Problem 21 - College Algebra

Question

Solve each system by addition. $$ \begin{aligned} -2 x+5 y &=-42 \\ 7 x+2 y &=30 \end{aligned} $$

Step-by-Step Solution

Answer

x = 6, y = -6.

Step 1: Align the Equation System

Given the system of linear equations: \[\begin{aligned} -2x + 5y &= -42 \ 7x + 2y &= 30 \end{aligned}\] The goal is to use the method of addition (also known as elimination) to eliminate one of the variables, preferably by ensuring the coefficients of either \(x\) or \(y\) are the same in both equations.

Step 2: Make the Coefficients of y Equal

To eliminate \(y\), let's make the coefficients of \(y\) in both equations equal. We do this by finding a common multiple of the coefficients 5 and 2, which is 10.Multiply the first equation by 2: \[\begin{aligned} (2)(-2x + 5y) &= (2)(-42) \ -4x + 10y &= -84 \end{aligned}\] Multiply the second equation by 5: \[\begin{aligned} (5)(7x + 2y) &= (5)(30) \ 35x + 10y &= 150 \end{aligned}\]

Step 3: Eliminate the y Variable

Now that the coefficients of \(y\) are equal, subtract the first modified equation from the second modified equation to eliminate \(y\):\[\begin{aligned} (35x + 10y) - (-4x + 10y) &= 150 + 84 \ 35x + 10y + 4x - 10y &= 150 + 84 \ 39x &= 234 \end{aligned}\]

Step 4: Solve for x

Divide both sides of the equation by 39 to solve for \(x\):\[ x = \frac{234}{39} \]Simplifying the fraction gives:\[ x = 6 \]

Step 5: Substitute x Back into One of the Original Equations

Now that we have \(x = 6\), substitute this value back into one of the original equations to solve for \(y\). Let's use the first original equation:\[-2(6) + 5y = -42\]Simplify and solve for \(y\):\[-12 + 5y = -42\] Add 12 to both sides:\[5y = -42 + 12\] \[5y = -30\] Divide both sides by 5:\[y = -6\]

Step 6: Conclusion

The solution to the system of equations is \(x = 6\) and \(y = -6\). You can check the solution by substituting back into the original equations to ensure both are satisfied.

Key Concepts

Elimination Method

The elimination method, also known as the addition method, is a powerful technique used to solve systems of linear equations. This method involves combining equations in a way that eliminates one variable, allowing you to solve for the other variable more easily. Here's how it works in simple terms:

When you have two equations, you can add or subtract them to cancel out one of the variables. This process typically requires making the coefficients of one of the variables (either \(x\) or \(y\)) equal in both equations.
Once that's achieved, adding or subtracting the equations will eliminate that variable, letting you solve for the remaining variable.

• The key steps involve aligning the system of equations.
• Modifying coefficients to match.
• Subtracting (or adding) the equations to get rid of one variable.

After isolating one variable, the solution remains to substitute back to find the other. The elimination method is especially handy when the coefficients can be easily manipulated to match.

Coefficients

Coefficients are the numerical factors of the variables in equations. They play a crucial role in solving linear equations, especially when using the elimination method.

Consider the example equations \(-2x + 5y = -42\) and \(7x + 2y = 30\). Here, \(-2\) and \(7\) are the coefficients of \(x\), and \(5\) and \(2\) are the coefficients of \(y\).

To use the elimination method effectively, focus on making the coefficients of either \(x\) or \(y\) equal in both equations.

• This can be done by finding a common multiple of the coefficients.
• Multiply each equation by a suitable number to achieve this balance.

Once achieved, combining the equations will cancel out one variable, allowing you to solve for the other. Understanding coefficients is essential as they dictate how equations can be manipulated to facilitate solving.

Substitution

Substitution is a method used to solve systems of equations, complementing techniques like the elimination method. After one variable is isolated through elimination, substitution allows for solving the system entirely.

Once you find the value of one variable, substitute it back into one of the original equations. This will allow you to find the value of the other variable.

For instance, using the example above: after finding \(x = 6\), substituting it into the first original equation \(-2x + 5y = -42\) allows solving for \(y\):

• Substitute \(x = 6\) into \(-2x + 5y = -42\).
• Solve for \(y\) by simplifying and isolating \(y\).

This method is straightforward once one variable is known, making it easier to determine the complete solution to the system of equations.

Fraction Simplification

Fraction simplification is a critical step in many mathematical processes, including solving linear equations. When you derive a solution that results in a fraction, simplifying it makes the solution clearer and more concise.

In the given solution, the fraction \(\frac{234}{39}\) is simplified to \(6\) through the following process:

• Check the greatest common divisor (GCD) of the numerator and the denominator.
• Divide both the numerator and the denominator by the GCD.

Simplifying fractions is not just about neatness; it's about accuracy and ensuring the solution is presented in its simplest form. This practice is crucial in verifying correctness and understanding the relationships between numbers and variables in equations.