Problem 26 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{l} 3 x-y=9 \\ 5 x+7 y=1 \end{array}\right) $$

Step-by-Step Solution

Answer

\( x = \frac{32}{13}, y = -\frac{21}{13} \).

Step 1: Choose a Method

To solve the system of equations, we need to choose between substitution and elimination. Since neither of the equations is easily isolated, the elimination method appears to be more straightforward.

Step 2: Align Equations for Elimination

We have the equations: \( 3x - y = 9 \) and \( 5x + 7y = 1 \). To eliminate one variable, we can manipulate the coefficients of either \( x \) or \( y \). Let's eliminate \( y \).

Step 3: Equalize Coefficients of y

Multiply the first equation by 7 to match the \( y \)-terms: \(21x - 7y = 63 \). Now, the system is: \[\begin{align*}21x - 7y &= 63 \5x + 7y &= 1 \end{align*} \]

Step 4: Add Equations

Add the two equations together to eliminate \( y \): \[\begin{align*}21x - 7y + 5x + 7y &= 63 + 1 \26x &= 64\end{align*} \]

Step 5: Solve for x

Divide both sides by 26 to find \( x \): \[ x = \frac{64}{26} = \frac{32}{13} \].

Step 6: Substitute x to Find y

Substitute \( x = \frac{32}{13} \) into the first original equation \( 3x - y = 9 \):\[ 3\left(\frac{32}{13}\right) - y = 9 \] which simplifies to \[ \frac{96}{13} - y = 9 \].

Step 7: Solve for y

Rearrange the equation to solve for \( y \):\[ y = \frac{96}{13} - 9 \]. Converting 9 to the same denominator, \( \frac{117}{13} \), gives us \( y = \frac{96}{13} - \frac{117}{13} = -\frac{21}{13} \).

Key Concepts

Elimination Method

The elimination method is a popular technique for solving systems of linear equations. It involves adding or subtracting equations to eliminate one variable, making it easier to solve the system. Here's how it works:

• First, arrange the equations so that like terms are aligned. This means ensuring that all the coefficients of the variables and constants are in the same vertical position in each equation.
• Next, manipulate the equations to obtain equal coefficients for one of the variables. This often involves multiplying one or both equations by suitable numbers.
• Once the coefficients are equal, add or subtract the equations to eliminate one of the variables. You will be left with an equation that contains only one variable.
• Solve this resulting equation to find the value of the variable.
• Finally, substitute back to find the value of the eliminated variable.

By following these steps, you can solve the system of equations without needing to perform complex algebraic manipulations.

Substitution Method

The substitution method is another effective approach for solving systems of equations, especially when one equation can be easily solved for one variable:

• Begin by solving one of the equations for either variable. Choose the equation where isolation is most straightforward.
• Once isolated, substitute this expression into the other equation. This replaces the variable with its representation in terms of the other variable.
• Solve the resulting equation. It will now contain only the remaining variable, making it simple to find a numerical solution.
• After finding the value of the second variable, substitute it back into the expression obtained earlier to find the first variable.
• Verify the solution by plugging both variables back into the original equations to ensure they satisfy both.

This method is particularly useful when dealing with equations that are easily reducible or already partially simplified.

Solving Linear Equations

Solving linear equations is the process of finding the values of variables that make each equation true. Here's a breakdown of how to approach this for a system of equations:

• Identify whether the system requires substitution, elimination, or another method. This depends on how the equations are structured.
• Write equations in standard form if necessary, aligning like terms vertically.
• Use algebraic manipulations like addition, subtraction, multiplication, or division to isolate variables.
• Ensure each step maintains equivalence between the original equation and the modified version.
• Check all variable solutions by substituting them back into the original equations to confirm they satisfy every equation in the system.

Linear equations often have one solution, no solution, or infinitely many solutions depending on their graphical intersection. Mastering these basic techniques offers a foundation for solving more complex mathematical problems.