Problem 47 - Beginning and Intermediate Algebra

Question

Let \(u\) represent \(1 / x, v\) represent \(1 / y,\) and \(w\) represent \(1 / z .\) Solve first for \(u, v,\) and \(w .\) Then solve the following system of equations: $$\begin{aligned} &\frac{2}{x}+\frac{2}{y}-\frac{3}{z}=3\\\ &\frac{1}{x}-\frac{2}{y}-\frac{3}{z}=9\\\ &\frac{7}{x}-\frac{2}{y}+\frac{9}{z}=-39 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the given system of equations is \(x = -\frac{7}{5}\), \(y = \frac{7}{3}\), and \(z = \frac{7}{13}\).

Step 1: Express the equations in terms of u, v, and w

By definition, let \(u = \frac{1}{x}\), \(v = \frac{1}{y}\), and \(w = \frac{1}{z}\). Now let's substitute these expressions into the given system of equations: $$\begin{aligned} &2u+2v-3w=3\\\ &u-2v-3w=9\\\ &7u-2v+9w=-39 \end{aligned}$$

Step 2: Solve the system of linear equations

Now, let's solve the system of linear equations, using the substitution method. From the first equation, we can express \(u\) in terms of \(v\) and \(w\): $$ u = \frac{3 + 3w - 2v}{2} $$ Now substitute this expression for \(u\) into the second and third equations: $$\begin{aligned} &\frac{3 + 3w - 2v}{2}-2v-3w=9\\\ &7\left(\frac{3 + 3w - 2v}{2}\right)-2v+9w=-39 \end{aligned}$$ Now, let's solve the first of these equations for \(v\): $$ (3 + 3w - 2v) - 4v - 6w = 18 $$ Simplifying the equation, we obtain: $$ v = \frac{1}{2}(3 + w) $$ Now substitute this expression for \(v\) into the second equation: $$ 7\left(\frac{3 + 3w - (3 + w)}{2}\right) + \frac{1}{2}(9 - w) + 9w = -39 $$ Simplify and solve for \(w\): $$ 21w - 39 = 0 $$ So, \(w = \frac{39}{21} = \frac{13}{7}\). Now, substitute the value of \(w\) back into the expression for \(v\): $$ v = \frac{1}{2}(3 + \frac{13}{7}) = \frac{3}{7} $$ And substitute both the values of \(w\) and \(v\) back into the expression for \(u\): $$ u = \frac{3 + 3(\frac{13}{7}) - 2(\frac{3}{7})}{2} = -\frac{5}{7} $$

Step 3: Find the values of x, y, and z

Now, let's find the values of \(x\), \(y\), and \(z\) from the values of \(u\), \(v\), and \(w\): $$ x = \frac{1}{u} = \frac{1}{-\frac{5}{7}} = -\frac{7}{5} $$ $$ y = \frac{1}{v} = \frac{1}{\frac{3}{7}} = \frac{7}{3} $$ $$ z = \frac{1}{w} = \frac{1}{\frac{13}{7}} = \frac{7}{13} $$ So, the solution to the given system of equations is \(x = -\frac{7}{5}\), \(y = \frac{7}{3}\), and \(z = \frac{7}{13}\).

Key Concepts

System of Equations

In linear algebra, a system of equations is a collection of two or more equations with a set of variables. Here, we dealt with a system of three equations involving three variables: \(x\), \(y\), and \(z\). A system like this represents multiple equations that must be solved simultaneously to find a common solution. When solving such systems, you are essentially looking for a set of values that satisfies all equations at the same time.
Different methods can be used to solve these systems, including graphing, substitution, and elimination. In this exercise, we transformed the variables into \(u\), \(v\), and \(w\) to ease the complex fractions and simplify the system of linear equations. Understanding the relationship between each equation is key to correctly solving such systems.

Substitution Method

The substitution method is an algebraic technique for solving a system of equations. It involves expressing one variable in terms of another and substituting this expression into the other equations. This replaces one equation's variable, making it easier to solve the system step-by-step.

• In our example, we expressed \(u\) in terms of \(v\) and \(w\) from the first equation. This simpler form allowed us to substitute \(u\) into the other two equations.
• The goal is to progressively reduce the number of variables and equations until you solve for one variable at a time.

Substitution is particularly useful when one of the equations is simple, involving a single variable, which can be neatly re-arranged. The technique simplifies complex, multi-variable systems to make solving them more manageable.

Variables Transformation

Variable transformation is a powerful tool in simplifying systems of equations. In our exercise, we redefined the original variables \(x\), \(y\), and \(z\) as \(u = \frac{1}{x}\), \(v = \frac{1}{y}\), and \(w = \frac{1}{z}\).
This transformation turns fractional terms in the equations into linear terms involving \(u\), \(v\), and \(w\). This adjustment not only simplifies the manipulation of equations but also can reveal underlying structure or solutions that are difficult to see with the original variables.
After solving the linear equations with the transformed variables, we were able to revert them back to obtain the values of the original variables \(x\), \(y\), and \(z\). This showcases how variable transformation can streamline the problem-solving process in linear algebra.

Linear Equations

Linear equations are equations of the first order, meaning each term is either a constant or the product of a constant and a single variable. The highest power of the variable(s) in linear equations is one, making them straightforward, yet pivotal in mathematics for modeling relationships between quantities.
In the provided system, once transformed using \(u\), \(v\), and \(w\), each equation became linear, taking the form \(a * variable + b * variable + c * variable = constant\). Solving these linear equations usually involves manipulating the equations to isolate the variables.
Techniques such as simplification, rearranging terms, and operations like addition or subtraction of entire equations help isolate each variable one by one. Understanding and mastering the handling of linear equations paves the way for tackling more advanced mathematical problems, as they form the foundation of numerous algebraic concepts.