Problem 34 - Intermediate Algebra

Question

Solve each system using substitution. $$ \left\\{\begin{array}{l} x+6 z=-36 \\ 5 x+3 y-2 z=-20 \\ y+4 z=-20 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \(x = -6\), \(y = 0\), \(z = -5\).

Step 1: Choose an equation for substitution

Choose an equation to express one variable in terms of another. The first equation is simpler, so let's solve for \(x\): \(x = -36 - 6z\).

Step 2: Substitute into the remaining equations

Substitute \(x = -36 - 6z\) into the other two equations. For the second equation, replace \(x\):\[5(-36 - 6z) + 3y - 2z = -20.\] This simplifies to \(-180 - 30z + 3y - 2z = -20.\)

Step 3: Simplify and solve for \(y\) and \(z\)

Combine like terms in the substituted equation: \(-180 + 3y - 32z = -20\), which simplifies to \(3y - 32z = 160\). For the third original equation, substitute \(y\):\[y + 4z = -20.\] Solve for \(y\) to get \(y = -20 - 4z\).

Step 4: Substitution for single-variable equation

Substitute \(y = -20 - 4z\) into the equation \(3y - 32z = 160\):\[3(-20 - 4z) - 32z = 160.\] Simplify to \(-60 - 12z - 32z = 160\), which results in \(-60 - 44z = 160.\)

Step 5: Solve for \(z\)

Rearrange the equation to solve for \(z\):\[ -44z = 160 + 60 \] gives \(-44z = 220\). Divide everything by \(-44\) to obtain \(z = -5.\)

Step 6: Solve for \(y\)

Use \(z = -5\) to find \(y\) using \(y = -20 - 4z\): \[ y = -20 - 4(-5) = -20 + 20 = 0. \]

Step 7: Solve for \(x\)

Use \(z = -5\) in the equation \(x = -36 - 6z\): \[ x = -36 - 6(-5) = -36 + 30 = -6. \]

Step 8: Verify solutions

Verify solutions in all original equations: 1) \(x + 6z = -36\): \(-6 + 6(-5) = -36\), 2) \(5x + 3y - 2z = -20\): \(5(-6) + 3(0) - 2(-5) = -20\), and 3) \(y + 4z = -20\): \(0 + 4(-5) = -20\). All equations hold true.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in linear algebra to solve systems of equations. This method focuses on expressing one variable in terms of another and then substituting this expression into the other equations in the system. This approach is particularly useful when we have at least one equation that is relatively simple and easy to solve for one variable.
First, let's see how this concept is applied in our exercise. We have a system of three equations:

• Equation 1: \(x + 6z = -36\)
• Equation 2: \(5x + 3y - 2z = -20\)
• Equation 3: \(y + 4z = -20\)

To start, we need to isolate one variable. The first equation is the simplest and can be easily rearranged to express \(x\) in terms of \(z\): \(x = -36 - 6z\). This is the key part of the substitution method—finding an expression for one variable guiding the rest of the solution steps. This expression is then used in the other equations to reduce the number of variables remaining.

Solving Systems of Equations

When dealing with systems of equations, our goal is to find values for each variable that satisfy all equations at the same time. For systems involving linear equations, such as the one in our exercise, these solutions often correspond to points of intersection between lines or planes represented by the equations.
In our exercise, after we derived \(x = -36 - 6z\), we substitute this expression into the other two equations. Substituting \(x\) in Equation 2 changes it into a two-variable equation involving \(y\) and \(z\). We then continue solving the system to reduce these two into a single variable equation.
This systematic substitution process eventually simplifies our set of equations, leading us to determine specific values for \(z\), \(y\), and \(x\). This multi-step procedure is a hallmark of solving systems of equations, ensuring that we find a unique solution, if it exists, that fits all parts of the system.

Step-by-Step Solution

A step-by-step solution to systems of equations is crucial for preventing mistakes and understanding each phase of the solution. Here is a refined walkthrough of the given exercise:

• **Step 1**: Begin with an easy equation and express one variable in terms of another. Here, \(x = -36 - 6z\) was drawn from the first equation.
• **Step 2**: Substitute this expression into the remaining equations. When substituted into Equation 2, we obtain: \(-180 + 3y - 32z = -20\).
• **Step 3**: Simplify and solve for one of the remaining two variables (\(y\) and \(z\)). We derive \(3y - 32z = 160\), and from Equation 3: \(y = -20 - 4z\).
• **Step 4**: Substitute again. Substitute \(y = -20 - 4z\) into the derived equation from Step 2 to simplify further: \(-60 - 44z = 160\).
• **Step 5**: Solve for \(z\). Rearrange to find \(z = -5\).
• **Step 6**: Utilize \(z = -5\) to find \(y\) from its expression. Here, \(y = 0\).
• **Step 7**: Finally, with \(z = -5\), determine \(x\). We get \(x = -6\).
• **Verification**: Substitute all obtained values back into the original equations to check correctness. All checks confirm the solution.

By progressing methodically through each of these steps, we accurately arrive at the intersection of the three planes represented by the equations, confirming our solution is valid.