { { ⎩ 𝑥 1 − 𝑥 3 = − 2 2 𝑥 2 − 𝑥 4 = 0 𝑥 1 − 2 𝑥 2 + 𝑥 3 = 0 − 𝑥 3 + 𝑥 4 = 1 ⎧ { { { ⎨ { { { ⎩ 𝑥 1 − 𝑥 2 − 5 𝑥 3 + 3 𝑥 4 = − 1 𝑥 1 + 𝑥 2 + 5 𝑥 3 − 3 𝑥 4 = 0 𝑥 2 + 5 𝑥 3 − 3 𝑥 4 = 1 𝑥 1 − 2 𝑥 2 − 1 0 𝑥 3 + 6 𝑥 4 = − 1 - Find two other forms of the parametric solution to Exercise 11 above by reorganizing the equations so that
or𝑥 can be the free variable.𝑦 - A local buffet charges
per person for the basic buffet and$ 7 . 5 0 for the deluxe buffet (which includes crab legs.) If 27 diners went out to eat and the total bill was$ 9 . 2 5 before taxes, how many chose the basic buffet and how many chose the deluxe buffet?$ 2 2 7 . 0 0 - At The Old Home Fill’er Up and Keep on a-Truckin’ Cafe, Mavis mixes two different types of coffee beans to produce a house blend. The first type costs $3 per pound and the second costs $8 per pound. How much of each type does Mavis use to make 50 pounds of a blend which costs $6 per pound?
- Skippy has a total of
10,000 to split between two investments. One account offers$ simple interest, and the other account offers3 % simple interest. For tax reasons, he can only earn8 % in interest the entire year. How much money should Skippy invest in each account to earn$ 5 0 0 in interest for the year?$ 5 0 0 - A
salt solution is to be mixed with pure water to produce 75 gallons of a1 0 % salt solution. How much of each are needed?3 % - 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?
- Discuss with your classmates how you would approach Exercise 32 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
acid solution using stock solutions at6 0 % and2 0 % , respectively, what would the triangular form of the resulting system look like? Explain.4 0 %
Answers
-
Consistent independent
Solution( 6 , − 1 2 ) -
Consistent independent
Solution( − 7 3 , − 3 ) -
Consistent independent
Solution( − 1 6 7 , − 6 2 7 ) -
Consistent independent
Solution( 4 9 1 2 , − 2 5 1 8 ) -
Consistent dependent
Solution( 𝑡 , 3 2 𝑡 + 3 )
for all real numbers𝑡 -
Consistent dependent
Solution( 6 − 4 𝑡 , 𝑡 )
for all real numbers𝑡 -
Inconsistent
No solution -
Inconsistent
No solution
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.
{ 𝑥 + 𝑦 = 5 𝑦 = 7 Consistent independent
Solution
( − 2 , 7 ) ⎧ { { { ⎨ { { { ⎩ 𝑥 − 5 3 𝑦 − 7 3 𝑧 = − 7 3 𝑦 + 5 4 𝑧 = 2 𝑧 = 0 Consistent independent
Solution
( 1 , 2 , 0 ) ⎧ { { { ⎨ { { { ⎩ 𝑥 − 1 4 𝑦 + 1 4 𝑧 = 5 4 𝑦 + 3 𝑧 = 1 5 0 = 0 Consistent dependent
Solution
( − 𝑡 + 5 , − 3 𝑡 + 1 5 , 𝑡 ) for all real numbers
𝑡 ⎧ { { { ⎨ { { { ⎩ 𝑥 − 1 4 𝑦 + 1 4 𝑧 = 5 4 𝑦 + 3 𝑧 = 1 5 0 = 1 Inconsistent
No solution
{ 𝑥 + 𝑦 + 𝑧 = − 1 7 𝑦 − 3 𝑧 = 0 Consistent dependent
Solution
( − 4 𝑡 − 1 7 , 3 𝑡 , 𝑡 ) for all real numbers
𝑡 ⎧ { { ⎨ { { ⎩ 𝑥 − 2 𝑦 + 3 𝑧 = 7 𝑦 − 1 1 5 𝑧 = − 1 6 5 𝑧 = 1 Consistent independent
Solution
( 2 , − 1 , 1 ) ⎧ { { ⎨ { { ⎩ 𝑥 + 𝑦 + 2 𝑧 = 0 𝑦 − 3 2 𝑧 = 6 𝑧 = − 2 Consistent independent
Solution
( 1 , 3 , − 2 ) ⎧ { { { ⎨ { { { ⎩ 𝑥 − 1 2 𝑦 + 1 2 𝑧 = − 1 2 𝑦 + 3 5 𝑧 = 3 5 0 = 1 Inconsistent
no solution
⎧ { { ⎨ { { ⎩ 𝑥 − 𝑦 + 𝑧 = − 4 𝑦 − 7 𝑧 = 1 7 𝑧 = − 2 Consistent independent
Solution
( 1 , 3 , − 2 ) ⎧ { { ⎨ { { ⎩ 𝑥 − 2 𝑦 + 2 𝑧 = − 2 𝑦 = 1 2 𝑧 = 1 Consistent independent
Solution( − 3 , 1 2 , 1 ) ⎧ { { { ⎨ { { { ⎩ 𝑥 − 1 2 𝑦 + 1 2 𝑧 = 1 2 𝑦 − 2 3 𝑧 = 0 𝑧 = 1 Consistent independent
Solution
( 1 3 , 2 3 , 1 ) ⎧ { { ⎨ { { ⎩ 𝑥 − 3 𝑦 − 4 𝑧 = 3 𝑦 + 1 1 1 3 𝑧 = 4 1 3 0 = 0 Consistent dependent
Solution
( 1 9 1 3 𝑡 + 5 1 1 3 , − 1 1 1 3 𝑡 + 4 1 3 , 𝑡 ) for all real numbers
𝑡 ⎧ { { ⎨ { { ⎩ 𝑥 + 𝑦 + 𝑧 = 4 𝑦 + 1 2 𝑧 = 3 2 0 = 1 Inconsistent
no solution
⎧ { { ⎨ { { ⎩ 𝑥 − 𝑦 + 𝑧 = 8 𝑦 − 2 𝑧 = − 5 𝑧 = 1 Consistent independent
Solution
( 4 , − 3 , 1 ) ⎧ { { { ⎨ { { { ⎩ 𝑥 − 3 2 𝑦 + 1 2 𝑧 = − 1 2 𝑦 + 𝑧 = − 1 1 2 0 = 0 Consistent dependent
Solution
( − 2 𝑡 − 3 5 4 , − 𝑡 − 1 1 2 , 𝑡 ) for all real numbers \(t\
⎧ { { { { ⎨ { { { { ⎩ 𝑥 1 + 2 3 𝑥 2 − 1 6 3 𝑥 3 − 𝑥 4 = 2 5 3 𝑥 2 + 4 𝑥 3 − 3 𝑥 4 = 2 0 = 0 0 = 0 Consistent dependent
Solution
( 8 𝑠 − 𝑡 + 7 , − 4 𝑠 + 3 𝑡 + 2 , 𝑠 , 𝑡 ) for all real numbers
and𝑠 𝑡 ⎧ { { { { ⎨ { { { { ⎩ 𝑥 1 − 𝑥 3 = − 2 𝑥 2 − 1 2 𝑥 4 = 0 𝑥 3 − 1 2 𝑥 4 = 1 𝑥 4 = 4 Consistent independent
Solution
( 1 , 2 , 3 , 4 ) ⎧ { { { ⎨ { { { ⎩ 𝑥 1 − 𝑥 2 − 5 𝑥 3 + 3 𝑥 4 = − 1 𝑥 2 + 5 𝑥 3 − 3 𝑥 4 = 1 2 0 = 1 0 = 0 Inconsistent
No solution
- If
is the free variable then the solution is𝑥 and if( 𝑡 , 3 𝑡 , − 𝑡 + 5 ) is the free variable then the solution is𝑦 .( 1 3 𝑡 , 𝑡 , − 1 3 𝑡 + 5 ) chose the basic buffet and1 3 chose the deluxe buffet.1 4 - Mavis needs 20 pounds of $3 per pound coffee and 30 pounds of $8 per pound coffee.
- Skippy needs to invest
6000 in the$ account and3 % 4000 in the$ account.8 % gallons of the2 2 . 5 solution and1 0 % gallons of pure water.5 2 . 5 pounds of Type I,4 3 − 1 2 𝑡 pounds of Type II and2 3 − 1 2 𝑡 pounds of Type III where𝑡 .0 ≤ 𝑡 ≤ 4 3