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Chapter 10: Problem 21

Find all solutions of each system. $$\left\\{\begin{aligned} 2 x-\quad y+ \quad z &=4 \\ x+3 y+2 z &=-1 \\ 7 x \quad\quad +5 z=11 \end{aligned}\right.$$

Short Answer

Expert verified
The solution is \( x = 3 \), \( y = 0 \), and \( z = -2 \).

Step by step solution

01

Write the System of Equations

We are given three equations: 1. \( 2x - y + z = 4 \)2. \( x + 3y + 2z = -1 \)3. \( 7x + 5z = 11 \). Let's work on solving this system using an elimination or substitution method.
02

Eliminate y from Equations 1 and 2

To eliminate \( y \), multiply Equation 1 by 3 to align the coefficient of \( y \) with Equation 2: \[ 6x - 3y + 3z = 12 \]Now subtract Equation 2:\[ (6x - 3y + 3z) - (x + 3y + 2z) = 12 - (-1) \]This simplifies to:\[ 5x + z = 13 \].Let's refer to this as Equation 4.
03

Eliminate z using Equations 3 and 4

From Equation 3: \( 7x + 5z = 11 \) and Equation 4: \( 5x + z = 13 \), use substitution.Solve Equation 4 for \( z \):\[ z = 13 - 5x \].Substitute into Equation 3:\[ 7x + 5(13 - 5x) = 11 \]\[ 7x + 65 - 25x = 11 \]\[ -18x = -54 \].This simplifies to:\[ x = 3 \].
04

Substitute x back to find z

Substitute \( x = 3 \) back into Equation 4 to find \( z \):\[ 5(3) + z = 13 \]\[ 15 + z = 13 \]\[ z = -2 \].
05

Substitute x and z back to find y

With \( x = 3 \) and \( z = -2 \), substitute back into Equation 1 to find \( y \):\[ 2(3) - y - 2 = 4 \]\[ 6 - y - 2 = 4 \]\[ -y = 0 \]\[ y = 0 \].
06

Final Solution Verification

Verify by substituting \( x = 3 \), \( y = 0 \), \( z = -2 \) in all original equations and confirm all are satisfied. For Equation 1: \( 2(3) - 0 - 2 = 4 \) 鉁旓笍For Equation 2: \( 3 + 0 - 4 = -1 \) 鉁旓笍For Equation 3: \( 21 - 10 = 11 \) 鉁旓笍All equations are correct. Thus, the solutions satisfy the system.

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Key Concepts

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

Substitution Method
The substitution method is a technique to solve systems of equations by replacing one variable with an expression obtained from one of the equations. It's a step-by-step process that allows us to express one variable in terms of the others, simplifying the overall problem. This method is particularly useful for systems where one equation can be easily solved for one of the variables.
  • Begin by solving one of the equations for one variable.
  • Substitute this expression into the other equations, effectively reducing the number of variables.
  • Solve the resulting equation to find the value of one variable.
  • Substitute this value back into one of the original equations to solve for the remaining variable(s).
In the given problem, after deriving Equation 4 \(5x + z = 13\), we used substitution by expressing \(z\) in terms of \(x\) as \(z = 13 - 5x\). This was then substituted into Equation 3, simplifying the search for \(x\). By handling one variable at a time, substitution provides a straightforward route to reach a solution.
Elimination Method
The elimination method involves removing one variable by aligning and combining equations. It is effective for solving systems of linear equations, especially when the coefficients of one variable are already lined up or can be manipulated easily.
  • Align the equations so the coefficients of one variable are equal or can be made equal.
  • Add or subtract the equations to eliminate that variable.
  • Continue the process with the simplified system until only one variable remains.
  • Solve for the remaining variables by back-substitution.
In our exercise, the elimination method was used to remove \(y\) between Equations 1 and 2 by multiplying Equation 1 by 3 and subtracting Equation 2 from the result. This strategic alignment led to a new equation, Equation 4 \(5x + z = 13\), thus simplifying the problem. The key to using elimination is strategically choosing which variable to eliminate and performing precise arithmetic without disturbing the balance of the equations.
Solution Verification
Solution verification is a crucial step in solving systems of equations, ensuring that the proposed values of variables satisfy all original equations. Without verification, errors may go unnoticed, leading to incorrect solutions.
  • Insert the found values of variables into each of the original equations.
  • Confirm that each equation holds true with these values.
  • If all equations are valid, then the solution is correct.
In this problem, we verified the solution \(x = 3\), \(y = 0\), and \(z = -2\) by substituting each back into the original equations. Each check confirmed that these values satisfy:- Equation 1: \(2(3) - 0 - 2 = 4\)- Equation 2: \(3 + 0 - 4 = -1\)- Equation 3: \(21 - 10 = 11\)This final step is essential in confirming that the calculations and logical deductions throughout the process were accurate.

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Most popular questions from this chapter

Solve the following system for \(x, y,\) and \(z\) in terms of \(p, q\) and \(r\) $$\left\\{\begin{array}{l}x(x+y+z)=p^{2} \\\y(x+y+z)=q^{2} \\\z(x+y+z)=r^{2}\end{array}\right.$$

The substitution method in Exercise 51 leads to a quadratic equation. Here is an alternative approach to solving that system; this approach leads to linear equations. Multiply equation (2) by 2 and add the resulting equation to equation (1). Now take square roots to conclude that \(x+y=\pm 3 .\) Next, multiply equation (2) by \(2,\) and subtract the resulting equation from equation (1). Take square roots to conclude that \(x-y=\pm 1 .\) You now have the following four linear systems, each of which can be solved (with almost no work) by the addition-subtraction method. $$\begin{array}{ll}\left\\{\begin{array}{l}x+y=3 \\\x-y=1\end{array}\right. & \left\\{\begin{array}{l}x+y=3 \\\x-y=-1\end{array}\right. \\\\\left\\{\begin{array}{l}x+y=-3 \\\x-y=1\end{array}\right.\end{array}\left\\{\begin{array}{l}x+y=-3 \\\x-y=-1\end{array}\right.$$ Solve these systems and compare your results with those obtained in Exercise 51

Let \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) and \(B=\left(\begin{array}{ll}e & f \\ g & h\end{array}\right) .\) Is it true that \(\operatorname{det}(A B)=(\operatorname{det}(A))(\operatorname{det}(B)) ?\)

The matrices \(A, B, C, D, E, F,\) and \(G\) are defined as follows: $$ A=\left(\begin{array}{rr} 2 & 3 \\ -1 & 4 \end{array}\right) \quad B=\left(\begin{array}{rr} 1 & -1 \\ 3 & 0 \end{array}\right) \quad C=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) $$ $$\begin{aligned} &D=\left(\begin{array}{rrr} -1 & 2 & 3 \\ 4 & 0 & 5 \end{array}\right) \quad E=\left(\begin{array}{rr} 2 & 1 \\ 8 & -1 \\ 6 & 5 \end{array}\right)\\\ &F=\left(\begin{array}{rr} 5 & -1 \\ -4 & 0 \\ 2 & 3 \end{array}\right) \quad G=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned}$$ In each exercise, carry out the indicated matrix operations if they are defined. If an operation is not defined, say so. $$A+(B+C)$$

A parabola \(y=a x^{2}+b x+c\) passes through the three points \((1,-2),(-1,0),\) and (2,3) (a) Write down a system of three linear equations that must be satisfied by \(a, b,\) and \(c\) (b) Solve the system in part (a). (c) With the values for \(a, b,\) and \(c\) that you found in part (b), use a graphing utility to draw the parabola \(y=a x^{2}+b x+c .\) Find a viewing rectangle that seems to confirm that the parabola indeed passes through the three given points. (d) Another way to check your result in part (b): Apply the quadratic regression option on a graphing utility after entering the three given data points \((1,-2),(-1,0),\) and \((2,3) .\) (This is a valid check because, in general, three noncollinear points determine a unique parabola.)
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