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Chapter 7: Problem 14

Use back-substitution to solve the system of linear equations. $$\left\\{\begin{aligned} x-y+2 z &=22 \\ 3 y-8 z &=-9 \\ z &=-3 \end{aligned}\right.$$

Short Answer

Expert verified
The solution for the system of equations is \( x = 17 \), \( y = -11 \), and \( z = -3 \).

Step by step solution

01

Start with the known variable

We start by identifying the isolated variable, which is already given in the equations. In this case, \( z = -3 \).
02

Substitute the known variable

Next, substitute the found value of \( z \) into the second equation to find the value of \( y \). This is done as follows: \( 3y -8(-3) = -9 \). So, \( 3y + 24 = -9 \). Isolating \( y \) on one side, it yields \( y \) as \( -11 \).
03

Substitute the found variables value into the remaining equation

Finally, substitute the found values of \( y \) and \( z \) into the first equation to find the value of \( x \). The first equation is: \( x - y + 2z = 22 \), substituting \( y \) and \( z \) we get: \( x - (-11) + 2(-3) = 22 \). Hence, \( x = 22 - 11 + 6 = 17 \).

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Key Concepts

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

Back-substitution
Back-substitution is a method commonly used to solve systems of linear equations. It is particularly effective when dealing with equations arranged in an upper triangular form, where one of the variables is already isolated in the last equation.
To utilize back-substitution, follow these general steps:
  • Start by solving the equation where one variable is already isolated.
  • Substitute the known variable into the preceding equations, working your way up.
  • This substitution helps in progressively finding values for the other variables.
In our exercise, the variable 'z' is already given as \(z = -3\). With 'z' known, we substitute it into the second equation to determine the value of 'y'. Once 'y' is found, both 'y' and 'z' help solve the first equation for 'x'.
This clear linear stepping ensures that each variable is solved systematically, ultimately finding the solution for the entire system.
Isolate variables
Isolating variables is a crucial technique for solving equations, which involves rearranging the equation until the variable you need is alone on one side.
This process simplifies calculations by focusing on one variable at a time. In the exercise, after finding \(z = -3\), it became necessary to isolate 'y' in the equation \(3y - 8z = -9\).
  • First, substitute the value of 'z'.
  • Rearrange the terms to bring 'y' to one side.
  • Simplify to solve for the isolated 'y'.
For example, by substituting \(z = -3\) into \(3y - 8(-3) = -9\), the equation is simplified to \(3y + 24 = -9\).
By gradually isolating 'y', we derive \(y = -11\). This precise simplification is key to unraveling systems of equations efficiently.
Solve for variables
Once you've isolated a variable, the next logical step is to solve for its value. This step is vital when working through systems of linear equations.
After isolating 'y', the exercise progresses by solving for this variable, which was deduced to be \(y = -11\). This solution requires arithmetic simplification.
  • Perform calculations to find the numeric value of the variable.
  • Repeat the substituting and solving for other variables as necessary.
In our exercise, once both 'y' and 'z' were found, these values needed to be substituted into the first equation \(x - y + 2z = 22\).
By plugging in \(y = -11\) and \(z = -3\), the equation solves cleanly for 'x' as \(x = 17\). Solving for each variable sequentially ensures concrete answers, making it an essential skill in algebra.

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Most popular questions from this chapter

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