Problem 35 - A Graphical Approach to College Algebra

Question

\(\begin{aligned} &\frac{2}{x}-\frac{2}{y}+\frac{1}{z}=-1\\\ &\frac{4}{x}+\frac{1}{y}-\frac{2}{z}=-9\\\ &\frac{1}{x}+\frac{1}{y}-\frac{3}{z}=-9 \end{aligned}\)

Step-by-Step Solution

Answer

x = -\frac{4}{5}, y = \frac{4}{7}, z = 0.1714

Step 1: Identify Given Equations

We have a system of three equations involving fractions:1. \( \frac{2}{x} - \frac{2}{y} + \frac{1}{z} = -1 \)2. \( \frac{4}{x} + \frac{1}{y} - \frac{2}{z} = -9 \)3. \( \frac{1}{x} + \frac{1}{y} - \frac{3}{z} = -9 \)

Step 2: Substitution Variables

To simplify the equations, we substitute:\( a = \frac{1}{x} \), \( b = \frac{1}{y} \), \( c = \frac{1}{z} \).Then the system becomes:1. \( 2a - 2b + c = -1 \)2. \( 4a + b - 2c = -9 \)3. \( a + b - 3c = -9 \)

Step 3: Simplify Equations

We work with the new system:1. \( 2a - 2b + c = -1 \)2. \( 4a + b - 2c = -9 \)3. \( a + b - 3c = -9 \)We can simplify equation 1 by dividing through by 2:\( a - b + \frac{c}{2} = -\frac{1}{2} \)

Step 4: Eliminate a or b

Using equation 1 (simplified): \( a - b + \frac{c}{2} = -\frac{1}{2} \)From equation 3: \( a + b - 3c = -9 \), we add equation 1 to eliminate \( b \):- Adding eq. 1 (\( 2a - 2b + c = -1 \)) with eq. 3 (\( a + b - 3c = -9 \)) gives:\[ 3a - 2b - 2c = -10 \]

Step 5: Use Two Equations to Solve for Two Variables

Let's pick equations 2 and 4:From eq. 2: \( 4a + b - 2c = -9 \)From summed eq. 4: \( 3a - 2b - 2c = -10 \)Subtract second line from first: \[ a + 3b = 1 \]

Step 6: Solve the Linear System

Now we have:1. \( a + 3b = 1 \)2. \( 3a - b - 5 = 0 \) (modifying eq. 5)We solve these two equations for \( a \) and \( b \).Add equations:\[ 4a + 2b = -5 \]\( a = -\frac{5}{4} \), \( b = \frac{7}{4} \)

Step 7: Solve for c

Substitute \( a = -\frac{5}{4} \), \( b = \frac{7}{4} \) into equation 2 or 3:\( a + b - 3c = -9 \) becomes \(-\frac{5}{4} + \frac{7}{4} - 3c = -9 \ \frac{2}{4} - 3c = -9 \ 3c = 8.5 + 9 \ 3c = 17.5 \ c = \frac{17.5}{3} = 5.833 \)

Step 8: Solve for x, y, z

Recall: \( a = \frac{1}{x} = -\frac{5}{4} \), so \( x = \frac{-4}{5} \)\( b = \frac{1}{y} = \frac{7}{4} \), so \( y = \frac{4}{7} \)\( c = \frac{1}{z} = 5.833 \), so \( z = \frac{1}{5.833} = 0.1714 \)

Key Concepts

Fractional equations

Fractional equations are mathematical expressions that include fractions. They typically involve variables in the denominator, which can complicate solving equations. In the original exercise, each equation contains terms like \(\frac{1}{x}\), \(\frac{1}{y}\), and \(\frac{1}{z}\). This indicates the presence of fractional equations, as these fractions represent the inverse of the variables.

To effectively solve these, we often need to clear the fractions. This involves finding a common multiple of the denominators or substituting the fractions with another variable to simplify the equations. This was done by substituting \(a = \frac{1}{x}\), \(b = \frac{1}{y}\), and \(c = \frac{1}{z}\). Doing so transforms the complex fractional equations into linear equations, facilitating easier manipulation and solution of the system.

A key to success with fractional equations is patience and careful manipulation. Always check your work at each step to avoid mistakes.

Substitution method

The substitution method is a powerful tool used to solve systems of equations, especially when dealing with fractional equations. It involves substituting one variable with another equivalent expression, simplifying the system into more manageable equations.

In the exercise, by substituting \(a = \frac{1}{x}\), \(b = \frac{1}{y}\), and \(c = \frac{1}{z}\), the equations are transformed from fractional to linear form, making them easier to solve. This method allows for direct manipulation of equations to eliminate variables step by step.

• First, simplify one of the equations by substituting a fraction.
• Next, use the simplified equations to write other equations in terms of fewer variables.
• Continue this process until you can easily solve for one of the variables.

By methodically eliminating variables, the substitution method streamlines the solving process. This technique is especially handy when one variable (like \(a, b,\) or \(c\)) can be expressed simply in terms of other variables or constants. Once one variable is found, it can be substituted back to solve for others.

Solving linear systems

Solving linear systems involves finding the values of unknown variables that satisfy all equations simultaneously. Once we've cleared the fractions in the original exercise by substitution, we're left with a linear system involving the variables \(a, b,\) and \(c\).

To solve, we first simplify expressions and use basic algebraic techniques to reduce to a smaller number of equations.

• Identify and manipulate simpler equations to express one variable in terms of others.
• Substitute expressions into other equations, reducing the quantity of unknowns with each step.
• Continue until only one unknown remains, solve, and back-substitute to find all variables' values.

For example, after substituting expressions and simplifying, we could isolate \(a\) and \(b\), then substitute back to find \(c\). The interaction of calculations and substitutions helps unravel complex linear systems.

Remember to always verify solutions by substituting them back into the original equations to ensure they satisfy all conditions. This double-checking is crucial for ensuring accuracy, especially when dealing with fractional equations initially.