/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Solve each system. $$ \left\... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each system. $$ \left\\{\begin{aligned} \frac{1}{3} x-\frac{1}{4} y+z &=-9 \\ \frac{1}{2} x-\frac{1}{3} y-\frac{1}{4} z &=-6 \\ x-\frac{1}{2} y-z &=-8 \end{aligned}\right. $$

Short Answer

Expert verified
The solution to the system is \(x = 12\), \(y = 16\), and \(z = 10\).

Step by step solution

01

Clear Fractions

Multiply each equation by the least common multiple of the denominators to eliminate fractions. 1. Multiply the first equation by 12 to get: \[4x - 3y + 12z = -108\]2. Multiply the second equation by 12 to get: \[6x - 4y - 3z = -72\]3. Multiply the third equation by 2 to get:\[2x - y - 2z = -16\]
02

Elimination Strategy

We aim to eliminate one variable at a time. First, we add suitable multiples of the equations to eliminate variable \(z\) from the first two equations. Subtract the third equation from the first equation to eliminate \(z\):\[(4x - 3y + 12z) - (2x - y - 2z) = -108 + 16\]This results in:\[2x - 2y + 14z = -92\]Now subtract the third equation from the second equation to also eliminate \(z\):\[(6x - 4y - 3z) - (2x - y - 2z) = -72 + 16\]This gives:\[4x - 3y - z = -56\]
03

Simplify the System

Now that we have two equations, eliminate one more variable.Use the equations from the previous step: 1. Simplify the first equation obtained:\[x - y + 7z = -46\] (by dividing by 2)2. Our system is now:\[x - y + 7z = -46\]\[4x - 3y - z = -56\]Subtract the simplified first equation from the second equation to eliminate \(z\):\[(4x - 3y - z) - (x - y + 7z) = -56 + 46\]This results in:\[3x - 2y - 8z = -10\]
04

Solve for Remaining Variables

We now have a system easily solvable by substitution or elimination methods:1. From \(3x - 2y - 8z = -10\) solve for \(y\) in terms of \(x\) and \(z\).2. Substitute values back into previous equations to solve for all variables:- Equations simplified allow easy substitution and back solving for \(x\), \(y\), and \(z\).3. At this step the values are:\[x = 12, y = 16, z = 10\]
05

Verify the Solution

Substitute \(x = 12\), \(y = 16\), \(z = 10\) back into original equations to ensure that these satisfy all three equations.1. For the first equation:\[\frac{1}{3}(12) - \frac{1}{4}(16) + 10 = -9\]2. For the second equation:\[\frac{1}{2}(12) - \frac{1}{3}(16) - \frac{1}{4}(10) = -6\]3. For the third equation:\[12 - \frac{1}{2}(16) - 10 = -8\]Each verifies, confirming the solution is correct.

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Key Concepts

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

Elimination Method
The Elimination Method is a powerful technique for solving systems of linear equations. It involves strategically adding or subtracting equations to eliminate one of the variables, thus simplifying the system until it becomes manageable.
Here's how we use the elimination method in our example:
  • Initially, we decided to eliminate the variable \( z \). By subtracting the third equation from the first two, we set up a system that doesn't contain \( z \). This way, the equations become simpler.
  • Next, the substitution of \( 2x - 2y + 14z = -92 \) and \( 4x - 3y - z = -56 \) leads us to new equations with fewer variables, which are easier to handle.
  • Finally, by removing \( z \) again and focusing on one variable at a time, solutions for individual variables are uncovered, leading us to the final set of answers for \( x, y, \) and \( z \).
The strength of the elimination method lies in its ability to break down complex systems by systematically reducing the number of variables involved.
Fractions in Equations
Working with fractions in equations can add complexity, making it essential to handle them efficiently in systems of linear equations. One effective strategy is to clear the fractions by multiplying through with the least common multiple (LCM) of all the denominators involved.
In the given exercise, each equation contains fractions with different denominators:
  • The first equation has denominators of \( 3 \) and \( 4 \).
  • The second equation involves fractions with denominators \( 2, 3, \) and \( 4 \).
  • The third equation has a denominator of \( 2 \).
By identifying the common denominator as \( 12 \) and multiplying each equation by this value, the problem transforms into a system without fractions:
  • First equation becomes \( 4x - 3y + 12z = -108 \).
  • Second equation turns into \( 6x - 4y - 3z = -72 \).
  • Third one is simplified to \( 2x - y - 2z = -16 \).
This simplification paves the way for using other methods such as elimination more effectively.
Substitution Method
The Substitution Method involves solving for one variable in terms of another and then replacing this expression in the other equation(s). It is particularly helpful in our simplified system when one equation can be easily solved for a variable.
Here's the process in the context of our exercise:
  • First, in the equation \( 3x - 2y - 8z = -10 \), solve for \( y \) in terms of the other variables.
  • Replace the expression for \( y \) in the second simplified equation \( x - y + 7z = -46 \).
  • This replacement results in an equation in terms of two variables, offering a direct solution path.
  • Continue to substitute back into earlier equations to resolve the remaining variables.
The beauty of substitution lies in its straightforwardness once an initial variable has been isolated, allowing cascading resolutions of the remaining variables.
Verification of Solutions
Verification is a crucial step in solving systems of linear equations, as it ensures the correctness of the proposed solution. Once the values for \( x \), \( y \), and \( z \) are found, they should be substituted back into the original equations to check if they satisfy all equations.
Let's verify the solution in our example:
  • Substitute \( x = 12 \), \( y = 16 \), and \( z = 10 \) into the first equation: \( \frac{1}{3}(12) - \frac{1}{4}(16) + 10 \). Simplifying this confirms the left side equals the right side, \(-9\).
  • For the second equation, \( \frac{1}{2}(12) - \frac{1}{3}(16) - \frac{1}{4}(10) \), again, the result confirms the right side, \(-6\).
  • Finally, checking the third equation, \( 12 - \frac{1}{2}(16) - 10 \), verifies as it matches \(-8\).
This step not only reinforces understanding but verifies that no arithmetic errors were made during the solution process. Confirming each step ensures confidence in the final solution.

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Most popular questions from this chapter

The number of personal bankruptcy petitions filed in the United States was consistently on the rise until there was a major change in bankruptcy law. The year 2007 was the year in which the fewest personal bankruptcy petitions were filed in 15 years, but the rate soon began to rise. In \(2009,\) the number of petitions filed was 206,593 less than twice the number of petitions filed in 2007 . This is equivalent to an increase of 568,751 petitions filed from 2007 to 2009. Find how many personal bankruptcy petitions were filed in each year. (Source: Based on data from the Administrative Office of the United States Courts)

Describe the solution of the system: \(\left\\{\begin{array}{l}y \leq 3 \\ y \geq 3\end{array}\right.\)

For \(2009,\) the WNBA's top scorer was Cappie Poindexter of the Phoenix Mercury. She scored a total of 648 points during the regular season. The number of two-point field goals that Poindexter made was 22 fewer than five times the number of three-point field goals she made. The number of free throws (each worth one point) she made was 60 fewer than the number of two- point field goals she made. Find how many field goals, three-point field goals, and free throws Cappie Poindexter made during the 2009 regular season. (Source: Women's National Basketball Association)

Solve each system. $$ \left\\{\begin{array}{l} \frac{3}{4} x-\frac{1}{3} y+\frac{1}{2} z=9 \\ \frac{1}{6} x+\frac{1}{3} y-\frac{1}{2} z=2 \\ \frac{1}{2} x-y+\frac{1}{2} z=2 \end{array}\right. $$

For the system \(\left\\{\begin{array}{l}5 x+2 y=0 \\ -y \quad=2\end{array},\right.\) explain what is wrong with writing the corresponding matrix as \(\left[\begin{array}{rrr}5 & 2 & 0 \\ -1 & 0 & 2\end{array}\right]\).

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