Problem 54 - Algebra A Combined Function

Question

Solving systems involving more than three variables can be accomplished with methods similar to those encountered in this section. Apply what you already know to solve each system of equations in four variables. $$ \left\\{\begin{aligned} 5 x+4 y &=29 \\ y+z-w &=-2 \\ 5 x+z &=23 \\ y-z+w &=4 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{49}{5}\), \(y = -5\), \(z = -26\), \(w = -23\).

Step 1: Identify Equations and Variables

We have a system of four equations that involves four variables: \(x\), \(y\), \(z\), and \(w\). The equations are: 1. \(5x + 4y = 29\) 2. \(y + z - w = -2\) 3. \(5x + z = 23\) 4. \(y - z + w = 4\).

Step 2: Eliminate a Variable

Subtract equation 2 from equation 4 to eliminate \(w\): \((y - z + w) - (y + z - w) = 4 - (-2)\), simplifying, we get: \(-2z + 2w = 6\). Simplify further: \(-z + w = 3\).

Step 3: Express One Variable in Terms of Others

From equation 3, we have \(5x + z = 23\). Solve for \(z\): \(z = 23 - 5x\). Substitute \(z = 23 - 5x\) in \(-z + w = 3\): \(- (23 - 5x) + w = 3\). This gives \(w = 3 + 23 - 5x = 26 - 5x\).

Step 4: Solve for Remaining Variables

Now we substitute \(z = 23 - 5x\) into equation 2: \(y + (23 - 5x) - (26 - 5x) = -2\). This simplifies to \(y + 3 = -2\), so \(y = -5\).

Step 5: Solve for x and Check Solutions

Substitute \(y = -5\) back into equation 1: \(5x + 4(-5) = 29\). So, \(5x - 20 = 29\), which gives \(5x = 49\). Solving for \(x\), we get \(x = rac{49}{5}\). Now using \(x = rac{49}{5}\), find \(z = 23 - 5x = 23 - 49 = -26\), and finally \(w = 26 - 5x = 26 - 49 = -23\).

Step 6: Verify the Solution

Substitute \(x = rac{49}{5}\), \(y = -5\), \(z = -26\), \(w = -23\) back into the original equations to ensure all equations are satisfied. Each substitution confirms that the computed values satisfy the four equations.

Key Concepts

Elimination Method

The elimination method is a popular technique for solving systems of equations. It works by reducing the number of variables in equations, typically through addition or subtraction of equations in order to 'eliminate' one of the variables.
This method simplifies the problem and allows you to solve for the other variables more easily.

In the given exercise, we started by identifying the equations and variables involved. We had four equations with four variables: \(x\), \(y\), \(z\), and \(w\).

• First, we used the elimination method by subtracting one equation from another.
• This helped us eliminate the variable \(w\), reducing the complexity of our system.

By rearranging the equations, we simplified them step by step, focusing on removing variables.The purpose of the elimination method is to make systems of equations manageable, especially when multiple variables are involved. The key is to strategically choose which variable to eliminate first, often based on simplicity or ease of breaking down the system.
Understanding the process of elimination can greatly assist to streamline the solving process of any complex algebraic problem.

Substitution Method

The substitution method is another effective approach for solving systems of equations. This technique involves solving one equation for one variable and then substituting that expression into the other equations.

In our example, we took this step by expressing one of the variables in terms of others.

• From one of the simplified equations, we solved for \(z\).
• Once \(z\) was expressed in terms of another variable, we substituted this expression into another equation to find \(w\).

This process facilitates solving the entire system by systematically reducing the number of variables in other equations.
Specifically, once variables are expressed in relation to one another, they can be easily managed and provide a clear path to a solution.

In summary, the substitution method is beneficial when dealing with equations where isolating a variable is straightforward. It creates a cascade where solving an equation for one variable can make other equations simpler, hence speeding up the solving process. Ensuring each substitution is logical and accurate is essential for obtaining the correct solution.

Variables in Algebra

Variables in algebra are symbols commonly represented by letters such as \(x\), \(y\), \(z\), and \(w\). These symbols stand in for unknown values and play a crucial role in forming algebraic equations and systems of equations.

Understanding how to work with variables is essential in algebra as they allow us to express complex relationships succinctly.

• When you solve an equation, you are essentially figuring out what number each variable represents that makes all the equations true at the same time.
• Variables can be manipulated using various algebraic techniques, such as substitution and elimination as discussed.

Different variables can represent different values or measurements, such as dimensions in a geometry problem or various quantities in word problems. In the given problem:

• We had a set of four variables which required us to deduce their values based on the given equations.
• Handling multiple variables often requires breaking down equations into simpler forms, highlighting their interconnected relationships.

It's important to consistently track each variable and ensure that operations performed on them are valid across all involved equations.
Mastering this skill allows budding mathematicians to tackle more intricate mathematical concepts where variables define the parts of real-world scenarios. The key is to practice and to understand the logic behind the manipulation of variables.