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Chapter 7: Problem 7

Checking Solutions In Exercises \(7 - 10\) , determine whether each ordered triple is a solution of the system of equations. $$\left\\{ \begin{aligned} 6 x - y + z & = - 1 \\ 4 x - 3 z & = - 19 \\ 2 y + 5 z & = 25 \end{aligned} \right.$$ $$\begin{array} { l l } { \text { (a) } ( 2,0 , - 2 ) } & { \text { (b) } ( - 3,0,5 ) } \\ { \text { (c) } ( 0 , - 1,4 ) } & { \text { (d) } ( - 1,0,5 ) } \end{array}$$

Short Answer

Expert verified
Of all the given ordered triples (-1,0,5) is a valid solution of the given system. The rest are not.

Step by step solution

01

Calculating for (2,0, -2)

Substitute the first ordered triple (2, 0, -2) for \(x\), \(y\), and \(z\) respectively in the original equations and check if they hold true. The equations become: \[\begin{align*}6(2) - 0 - 2 & = -1 \4(2) - 3(-2) & = -19 \2(0) + 5(-2) & = 25\end{align*}\]After simplifying, you find: \[\begin{align*}12 - 2 & = -1 \8 + 6 & = -19 \0 - 10 & = 25\end{align*}\]Clearly, none of these equations hold true. So, (2,0,-2) is not a solution.
02

Calculating for (-3,0,5)

Do the same substitution as in step 1, but with the numbers (-3,0,5). This gives: \[\begin{align*}6(-3) - 0 + 5 & = -1 \4(-3) - 3(5) & = -19 \2(0) + 5(5) & = 25\end{align*}\]After simplifying, these equations are:\[\begin{align*}-18 + 5 & = -1 \-12 - 15 & = -19 \0 + 25 & = 25\end{align*}\]The simplified equations are false, so (-3, 0, 5) is also not a solution.
03

Calculating for (0,-1,4)

By doing the same substitution as in the previous steps for (0,-1,4) we get: \[\begin{align*}6(0) - (-1) + 4 & = -1 \4(0) - 3(4) & = -19 \2(-1) + 5(4) & = 25\end{align*}\]Similar simplifying leads to: \[\begin{align*}0 + 1 + 4 & = -1 \0 - 12 & = -19 \-2 + 20 & = 25\end{align*}\]Again, all simplified equations are false, which means (0,-1,4) is also not a solution.
04

Calculating for (-1,0,5)

The final calculation involves substituting (-1,0,5) into the three original equations: \[\begin{align*}6(-1) - 0 + 5 & = -1 \4(-1) - 3(5) & = -19 \2(0) + 5(5) & = 25\end{align*}\]After simplification, the equations look like this: \[\begin{align*}-6 + 5 & = -1 \-4 - 15 & = -19 \0 + 25 & = 25\end{align*}\]This time, all three equations hold true, which means (-1,0,5) is a valid solution of the given system.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ordered Triple
In the context of systems of equations, an "ordered triple" is a set of three numbers that represent a potential solution to a system of three equations. These numbers correspond to the variables in the system, usually denoted as \(x\), \(y\), and \(z\). An ordered triple is written in the form \((x, y, z)\). Each value in the ordered triple is plugged into the respective variable in each equation to check if they satisfy all the equations simultaneously.

For example, when working with the ordered triple \((2, 0, -2)\):
  • \(x = 2\)
  • \(y = 0\)
  • \(z = -2\)
The goal is to determine whether substituting these values into all the equations makes them true. If they do, the ordered triple is a valid solution. Each triple must satisfy every equation in the system to be considered a solution.
Solution Verification
"Solution verification" is the process of substituting an ordered triple into a system of equations to check if it satisfies each equation. This involves plugging in the numbers from the ordered triple into the system and simplifying to see if each equation holds true. This method helps confirm whether the ordered triple is indeed a solution to the given system.

Consider the system of equations provided:
1. \(6x - y + z = -1\)
2. \(4x - 3z = -19\)
3. \(2y + 5z = 25\)

To verify a solution, such as \((-1, 0, 5)\):
  • Substitute \(x = -1\), \(y = 0\), \(z = 5\) into each equation.
  • Simplify each equation to see if they are true.
If all equations are true after substitution, then \((-1, 0, 5)\) is a solution. This process helps spot errors and confirms the correctness of a solution.
Substitution Method
The "substitution method" involves solving a system of equations by expressing one variable in terms of the others and substituting this expression into the remaining equations. This method simplifies the system and reduces the number of variables, making it easier to solve.

Here's how to approach a system using substitution:
  • Pick one equation and solve it for one variable. This creates a simple expression.
  • Substitute this expression into the other equations to eliminate that variable.
  • Continue solving with fewer variables.
For instance, if you have a simpler two-variable equation after substitution, solve for one of these variables and backtrack by substituting again if needed. The substitution method is especially useful when dealing with more complex variables or when equations are easily manipulated to isolate variables.

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Most popular questions from this chapter

Break-Even Analysis In Exercises 57 and 58 , find the sales necessary to break even \((R=C)\) for the total cost \(C\) of producing \(x\) units and the revenue \(R\) obtained by selling \(x\) units. (Round to the nearest whole unit.) $$C=8650 x+250,000, \quad R=9950 x$$

Acid Mixture Thirty liters of a 40\(\%\) acid solution is obtained by mixing a 25\(\%\) solution with a 50\(\%\) solution. (a) Write a system of equations in which one equation represents the amount of final mixture required and the other represents the percent of acid in the final mixture. Let \(x\) and \(y\) represent the amounts of the 25\(\%\) and 50\(\%\) solutions, respectively. (b) Use a graphing utility to graph the two equations in part (a) in the same viewing window. As the amount of the 25\(\%\) solution increases, how does the amount of the 50\(\%\) solution change? (c) How much of each solution is required to obtain the specified concentration of the final mixture?

Fitting a Line to Data To find the least squares regression line \(y=a x+b\) for a set of points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) you can solve the following system for \(a\) and \(b\) . $$ \left\\{\begin{array}{c}{n b+\left(\sum_{i=1}^{n} x_{i}\right) a=\left(\sum_{i=1}^{n} y_{i}\right)} \\ {\left(\sum_{i=1}^{n} x_{i}\right) b+\left(\sum_{i=1}^{n} x_{i}^{2}\right) a=\left(\sum_{i=1}^{n} x_{i} y_{i}\right)}\end{array}\right. $$ In Exercises 55 and \(56,\) the sums have been evaluated. Solve the given system for \(a\) and \(b\) to find the least squares regression line for the points. Use a graphing utility to confirm the result. $$ \left\\{\begin{aligned} 5 b+10 a &=20.2 \\ 10 b+30 a &=50.1 \end{aligned}\right. $$

True or False? In Exercises 59 and 60 , determine whether the statement is true or false. Justify your answer. If two lines do not have exactly one point of intersection. then they must be parallel.

Finding Minimum and Maximum Values, find the minimum and maximum values of the objective function and where they occur, subject to the constraints \(x \geq 0, y \geq 0,3 x+y \leq 15\) and \(4 x+3 y \leq 30 .\) $$ z=3 x+y $$
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