Problem 25 - Algebra A Combined Function

Question

Solve each system. $$ \left\\{\begin{array}{rr} x+2 y-z= & 5 \\ -3 x-2 y-3 z= & 11 \\ 4 x+4 y+5 z= & -18 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = 1\), \(y = -0.25\), \(z = -4.5\).

Step 1: Choose Variables to Eliminate

Look at the three equations and decide which variable to eliminate first. It's often easiest to choose the variable that has equal coefficients in multiple equations. Here, we can choose to eliminate the variable \(y\) by adding equations 1 and 2.

Step 2: Eliminate the First Variable

Add equations 1 and 2 to eliminate \(y\):\[(x + 2y - z) + (-3x - 2y - 3z) = 5 + 11\]Simplify the resulting equation:\(-2x - 4z = 16\)Divide every term by 2 for simplicity:\(-x - 2z = 8\) (Equation 4)

Step 3: Eliminate the Same Variable Again

Add equations 1 and 3 to eliminate \(y\):\[(x + 2y - z) + (4x + 4y + 5z) = 5 - 18\]Simplify the resulting equation:\(5x + 4z = -13\) (Equation 5)

Step 4: System of Two Equations

Now solve the system of two unknowns using Equation 4: \(-x - 2z = 8\) and Equation 5: \(5x + 4z = -13\).Solve Equation 4 for \(x\): \(x = -8 - 2z\).

Step 5: Substitute and Solve for z

Substitute \(x = -8 - 2z\) into Equation 5:\[5(-8 - 2z) + 4z = -13\]Simplify and solve for \(z\):\[-40 - 10z + 4z = -13\]Combine like terms:\[-6z = 27\]\(z = -4.5\).

Step 6: Find x

Substitute \(z = -4.5\) back into \(x = -8 - 2z\):\[x = -8 - 2(-4.5)\]Simplify:\[x = -8 + 9 = 1\].

Step 7: Solve for y

Substitute \(x = 1\) and \(z = -4.5\) into the first equation to solve for \(y\):\[1 + 2y - (-4.5) = 5\]Simplify:\[1 + 2y + 4.5 = 5\]\[2y + 5.5 = 5\]\[2y = -0.5\]\(y = -0.25\).

Step 8: Verify the Solution

Verify the values \(x = 1\), \(y = -0.25\), \(z = -4.5\) satisfy all original equations.1. Equation 1: \(1 + 2(-0.25) - (-4.5) = 5\) is true.2. Equation 2: \(-3(1) - 2(-0.25) - 3(-4.5) = 11\) simplifies correctly.3. Equation 3: \(4(1) + 4(-0.25) + 5(-4.5) = -18\) checks out as well.

Key Concepts

Variable Elimination

When solving a system of linear equations, variable elimination is a straightforward method where we aim to cancel out a variable by adding or subtracting equations. This simplification helps reduce the number of equations you need to work with, thus making the system easier to solve.

In the provided exercise, we decided to eliminate the variable \( y \) first because it appears with equal coefficients in multiple equations, specifically equations 1 and 2. By adding these two equations, the \( y \) terms cancel each other out.

• Addition and subtraction are used to strategically eliminate one variable, thus simplifying the original system of equations.
• Choose the variable that makes the elimination process the simplest, usually the one with coefficients that are easy to manage.

This approach sets the stage for reducing a three-equation system into a two-equation system, allowing easier solution of the remaining variables.

Substitution Method

After reducing the system of equations by elimination, the substitution method comes into play. This method involves solving one of the resulting simpler equations for a single variable and then substituting this expression into another equation to find another variable.

Once we eliminated the \( y \) variable and obtained Equations 4: \(-x - 2z = 8 \) and 5: \( 5x + 4z = -13 \), we solved Equation 4 for \( x \):

• First, express one variable in terms of another, for example, \( x = -8 - 2z \).
• Substitute this expression into another equation to solve for a different variable, in this case \( z \).

This method allows us to sequentially find each variable's value, thereby solving the system step by step.

Linear Equations

Linear equations are the backbone of systems of equations. They are equations of the first degree, meaning each term is either a constant or the product of a constant and a single variable.

Our system of equations can be represented in the matrix form, which highlights their linearity:

• Equation 1: \( x + 2y - z = 5 \)
• Equation 2: \(-3x - 2y - 3z = 11 \)
• Equation 3: \(4x + 4y + 5z = -18 \)

The uniform structure of these equations indicates that they are linear, as there are no exponents on any variable. Linear equations typically allow for straightforward solutions using methods such as elimination and substitution. The simplicity of linear equations—establishing a straight line when graphed—makes them fundamental in algebra.

Verifying Solutions

Once the values for all variables in a system of equations are found, verifying solutions is an essential step to confirm their correctness. This involves substituting the found values back into the original equations to ensure each equation holds true.

In our exercise, we verified the solution \( x = 1 \), \( y = -0.25 \), and \( z = -4.5 \) by checking each original equation:

• Substitute into Equation 1: Does \( 1 + 2(-0.25) + 4.5 = 5 \) hold? Yes, it does.
• Substitute into Equation 2: \(-3(1) - 2(-0.25) - 3(-4.5) = 11\) remains correct upon simplification.
• Substitute into Equation 3: \( 4(1) + 4(-0.25) + 5(-4.5) = -18 \), and it checks out perfectly.

Verifying solutions ensures accuracy and strengthens confidence in the methods used to solve the system. It is a crucial final step in problem-solving, confirming the solutions are indeed correct.