Problem 48 - Elementary and Intermediate Algebra

Question

Solve using elimination (addition): $$\left\\{\begin{array}{l}5 x-4 y=19 \\\3 x+2 y=7\end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((x, y) = (3, -1)\).

Step 1: Align the Equations

Write down the given system of equations: 1. \(5x - 4y = 19\)2. \(3x + 2y = 7\)Our goal is to eliminate one of the variables by adding the equations. Align the equations vertically to get ready for elimination.

Step 2: Make the Coefficients Opposite

To eliminate \(y\), ensure the coefficients of \(y\) in both equations are opposites. Currently, the coefficients are \(-4\) and \(2\). Multiply the second equation by \(2\) to achieve opposite coefficients of \(y\). This transforms the system:1. \(5x - 4y = 19\)2. \(6x + 4y = 14\)

Step 3: Add the Equations to Eliminate y

Add the two equations from Step 2 to eliminate the variable \(y\).\[(5x - 4y) + (6x + 4y) = 19 + 14\]This simplifies to:\[11x = 33\]

Step 4: Solve for x

Isolate \(x\) by dividing both sides of the equation by 11:\[x = \frac{33}{11} = 3\]

Step 5: Substitute x Back to Find y

Substitute \(x = 3\) back into one of the original equations to find \(y\). Using the second equation:\[3(3) + 2y = 7\]Simplify the left side:\[9 + 2y = 7\]Subtract 9 from both sides:\[2y = -2\]

Step 6: Solve for y

Divide both sides of the equation by 2 to solve for \(y\):\[y = \frac{-2}{2} = -1\]

Step 7: Conclusion

We found \(x = 3\) and \(y = -1\). Therefore, the solution to this system of equations is \((x, y) = (3, -1)\).

Key Concepts

Systems of Equations

A system of equations consists of two or more equations with the same set of variables. They are written in the form:

• \(ax + by = c\)
• \(dx + ey = f\)

The goal of solving a system is to find the values of the variables that satisfy all equations simultaneously. In our exercise, we're using a system of two linear equations:

• \(5x - 4y = 19\)
• \(3x + 2y = 7\)

We can approach systems of equations using various methods, like substitution and graphing. However, today, we focus on the elimination method. This involves manipulating the equations so that adding or subtracting them will cancel out one of the variables and make it easier to solve.

Solving Linear Equations

Linear equations involve variables raised only to the first power. They have a straightforward form: \(ax + by = c\). To solve these equations, we make them as simple as possible to isolate the variable. For example, when we worked on eliminating \(y\) in our system:
The first step involved transforming the second equation \(3x + 2y = 7\) into \(6x + 4y = 14\). The coefficients of \(y\) become opposites, allowing us to add equations to cancel out \(y\).Let's detail the process of solving after elimination:

• Add aligned equations to cancel \(y\): \[5x - 4y + 6x + 4y = 19 + 14\]Resulting in:
• \[11x = 33\]
• Solve for \(x\) by isolating it on one side:
• \[x = \frac{33}{11} = 3\]

Once \(x\) is determined, substitute it back into any original equation and solve for \(y\). This isolation process is key in solving linear equations.

Algebraic Techniques

Algebraic techniques are crucial for manipulating equations to find solutions. One common technique is multiplying equations to create opposite coefficients, as seen in this exercise. Here are some additional pointers to remember:

• When coefficients are not opposites, multiply one or both equations by necessary values to create opposites. This method is essential when using the elimination method.
• Be sure to perform operations on both sides of an equation evenly. This keeps the equations balanced and valid.
• After finding one variable, substitution back is key to finding the second variable.

In our exercise, transforming \(3x + 2y = 7\) into \(6x + 4y = 14\) was crucial. The elimination method relies heavily on these algebraic techniques to seamlessly find solutions.