Problem 18 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use back-substitution to solve the system of linear equations. $$\left\\{\begin{aligned} 5 x-8 z &=22 \\ 3 y-5 z &=10 \\ z &=-4 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -2\), \(y = -10/3\), \(z = -4\).

Step 1: Substitution of \(z\) in first equation

Substitute \(z = -4\) into the first equation. We get \[5x - 8*(-4) = 22\]. Solving this equation for \(x\) will give us the required value for \(x\).

Step 2: Calculation of \(x\)

Solving the equation from step 1, we get \[5x + 32 = 22\]. Subtracting 32 from both sides gives \(5x = 22 - 32 = -10\). Dividing both side by 5 we get \(x = -10 / 5 = -2\).

Step 3: Substitution of \(z\)

Substitute \(z = -4\) into the second equation. This gives us \[3y - 5*(-4) = 10\]. Solving this equation for \(y\) will give us the required value for \(y\).

Step 4: Calculation of \(y\)

Solving the equation from step 3, we get \[3y + 20 = 10\]. Subtracting 20 from both sides gives \(3y = 10 - 20 = -10\). Dividing both side by 3 we get \(y = -10 / 3 = -10/3\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Each equation in the system is in the form of a straight line when graphed. In this problem, we have three equations, each involving different variables, but they are related:

• First equation: \(5x - 8z = 22\)
• Second equation: \(3y - 5z = 10\)
• Third equation: \(z = -4\)

Notice that each of these equations is linear, as the exponents of each variable are 1. Solving a system of linear equations usually involves finding the values of the variables that satisfy all the equations simultaneously. This process can be visualized as finding the point or points where all lines (inside n-dimensional space) intersect. In our specific example, we are focusing on simplifying the system through already given situations (like \(z = -4\)) to find clear solutions for the remaining variables.

Substitution Method

The substitution method is a technique used to solve a system of equations. It involves solving one of the equations for one of the variables, and then substituting that expression into the other equations. This method is particularly useful when one of the equations is easy to solve, such as when one variable is already isolated, as with \(z = -4\) in our solution.

In the given exercise, we first solve the third equation, which has already been simplified for \(z\):

• We have \(z = -4\). This substitutes \(-4\) into the remaining equations, simplifying them to make \(x\) and \(y\) easier to find.

By choosing this approach, the other two equations unconsciously transform from involving two variables to single-variable equations, making our task much simpler. The essential steps for substitution then include identifying an equation suitable for substitution, solving for a variable, and then substituting into other equations to reduce their complexity.

Solving Equations Step-by-Step

Solving equations step-by-step is a detailed approach that ensures clarity and accuracy, especially in complex systems like these. Here, we use a systematic, logical approach to progress from having multiple unknowns to clear, definitive solutions.

Let's go through the key steps of solving the system step-by-step:

• **Identify a Simple Substitution**: Start with the known value, \(z = -4\), and substitute it into the other equations.
• **Simplified Substitution**: Substitute \(z = -4\) into the first equation: \(5x - 8(-4) = 22\), simplifying to \(5x + 32 = 22\). Solve for \(x\) to get \(x = -2\).
• **Next Equation Substitution**: Use \(z = -4\) in the second equation: \(3y - 5(-4) = 10\), simplifying to \(3y + 20 = 10\). Solve for \(y\) to find \(y = -\frac{10}{3}\).

Following these steps ensures that each part of the system is addressed correctly, allowing us to confidently arrive at the solution \((-2, -\frac{10}{3}, -4)\). This systematic way of breaking down problems can be applied in many other mathematical situations beyond just linear equations.