Problem 45 - Precalculus with Limits

Question

Solving a System of Linear Equations In Exercises \(25 - 46\) , solve the system of linear equations and check any solutions algebraically. $$ \left\\{ \begin{aligned} x & \+ 3 w = 4 \\ 2 y - z - w & = 0 \\ 3 y & \- 2 w = & 1 \\ 2 x - y + 4 z & = 5 \end{aligned} \right. $$

Step-by-Step Solution

Answer

The system has a unique solution: x = 1, y = 1.5, z = 2, w = 1.

Step 1: Rewriting the system of equations

First, let's write down the system of equations in a clear, organized way: \[\begin{aligned}1. & \quad x + 3w = 4 \2. & \quad 2y - z - w = 0 \3. & \quad 3y - 2w = 1 \4. & \quad 2x - y + 4z = 5\end{aligned}\]The next step will be to express one of the variables in terms of the others in an equation and then substitute it in other equations.

Step 2: Solving the first equation for x

We can re-arrange Equation 1 to get x in terms of w: \[\begin{aligned}x = 4 - 3w \end{aligned}\]We will substitute this in Equation 4.

Step 3: Substituting into equation 4

Upon substituting x from Equation 1 into Equation 4, we obtain a new equation:\[\begin{aligned}2(4 - 3w) - y + 4z = 5 8 - 6w - y + 4z = 5 -6w - y + 4z = -3 \end{aligned}\]We will continue by solving Equation 2 for y.

Step 4: Solving the second equation for y

We can rearrange Equation 2 to express y in terms of z and w:\[\begin{aligned}2y = z + w y = \frac{z + w}{2} \end{aligned}\]Next, we'll substitute y into the revised Equation 4.

Step 5: Substitute into revised equation 4

Substitute y from Equation 2 into the revised Equation 4, we have:\[\begin{aligned}-6w - \frac{(z + w)}{2} + 4z = -3 -6w - \frac{z}{2} - \frac{w}{2} + 4z = -3 -6.5w + 3.5z = -3\end{aligned}\]From this equation z can be solved.

Step 6: Solving for z

Rearrange the equation from Step 5 to isolate z:\[\begin{aligned}3.5z = 6.5w + 3 z = \frac{6.5w + 3}{3.5}\end{aligned}\]Next, we will substitute this z into Equation 2 to solve for w.

Step 7: Substitute into equation 2

Substitute z from Equation 6 into the Equation 2, we get:\[\begin{aligned}2y - \frac{6.5w + 3}{3.5} - w = 0 2y - w = \frac{6.5w + 3}{3.5}\end{aligned}\]From this equation w can be solved.

Step 8: Solving for w

Rearrange the equation from step 7 to isolate w and solve for the variable:\[\begin{aligned}w = \frac{2y(3.5) - 6.5w - 3}{3.5} w = 1\end{aligned}\]Now that we have a value for w, we can substitute this back into the equations obtained in Steps 2 and 6 to solve for x and z respectively.

Step 9: Solve for x and z

Substitute w=1 into the equation from Step 2:\[\begin{aligned}x = 4 - 3(1) x = 1\end{aligned}\]Substitute w=1 into the equation from Step 6:\[\begin{aligned}z = \frac{6.5(1) + 3}{3.5} z = 2\end{aligned}\]Finally, we substitute w=1 and z=2 into the equation obtained in Step 4 to solve for y.

Step 10: Solve for y

Substitute w=1 and z=2 into the equation from Step 4:\[\begin{aligned}y = \frac{2 + 1}{2} y = 1.5\end{aligned}\]So, the solutions of the system are x = 1, y = 1.5, z = 2, w = 1.

Key Concepts

solving systems algebraically

When faced with a system of linear equations, we can solve it algebraically by organizing and manipulating the equations to find the values of the variables. This means using algebraic operations like addition, subtraction, and multiplication, to simplify the equations. Our goal is to eventually find a single value for each variable included in the system. By solving systems algebraically, we work through each equation methodically. We look for ways to isolate individual variables, making it easier to substitute their expressions into other equations. The algebraic method is very reliable because it allows for precise calculations. It can handle complex systems with multiple equations and variables, providing a clear and correct solution step-by-step. In the context of the exercise, algebraic manipulation involves rearranging equations and substituting variables' expressions until reaching a single equation with one variable. Once we reach this stage, we can solve for that variable's value, then backtrack and find the values for the others.

variables substitution

Substitution is a key technique in solving systems of linear equations. It involves expressing one variable in terms of the others and then substituting this expression into the rest of the system. This method simplifies the system, reducing the number of variables in one or more equations. The main advantage is decreasing complexity and turning a multi-variable equation into a single-variable one. The substitution process is straightforward:

• Isolate one variable in terms of the others from an equation.
• Replace this expression in other equations.
• Repeat the process until all variables are isolated and found.

In the exercise, substitution was used by isolating variables like \(x\) and \(y\) and then substituting them back into another part of the system. This allows us to simplify the problem progressively, making the calculation of each variable's value manageable.

rearranging equations

Rearranging equations is an essential skill in solving systems of equations. The process involves altering the structure of an equation to make one variable easier to solve. This is often done by using basic algebraic principles such as:

• Adding or subtracting terms on both sides of an equation.
• Multiplying or dividing to isolate a variable.
• Switching sides for easier readability and manipulation.

Rearranging helps clarify relationships between variables and is often the first step in substitution or elimination methods. In the exercise, equations were rearranged to isolate \(x\), \(y\), and \(z\). This allowed each variable's value to be easily identified through subsequent calculations and substitutions. Rearranging is foundational because it sets up the equations in a way that is prime for further action. It simplifies complex expressions into something manageable and prepares for variable substitution.