Problem 4 - Understanding Elementary Algebra with Geometry

Question

$$\text { In Exercises } 1-14, \text { solve the system of equations using the elimination method.}$$ $$\left\\{\begin{aligned} -x+3 y &=8 \\ x-2 y &=-6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \((-2, 2)\).

Step 1: Write down the system of equations

The system of equations is:\[-x + 3y = 8 ewline x - 2y = -6\]

Step 2: Add the equations

To eliminate one of the variables, add the two equations together:\[(-x + 3y) + (x - 2y) = 8 + (-6)\]Simplify to combine like terms:\[(-x + x) + (3y - 2y) = 8 - 6\]This simplifies to:\[y = 2\]

Step 3: Substitute y back into one of the original equations

We now substitute \(y = 2\) into one of the original equations. Using the second equation:\[x - 2(2) = -6\]Simplify and solve for \(x\):\[x - 4 = -6 ewline x = -6 + 4 ewline x = -2\]

Step 4: Write the solution

The solution to the system of equations is \(x = -2\) and \(y = 2\). Thus, the solution is \((-2, 2)\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. In this case, the given system has two equations and two variables, x and y. The goal is to find the values of these variables that satisfy both equations simultaneously. Systems of equations can be solved using various methods, including substitution, elimination, and graphing. Each method has its own advantages depending upon the specific system you are trying to solve. Here, we will focus on the elimination method, which is efficient for systems where it is easy to cancel out a variable by adding or subtracting equations.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. In our problem, we start by writing down the given system: \[ \begin{aligned} -x + 3y &= 8 \ x - 2y &= -6 \end{aligned} \] To eliminate one of the variables, we add the two equations. By doing this, we aim to cancel out one of the variables, making it simpler to solve. When we add \( -x + 3y = 8 \) and \( x - 2y = -6 \), the 'x' terms cancel out: \[ (-x + 3y) + (x - 2y) = 8 + (-6) \ \ -x + x + 3y - 2y = 2 \ y = 2 \] With y found, we substitute back into one of the original equations to solve for x: \[ x - 2(2) = -6 \ x - 4 = -6 \ x = -2 \] Thus, the solution is \( x = -2 \) and \( y = 2 \), or \((-2, 2)\).

Substitution Method

The substitution method is another way to solve systems of equations. This method involves solving one of the equations for one variable and then substituting that expression into the other equation. This can often be simpler than the elimination method, especially if one equation is already solved for a variable. Though we didn't use the substitution method for our problem, let's see how it would work here. First, solve the second equation for \( x \): \( x - 2y = -6 \) Solving for x gives: \[ x = -6 + 2y \] Substitute this into the first equation: \[ -(-6 + 2y) + 3y = 8 \ 6 - 2y + 3y = 8 \ y = 2 \] With y found, substitute back to find x: \ x = -6 + 2(2) = -2 \ So, \( x = -2 \) and \( y = 2 \), or \( (-2, 2) \) again. This method, like elimination, helps in accurately finding the solution by systematically eliminating variables.