Problem 15 - Precalculus with Limits

Question

Using Back-Substitution In Exercises \(11 - 16 ,\) use back-substitution to solve the system of linear equations. $$\left\\{ \begin{aligned} 4 x - 2 y + z & = 8 \\ - y + z & = 4 \\ z & = 11 \end{aligned} \right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 1\), \(y = 7\), and \(z = 11\).

Step 1: Find the value of z

The last given equation is \(z = 11\). So, we have our first variable, z, equals to 11.

Step 2: Substituting z into the second equation

Having found the value of z, substitute it into the second equation, \(-y + z = 4\). Hence, substituting \(z = 11\) into the equation, we get \(-y + 11 = 4\). Solving for y, \(-y = 4 - 11\), we find that \(y = -(-7)\), therefore \(y = 7\).

Step 3: Substituting values of y and z into the first equation

After getting the values of z and y, we substitute them into the first equation \(4x - 2y + z = 8\). By inserting \(z = 11\) and \(y = 7\) to the equation, we get \(4x - 2*7 + 11 = 8\), thereby, the equation simplifies to \(4x = 8 - 11 +14\), which implies that \(x = 1\).

Key Concepts

Understanding Systems of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. For example, the system given in the exercise includes three equations with variables \(x\), \(y\), and \(z\). Linear equations graphically represent straight lines, and a system of these equations determines where the lines intersect, if at all. Our goal is to find a solution, which is a set of values for the variables that satisfy all equations in the system simultaneously. The system in the exercise is:

• \(4x - 2y + z = 8\)
• \(-y + z = 4\)
• \(z = 11\)

This particular system is organized in a kind of triangular form, which makes it suitable for a method called back-substitution, starting with the simplest equation to solve and working backwards.

Solving Equations with Back-Substitution

Back-substitution is a method used when systems of equations have been arranged in a step-like pattern, usually from easiest to hardest. Here, our goal is to find solutions incrementally by starting with the simplest equation. Given the system, we start by solving the last equation \(z = 11\), giving us a basis to substitute back into previously unsolved equations. Using this value, we progress to solve \(-y + z = 4\), allowing us to find \(y\). Finally, we substitute both the known \(z\) and \(y\) back into the first equation \(4x - 2y + z = 8\) to solve for \(x\). This cascade of solving from simplest to most complex creates a clear path to the solution of the whole system. Each step builds on the previous, ensuring consistency and accuracy in solving the system.

Mastering Algebraic Manipulation

Algebraic manipulation involves the use of various techniques to rearrange and simplify equations, which is vital when solving systems of linear equations. Let's break down the steps involved:

• Substitution: After determining \(z = 11\), we substitute this value into the second equation \(-y + z = 4\). This substitution step is fundamental, boiling down complex systems into manageable pieces.
• Simplification: Once substitution is applied, simplifying the equations becomes key. For example, when we found \(-y + 11 = 4\), simplifying this produced \(y = 7\).
• Rearrangement: Finally, using our values for \(y\) and \(z\), we substitute these into \(4x - 2y + z = 8\). Properly rearranging and simplifying this equation led to solving \(x\).

Each manipulation step ensures our pathway to solving for each variable is clear and logical, paving the way for systematically finding \(x\), \(y\), and \(z\) with ease using back-substitution.