Problem 4 - Elementary Algebra

Question

Use the elimination-by-addition method to solve each system. $$\left(\begin{array}{l}5 x+2 y=-4 \\ 5 x-3 y=6\end{array}\right)$$

Step-by-Step Solution

Answer

The solution is \( x = 0 \) and \( y = -2 \).

Step 1: Setup the Equations

We have two equations: 1. \( 5x + 2y = -4 \) 2. \( 5x - 3y = 6 \). To solve using the elimination-by-addition method, we need to eliminate one variable by adding the equations. Notice that the coefficients of \( x \) are the same in both equations.

Step 2: Subtract the Equations

Subtract the second equation from the first: \[ (5x + 2y) - (5x - 3y) = -4 - 6 \]This simplifies to eliminating the \( x \)-terms:\( 2y + 3y = -10 \) which becomes \( 5y = -10 \).

Step 3: Solve for \( y \)

Divide both sides of the equation \( 5y = -10 \) by 5 to solve for \( y \):\[ y = \frac{-10}{5} = -2 \].

Step 4: Substitute \( y \) Into One Equation

Substitute \( y = -2 \) back into the first equation \( 5x + 2y = -4 \): \[ 5x + 2(-2) = -4 \]This simplifies to:\[ 5x - 4 = -4 \]

Step 5: Solve for \( x \)

Solve for \( x \) by adding 4 to both sides of the equation:\[ 5x = 0 \]Now, divide by 5:\[ x = 0 \]

Step 6: Final Solution

The solution to the system of equations is \( x = 0 \) and \( y = -2 \). This means the point (0, -2) satisfies both equations.

Key Concepts

Solving Systems of Equations

Solving systems of equations involves finding values for variables that satisfy more than one equation simultaneously. When dealing with systems of equations, it means you have two or more equations working together. In our example, the two equations are:

• \( 5x + 2y = -4 \)
• \( 5x - 3y = 6 \)

Our goal is to find a pair \((x, y)\) that works in both equations at the same time. There are several methods to solve systems of equations, including graphing, substitution, and elimination. The elimination method, which we use here, helps in removing one of the variables to simplify our calculations and make it easier to find the solution for the remaining variable. Remember, the solution to a system of equations in two variables is a point represented by \((x, y)\), which lies on the lines of both equations.

Linear Equations

Linear equations form the backbone of many algebraic concepts. In simple words, a linear equation is an equation involving only first-degree terms (no squares, cubes, etc.). They typically appear in the general format of \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

• Linear equations represent straight lines when graphed on a coordinate plane.
• They are the simplest form of equations and a crucial part of algebraic studies.

In our example, both \(5x + 2y = -4\) and \(5x - 3y = 6\) are linear equations. Because both have the same coefficient for \(x\), it presents a perfect case to apply the elimination method effectively. Solving these equations means finding where these lines intersect, giving us the exact \((x, y)\) point that satisfies both equations.

Algebraic Techniques

To deal with systems of equations using algebraic techniques, especially elimination, there are systematic steps we should follow. The aim here is to eliminate one of the variables by making their coefficients equal so that when the equations are added or subtracted, one variable disappears.**Key steps in the elimination method include:** - Aligning the equations vertically to spot variables easily.- Deciding which variable to eliminate, ideally looking for coefficients that match or can be quickly made to match. In our solution:

• We have the same coefficient for \(x\) in both equations, which allows us to directly subtract one equation from the other.
• This cancellation technique leads us to the equation \(5y = -10\) after the subtraction, making it straightforward to solve for \(y\).
• Once \(y\) is known, substituting back into one of the original equations solves for \(x\).

Using algebraic elimination efficiently streamlines the process and opens pathways to solve more complex systems of equations.