Intermediate Algebra: Functions and Graphs

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 \(y\text{.}\)

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

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

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).

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).

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

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).

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).

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

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

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.

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.

• 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°.