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Section 2.4 Gaussian Reduction

Subsection 2.4.1 \(3\times 3\) Linear Systems

A solution to an equation in three variables, such as

\begin{equation*} x + 2y - 3z = -4 \end{equation*}

is an ordered triple of numbers that satisfies the equation. For example, \((0, -2, 0)\) and \((-1, 0, 1)\) are solutions to the equation above, but \((1, 1, 1)\) is not. You can verify this by substituting the coordinates into the equation to see if a true statement results.

\begin{alignat*}{5} \amp\text{For }(0,-2, 0):{}\amp \alert{0} {}+{}\amp 2(\alert{-2}) \amp{}-{}\amp 3(\alert{0}) \amp = -4{}\amp\hphantom{blank} \blert{\text{True}}\amp\\ \amp\text{For }(-1, 0, 1):{}\amp \alert{-1} {}+{}\amp 2(\alert{0}) \amp{}-{}\amp 3(\alert{1}) \amp = -4{}\amp\hphantom{blank} \blert{\text{True}}\amp\\ \amp\text{For }(1, 1, 1):{}\amp \alert{1} {}+{}\amp 2(\alert{1}) \amp{}-{}\amp 3(\alert{1}) \amp = -4{}\amp\hphantom{blank} \blert{\text{Not true}}\amp \end{alignat*}

A linear equation in three variables has infinitely many solutions.

An ordered triple \((x, y, z)\) can be represented geometrically as a point in space using a three-dimensional Cartesian coordinate system, as shown in the figure. The graph of a linear equation in three variables is a plane. A solution to a system of three linear equations in three variables is an ordered triple that satisfies each equation in the system. That triple represents a point where all three planes intersect.

point in 3D space
  • If the planes intersect in a single point, the system has a unique solution.

  • If there is no point that lies on all three planes (for instance, if at least two of the planes are parallel), the system is inconsistent.

  • If all three planes are the same, or if they intersect in a line, the system is dependent.

It is impractical to solve \(3\times 3\) systems by graphing. Even when technology for producing three-dimensional graphs is available, we cannot read coordinates on such graphs with any confidence. Thus, we will restrict our attention to algebraic methods of solving such systems.

  1. A solution to an equation in three variables is an

    • A) ordered pair that satisfies the equation.

    • B) ordered triple that satisfies the equation

    • C) arbitrary number that satisfies the equation

    .

  2. The graph of a linear equation in three variables is a

    • point

    • line

    • plane

    • parabola

    .

  3. A linear \(3 \times 3\) system has a unique solution if the three planes intersect

    • a single point

    • a line

    • a plane

    .

  4. Three planes may also intersect

    • in exacty 2 points

    • in exactly 3 points

    • in a line

    or
    • be the same plane

    • in a parabola

    • in a circle

    .

Answer 1.

\(\text{B) ... the equation}\)

Answer 2.

\(\text{plane}\)

Answer 3.

\(\text{a single point}\)

Answer 4.

\(\text{in a line}\)

Answer 5.

\(\text{be the same plane}\)

Solution.
  1. ordered triple that satisfies the equation

  2. plane

  3. a single point

  4. in a line; be the same plane

Subsection 2.4.2 Back-Substitution

The strategy for solving a \(3 \times 3\) system is the same as the strategy for \(2 \times 2\) systems: we would like to reduce the system to an equation in a single variable. Once we have found the value for that variable, we substitute its value into the other equations to find the remaining unknowns.

A special case of this technique is called back-substitution. It works when one of the equations involves exactly one variable, and a second equation involves that same variable and just one other variable. A \(3 \times 3\) system with these properties is said to be in .

Solve the system

\begin{alignat*}{3} x +{} 2y \amp {}+{} \amp 3z \amp {}={} \amp 2\\ -2y \amp {}-{} \amp 4z \amp {}={} \amp -2\\ \amp {}{} \amp 3z \amp {}={} \amp -3 \end{alignat*}
Solution.

We begin by solving the third equation to find \(z = -1\text{.}\) Then we substitute \(\alert{-1}\) for \(z\) in the second equation and solve for \(y\text{.}\)

\begin{align*} -2y - 4(\alert{-1}) \amp = -2\\ -2y + 4 \amp= -2\\ -2y \amp = -6\\ y\amp=3 \end{align*}

Finally, we substitute \(\alert{-1}\) for \(z\) and \(\alert{3}\) for \(y\) into the first equation to find \(x\text{.}\)

\begin{align*} x + 2(\alert{3}) + 3(\alert{-1}) \amp= 2\\ x + 6 - 3 \amp = 2\\ x\amp=-1 \end{align*}

The solution is the ordered triple \((-1, 3, -1)\text{.}\) You should verify that this triple satisfies all three equations of the system.

Use back-substitution to solve the system

\begin{equation*} \begin{alignedat}{3} 2x + 2y \amp {}+{} \amp z \amp {}={} \amp 10\\ y \amp {}-{} \amp 4z \amp {}={} \amp 9\\ \amp\amp 3z \amp {}={} \amp -6 \end{alignedat} \end{equation*}
Answer.

\(\left(5,1,-2\right)\)

Solution.

\({\left(5,1,-2\right)}\) satisfies all three equations.

Subsection 2.4.3 Gaussian Reduction

The method for solving a general \(3 \times 3\) linear system is called Gaussian reduction, after the German mathematician Carl Gauss. We use linear combinations to reduce the system to triangular form, and then use back-substitution to find the solutions.

  1. The method for solving a \(3 \times 3\) linear system is called

    • completing the square.

    • Gaussian reduction.

    • induction method.

    .

  2. A special case of this method is called

    • back-substitution.

    • negative reciprocals.

    • linear regression.

    • point-slope.

    .

  3. The special case works on systems

    • A) when one equation involves exactly one variable, and a second equation involves that same variable and just one other variable.

    • B) with more equations than unknowns.

    • C) having only two unknowns.

    .

  4. To reduce a system to the special form, we use

    • guess-and-check.

    • linear combinations.

    • the discriminant.

    .

Answer 1.

\(\text{Gaussian reduction.}\)

Answer 2.

\(\text{back-substitution.}\)

Answer 3.

\(\text{A) when ... other variable.}\)

Answer 4.

\(\text{ linear combinations.}\)

Solution.
  1. Gaussian reduction

  2. back-substitution

  3. when one equation involves exactly one variable, and a second equation involves that same variable and just one other variable

  4. linear combinations

To obtain the triangular form, we eliminate one of the variables from each of thethree equations by considering them in pairs. This results in a \(2\times 2\) system that we can solve using elimination.

Solve the system:

\begin{alignat*}{4} x \amp {}+{} \amp 2y \amp {}-{} \amp 3z \amp {}={} \amp -4\amp\hphantom{blankblank} (1)\\ 2x \amp {}-{} \amp y \amp {}+{} \amp z \amp {}={} \amp 3\amp\hphantom{blankblank} (2)\\ 3x \amp {}+{} \amp 2y \amp {}+{} \amp z \amp {}={} \amp 10\amp\hphantom{blankblank} (3) \end{alignat*}
Solution.

We can choose any one of the three variables to eliminate first. For this example, we will eliminate \(x\text{.}\) Next, we choose two of the equations, say (1) and (2), and use a linear combination: We multiply Equation (1) by \(-2\) and add the result to Equation (2) to produce Equation (4).

\begin{alignat*}{4} -2x \amp {}-{} \amp 4y \amp {}+{} \amp 6z \amp {}={} \amp 8\amp\hphantom{blankblank} (1\text{a})\\ \underline{\hphantom{-}2x\vphantom{y}} \amp \underline{{}-{}\vphantom{y}} \amp \underline{\hphantom{-5}y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{\hphantom{6}z\vphantom{y}} \amp {}={} \amp \underline{\hphantom{1}\vphantom{y} 3}\amp\hphantom{blankblank} (2)\\ \amp \amp{-5y} \amp {}+{} \amp 7z \amp {}={} \amp 11\amp\hphantom{blankblank} (4) \end{alignat*}

Now we have an equation involving only two variables. But we need two equations in two unknowns to find the solution. So we choose a different pair of equations, say (1) and (3), and eliminate \(x\) again. We multiply Equation (1) by \(-3\) and add the result to Equation (3) to obtain Equation (5).

\begin{alignat*}{4} -3x \amp {}-{} \amp 6y \amp {}+{} \amp 9z \amp {}={} \amp 12\amp\hphantom{blankblank} (1\text{b})\\ \underline{\hphantom{-}3x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{\hphantom{-}2y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{\hphantom{10}z\vphantom{y}} \amp {}={} \amp \underline{\vphantom{y}10}\amp\hphantom{blankblank} (3)\\ \amp \amp{-4y} \amp {}+{} \amp 10z \amp {}={} \amp 22\amp\hphantom{blankblank} (5) \end{alignat*}

We now form a \(2\times 2\) system with our new Equations (4) and (5).

\begin{alignat*}{2} -5y\amp + \amp 7z \amp = 11 \hphantom{blankblank} (4)\\ -4y\amp + \amp 10z \amp = 22 \hphantom{blankblank} (5) \end{alignat*}

Finally, we eliminate one of the remaining variables to obtain an equation in a single variable. We choose to eliminate \(y\text{,}\) so we add 4 times Equation (4) to \(-5\) times Equation (5) to obtain Equation (6).

\begin{alignat*}{2} -20y~\amp {}+{} \amp 28z \amp =\hphantom{-1} 44 \hphantom{blankblank} (4\text{a})\\ \underline{\hphantom{-}20y~}\amp\underline{{}-{}\vphantom{{}+ y~}} \amp \underline{\hphantom{-}50z\vphantom{ -20y~}} \amp = \underline{-110\vphantom{y}} \hphantom{blankblank} (5\text{a})\\ \amp \amp -22z \amp = \hphantom{1}{-66} \hphantom{blankblankl} (6) \end{alignat*}

Now we are ready to form a triangular system. We choose one of the original equations (in three variables), one of the equations from our \(2\times 2\) system, and our final equation in one variable. We choose Equations (1), (4), and (6).

\begin{alignat*}{4} x \amp {}+{} \amp 2y \amp {}-{} \amp 3z \amp {}={} \amp -4\amp\hphantom{blankblank} (1)\\ \amp {}{} \amp -5 y \amp {}+{} \amp 7z \amp {}={} \amp 11\amp\hphantom{blankblank} (4)\\ \amp \amp \amp {}{} \amp -22z \amp {}={} \amp -66\amp\hphantom{blankblank} (6) \end{alignat*}

This new system has the same solutions as the original system, and we can solveit by back-substitution. We first solve Equation (6) to find \(z = 3\text{.}\) Substituting 3 for \(z\) in Equation (4), we find

\begin{align*} -5y+7(\alert{3}) \amp=11 \\ -5y+21 \amp=11 \\ -5y \amp=-10 \end{align*}

So \(y = 2\text{.}\) Finally, we substitute \(\alert{3}\) for \(z\) and \(\alert{2}\) for \(y\) into Equation (1) to find

\begin{align*} x+2(\alert{2})-3(\alert{3}) \amp= -4 \\ x+4-9 \amp= -4 \\ x \amp= 1 \end{align*}

The solution to the system is the ordered triple \((1, 2, 3)\text{.}\) You should verify that this triple satisfies all three of the original equations.

We summarize the method for solving a \(3\times 3\) linear system as follows.

Steps for Solving a \(3\times 3\) Linear System.

  1. Clear each equation of fractions and put it in standard form.

  2. Choose two of the equations and eliminate one of the variables by forming a linear combination.

  3. Choose a different pair of equations and eliminate the same variable.

  4. Form a \(2\times 2\) system with the equations found in steps (2) and (3). Eliminate one of the variables from this \(2\times 2\) system by using a linear combination.

  5. Form a triangular system by choosing among the previous equations. Use back-substitution to solve the triangular system.

Suppose you eliminate \(y\) from two equations as Step 2 of Gaussian reduction. What should you do for Step 3?

  • Form a triangular system.

  • Eliminate either \(x\) or \(z\text{.}\)

  • Eliminate \(y\) again from a different pair of equations.

  • Substitute the value for \(y\) into the equations.

Answer.

\(\text{Choice 3}\)

Solution.

Eliminate \(y\) again from a different pair of equations.

Use Gaussian reduction to solve the system

\begin{equation*} \begin{aligned} x-2y+z \amp= -1\qquad\qquad \amp\amp (1)\\ \frac{2}{3}x +\frac{1}{3}y - z \amp= -1 \amp\amp (2)\\ 3+3y-2z \amp= 10 \amp\amp (3) \end{aligned} \end{equation*}

Follow the steps:

  • Step 1: Clear the fractions from Equation (2).

  • Step 2: Eliminate \(z\) from Equations (1) and (2).

  • Step 3: Eliminate \(z\) from Equations (1) and (3).

  • Step 4: Eliminate \(x\) from your new \(2\times 2\) system.

  • Step 5: Form a triangular system and solve by back-substitution.

Answer.

\(\left(2,2,1\right)\)

Solution.

\({\left(2,2,1\right)}\)

Subsection 2.4.4 Inconsistent and Dependent Systems

If, at any step in forming linear combinations, we obtain an equation of the form

\begin{gather*} 0x + 0y + 0z = k, \qquad (k \ne 0) \end{gather*}

then the system is inconsistent and has no solution. If we don’t obtain such an equation, but we do obtain one of the form

\begin{gather*} 0x + 0y + 0z = 0 \end{gather*}

then the system is dependent and has infinitely many solutions.

Solve the systems.

  1. \(\displaystyle \begin{alignedat}[t]{5} 3x\amp{}+{}\amp y\amp{}-{}\amp 2z\amp{}={}1 \amp\qquad\amp (1)\\ 6x\amp{}+{}\amp 2y\amp{}-{}\amp 4z\amp{}={}5 \amp\qquad\amp (2)\\ 2x\amp{}-{}\amp y\amp {}+{}\amp 3z\amp{}={}-1 \amp\qquad\amp (3) \end{alignedat} \)

  2. \(\displaystyle \begin{alignedat}[t]{5} -x\amp{}+{}\amp 3y\amp{}-{}\amp z\amp{}={}-2 \amp\qquad\amp (1)\\ 2x\amp{}+{}\amp y\amp{}-{}\amp 4z\amp{}={}-1 \amp\qquad\amp (2)\\ 2x\amp{}-{}\amp 6y\amp {}+{}\amp 2z\amp{}={}4 \amp\qquad\amp (3) \end{alignedat} \)

Solution.
  1. To eliminate \(y\) from Equations (1) and (2), we multiply Equation (1) by \(-2\) and add the result to Equation (2).

    \begin{alignat*}{4} -6x \amp {}-{} \amp 2y \amp {}+{} \amp 4z \amp {}={} \amp -2\\ \underline{\hphantom{-}6x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{2y} \amp \underline{{}-{}\vphantom{y} }\amp \underline{4z\vphantom{y}} \amp {}={} \amp \underline{\hphantom{-}5}\\ 0x\amp {}+{}\amp{0y} \amp {}+{} \amp 0z \amp {}={} \amp 3~ \end{alignat*}

    Because the resulting equation has no solution, the system is inconsistent.

  2. To eliminate \(x\) from Equations (1) and (3), we multiply Equation (1) by \(2\) and add Equation (3).

    \begin{alignat*}{4} -2x \amp {}+{} \amp 6y \amp {}-{} \amp 2z \amp {}={} \amp -4\\ \underline{\hphantom{-}2x\vphantom{y}} \amp \underline{{}-{}\vphantom{y}} \amp \underline{6y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{2z\vphantom{y}} \amp {}={} \amp \underline{\hphantom{-}4}\\ 0x\amp {}+{}\amp{0y} \amp {}+{} \amp 0z \amp {}={} \amp 0~ \end{alignat*}

    Because the resulting equation vanishes, the system is dependent and has infinitely many solutions.

Decide whether the system is inconsistent, dependent, or consistent andindependent.

\begin{equation*} \begin{aligned} x+3y-z\amp=4\\ -2x-6y+2z\amp=4\\ x+2y-z\amp=3 \end{aligned} \end{equation*}

  • Inconsistent

  • Dependent

  • Consistent and independent

Answer.

\(\text{Inconsistent}\)

Solution.

Inconsistent

True or false.

  1. If a system is dependent, it has no solutions.

    • True

    • False

  2. The equation \(0x + 0y + 0z = 0\) has no solution

    • True

    • False

  3. If a \(3 \times 3\) system is dependent, all three equations are the same..

    • True

    • False

  4. Gaussian reduction will reveal whether a system is dependent or inconsistent.

    • True

    • False

Answer 1.

\(\text{False}\)

Answer 2.

\(\text{False}\)

Answer 3.

\(\text{False}\)

Answer 4.

\(\text{True}\)

Solution.
  1. False

  2. False

  3. False

  4. True

Subsection 2.4.5 Applications

Here are some problems that can be modeled by a system of three linear equations.

One angle of a triangle measures \(4\degree\) less than twice the second angle, and the third angle is \(20\degree\) greater than the sum of the first two. Find the measure of each angle.

Solution.
  • Step 1:.

    We represent the measure of each angle by a separate variable.

    \begin{align*} \amp\text{First angle: }\amp\amp x\\ \amp\text{Second angle: }\amp\amp y\\ \amp\text{Third angle:}\amp\amp z \end{align*}
  • Step 2:.

    We write the conditions stated in the problem as three equations.

    \begin{align*} \amp x\text{ is } 4\degree \text{ less than }2 \text{ times } y: \amp\amp x = 2y - 4\\ \amp z\text{ is } 20\degree \text{ more than } x+ y: \amp\amp z = x + y + 20\\ \amp \text{the sum of the angles of a triangle is } 180\degree : \amp\amp x + y + z =180 \end{align*}
  • Step 3:.

    We follow the steps for solving a \(3\times 3\) linear system.

    1. We write the three equations in standard form.

    \begin{alignat*}{5} x\amp {}-{}\amp 2y\amp \amp \amp =\amp -4\amp\hphantom{blank} \amp(1)\\ x\amp {}+{}\amp y\amp {}-{}\amp z\amp =\amp -20\amp\hphantom{blank} \amp(2)\\ x\amp {}+{}\amp y\amp {}+{}\amp z\amp =\amp 180\amp\hphantom{blank} \amp(3) \end{alignat*}

    2–3. Because Equation (1) has no \(z\)-term, it will be most efficient to eliminate \(z\) from Equations (2) and (3). We add these two equations.

    \begin{alignat*}{5} x \amp {}+{} \amp y \amp {}-{} \amp z \amp {}={} \amp -20\qquad\amp\amp (2)\\ \underline{x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{\hphantom{2}y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{\hphantom{~}\vphantom{y}z} \amp {}={} \amp \underline{180\vphantom{y+y}}\qquad\amp\amp (3)\\ 2x\amp {}+{}\amp{2y} \amp \amp \amp {}={} \amp 160\qquad\amp\amp (4) \end{alignat*}

    4. We form a \(2\times 2\) system from Equations (1) and (4). We add the two equations to eliminate the variable \(y\text{,}\) yielding

    \begin{alignat*}{4} x \amp {}-{} \amp 2y \amp {}={} \amp -4\qquad\amp\amp (1)\\ \underline{2x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{2y} \amp {}={} \amp \underline{160\vphantom{y+y}}\qquad\amp\amp (4)\\ 3x\amp\amp \amp {}={} \amp 156\qquad\amp\amp (5) \end{alignat*}

    5. We form a triangular system using Equations (2), (1), and (5). We use back-substitution to complete the solution.

    \begin{alignat*}{4} x\amp {}+{}\amp y\amp {}+{}\amp z\amp =\amp 180\hphantom{blankblank} \amp(2)\\ x\amp {}-{}\amp 2y\amp \amp\amp =\amp -4\hphantom{blankblank} \amp(1)\\ 3x\amp\amp\amp\amp\amp = \amp156\hphantom{blankblank} \amp(5) \end{alignat*}

    We divide both sides of Equation (5) by \(3\) to find \(x = 52\text{.}\) We substitute \(52\) for \(x\) in Equation (1) and solve for \(y\) to find

    \begin{align*} \alert{52} - 2y \amp = -4\\ y \amp= 28 \end{align*}

    We substitute \(\alert{52}\) for \(x\) and \(\alert{28}\) for \(y\) in Equation (3) to find

    \begin{align*} \alert{52} + \alert{28} + z \amp= 180\\ z\amp=100 \end{align*}
  • Step 4:.

    The angles measure 52°, 28°, and 100°.

Exercises 2.4.6 Problem Set 2.4

Warm Up

Exercise Group.

For Problems 1 and 2, solve the system by elimination.

1.
\begin{align*} 4x-3\amp=3y\\ 25+5x\amp =-2y \end{align*}
Answer.

\((-3,-5) \)

2.
\begin{align*} \dfrac{2x}{3} + \dfrac{8y}{9} \amp=\dfrac{4}{3}\\ \dfrac{x}{2} + 2 = \amp\dfrac{y}{3} \end{align*}
3.

Karen has $2000, part of it invested in bonds paying 10%, and the rest in a certificate account at 8%. Her annual income from the two investments is $184. How much did Karen invest at each rate?

  1. Choose variables for the unknown quantities, and fill in the table.

    Principal Interest rate Interest
    Bonds
    Certificate
    Total ——
  2. Write one equation about the amount Karen invested.

  3. Write a second equation about Kaaren's annual interest.

Answer.
  1. Principal Interest rate Interest
    Bonds \(x\) \(0.10\) \(0.10x\)
    Certificate \(y\) \(0.08\) \(0.08y\)
    Total \(2000\) —— \(184\)
  2. \(\displaystyle x+y=2000\)

  3. \(\displaystyle 0.10x+0.08y=184\)

4.

The pharmacist at Glenoaks Hospital was asked to supply 10 liters of a 20% solution of iodine. She has on hand iodine solutions in 15% strength and 40% strength. How much of each should she mix to make the required solution?

  1. Choose variables for the unknown quantities, and fill in the table.

    Liters Strength Amount of iodine
    15% Solution
    40% solution
    20% solution
  2. Write one equation about the number of liters of solution.

  3. Write a second equation about the amount of iodine.

Skills Practice

Exercise Group.

For Problems 5 and 6, use back-substitution to solve the system.

5.

\(\begin{alignedat}[t]{5} 2x\amp{}+{}\amp 3y\amp{}-{}\amp z\amp{}={}-7\\ \amp \amp y\amp{}-{}\amp 2z\amp{}={}-6\\ \amp \amp \amp \amp 5z\amp{}={}15 \end{alignedat}\)

Answer.

\((-2,0,3) \)

6.

\(\begin{aligned}[t] 2x + z\amp = 5\\ 3y + 2z \amp =6\\ 5x \amp = 20 \end{aligned}\)

Exercise Group.

For Problems 7–12, use Gaussian reduction to solve the system.

7.

\(\begin{alignedat}[t]{5} x\amp{}+{}\amp y\amp{}+{}\amp z\amp{}={}0\\ 2x \amp{}-{} \amp 2y\amp{}+{}\amp z\amp{}={}8\\ 3x \amp{}+{} \amp 2y\amp {}+{}\amp z\amp{}={}2 \end{alignedat}\)

Answer.

\((2,-2,0) \)

8.

\(\begin{alignedat}[t]{5} x\amp{}-{}\amp 2y\amp{}+{}\amp 4z\amp{}={}-3\\ 3x \amp{}+{} \amp y\amp{}-{}\amp 2z\amp{}={}12\\ 2x \amp{}+{} \amp y\amp {}-{}\amp z\amp{}={}11 \end{alignedat}\)

9.

\(\begin{alignedat}[t]{5} 3x\amp{}-{}\amp 4y\amp{}+{}\amp 2z\amp{}={}20\\ 4x \amp{}+{} \amp 3y\amp{}-{}\amp 3z\amp{}={}-4\\ 2x \amp{}-{} \amp 5y\amp {}+{}\amp 5z\amp{}={}24 \end{alignedat}\)

Answer.

\((2,-3,1) \)

10.

\(\begin{aligned}[t] 4x + z\amp = 3\\ 2x - y \amp =2\\ 3y+2z \amp = 0 \end{aligned}\)

11.

\(\begin{alignedat}[t]{5} 4x\amp{}+{}\amp 6y\amp{}+{}\amp 3z\amp{}={-3} \\ 2x \amp{}-{} \amp 3y\amp{}-{}\amp 2z\amp{}=5 \\ -6x \amp{}+{} \amp 6y\amp {}+{}\amp 2z\amp{}={-5} \end{alignedat} \)

Answer.

\(\left(\dfrac{1}{2}, \dfrac{2}{3},-3 \right) \)

12.

\(\begin{alignedat}[t]{5} x\amp{}-{}\amp \dfrac{1}{2} y\amp{}-{}\amp \dfrac{1}{2}z\amp{}={}4 \\ x \amp{}-{} \amp \dfrac{3}{2}y\amp{}-{}\amp 2z\amp{}={}3 \\ \dfrac{1}{4}x \amp{}+{} \amp \dfrac{1}{4}y\amp {}-{}\amp \dfrac{1}{4}z\amp{}={}0 \end{alignedat} \)

Exercise Group.

For Problems 13 and 14, when you decide which variable to eliminate first, take advantage of the fact that one of the variables is already missing from each equation.

13.

\(\begin{alignedat}[t]{5} x\amp {}={} {-y} \\ x \amp{}+{} z=\dfrac{5}{6} \\ y \amp{}-{} 2z = -\dfrac{7}{6} \end{alignedat} \)

Answer.

\(\left(\dfrac{1}{2}, -\dfrac{1}{2},\dfrac{1}{3} \right) \)

14.

\(\begin{aligned}[t] x\amp = y+\dfrac{1}{2} \\ y \amp =z+\dfrac{5}{4} \\ 2z \amp = x-\dfrac{7}{4} \end{aligned} \)

Exercise Group.

For Problems 15 and 16, decide whether the system is inconsistent or dependent.

15.

\(\begin{alignedat}[t]{5} 3x\amp{}-{}\amp 2y\amp{}+{}\amp z\amp{}={6} \\ 2x \amp{}+{} \amp y\amp{}-{}\amp z\amp{}=2 \\ 4x \amp{}+{} \amp 2y\amp {}-{}\amp 2z\amp{}={3} \end{alignedat} \)

Answer.

Dependent

16.

\(\begin{aligned}[t] x\amp = 2y -7 \\ y \amp =4z+3 \\ x \amp - 8z = -1 \end{aligned} \)

Applications

Exercise Group.

Use a system of equations to solve Problems 17-22.

17.

The perimeter of a triangle is 155 inches. Side \(x\) is 20 inches shorter than side \(y\text{,}\) and side \(y\) is 5 inches longer than side \(z\text{.}\) Find the lengths of the sides of the triangle.

Answer.

\(x=40\) in, \(y=60\) in, \(z=55\) in

18.

One angle of a triangle measures 10° more than a second angle, and the third angle is 10° more than six times the measure of the smallest angle. Find the measure of each angle.

19.

The Java Shoppe sells a house brand of coffee that is only 2.25% caffeine for $6.60 per pound. The house brand is a mixture of Colombian coffee that sells for $6 per pound and is 2% caffeine, French roast that sells for $7.60 per pound and is 4% caffeine, and Sumatran that sells for $6.80 per pound and is 1% caffeine. How much of each variety is in a pound of house brand?

Answer.

\(\dfrac{1}{2} \) lb Colombian, \(\dfrac{1}{4} \) lb French, \(\dfrac{1}{4} \) Sumatran

20.

Vegetable Medley is made of carrots, green beans, and cauliflower. The package says that 1 cup of Vegetable Medley provides 29.4 milligrams of vitamin C and 47.4 milligrams of calcium. One cup of carrots contains 9 milligrams of vitamin C and 48 milligrams of calcium. One cup of green beans contains 15 milligrams of vitamin C and 63 milligrams of calcium. One cup of cauliflower contains 69 milligrams of vitamin C and 26 milligrams of calcium. How much of each vegetable is in 1 cup of Vegetable Medley?

21.

Reliable Auto Company wants to ship 1700 Status Sedans to three major dealers in Los Angeles, Chicago, and Miami. From past experience Reliable figures that it will sell twice as many sedans in Los Angeles as in Chicago. It costs $230 to ship a sedan to Los Angeles, $70 to Chicago, and $160 to Miami. If Reliable Auto has $292,000 to pay for shipping costs, how many sedans should it ship to each city?

22.

A farmer has 1300 acres on which to plant wheat, corn, and soybeans. The seed costs $6 for an acre of wheat, $4 for an acre of corn, and $5 for an acre of soybeans. An acre of wheat requires 5 acre-feet of water during the growing season, an acre of corn requires 2 acre-feet, and an acre of soybeans requires 3 acre-feet. If the farmer has $6150 to spend on seed and can count on 3800 acre-feet of water, how many acres of each crop should she plant in order to use all her resources?

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