Problem 5 - Understanding Elementary Algebra with Geometry

Question

$$\text { In Exercises } 1-14, \text { solve the system of equations using the elimination method.}$$ $$\left\\{\begin{array}{l} 7 x+2 y-15=0 \\ 3 x-2 y+5=0 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((1, 4)\).

Step 1: Write down the given system of equations

The system of equations is: \(7x + 2y - 15 = 0\) and \(3x - 2y + 5 = 0\)

Step 2: Add the equations to eliminate y

Add the two equations to eliminate \(y\). Let's add: \(7x + 2y - 15 = 0\) and \(3x - 2y + 5 = 0\). This results in: \(7x + 2y - 15 + 3x - 2y + 5 = 0\). Simplifying, we get: \(10x - 10 = 0\).

Step 3: Solve for x

From \(10x - 10 = 0\), add 10 to both sides to get \(10x = 10\). Then, divide both sides by 10 to get \(x = 1\).

Step 4: Substitute x back into either original equation to solve for y

Substitute \(x = 1\) into the first equation: \(7(1) + 2y - 15 = 0\). This simplifies to \(7 + 2y - 15 = 0\). Solving for \(y\), we get: \(2y - 8 = 0\), thus \(2y = 8\), and \(y = 4\).

Step 5: Write the solution as an ordered pair

The solution to the system of equations is \((x, y) = (1, 4)\).

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. In this case, we have two equations involving the variables \(x\) and \(y\). The goal is to find values for these variables that satisfy both equations simultaneously.

When dealing with systems of equations, we use various methods to find the solution. The solution can be:

• One unique solution
• No solution
• Infinitely many solutions

The solution is represented as an ordered pair \((x, y)\). In our exercise, our system of equations is:

\[ \begin{array}{l} 7x + 2y - 15 = 0 \ 3x - 2y + 5 = 0 \end{array} \]

Approaching systems of equations involves understanding some core methods for solving them. One common method is the elimination method, which simplifies the system by removing one variable.

Solving Linear Equations

Linear equations are equations of the first degree, meaning they have the highest exponent of the variable as 1. For example, in the equations \(7x + 2y - 15 = 0\) and \(3x - 2y + 5 = 0\), both are linear equations.

To solve linear equations, follow these steps:

• Isolate the variable you want to solve for.
• Simplify the equation by combining like terms.
• Use basic algebraic operations like addition, subtraction, multiplication, or division to solve for the variable.

In the given problem, once we have eliminated one variable, the remaining equation is linear. From our equations, we added them to eliminate \(y\) and got:

\[10x - 10 = 0\]

Solving this, we first added 10 to both sides, resulting in:

\[10x = 10\]

Then, we divided both sides by 10 to isolate \(x\):

\[ x = 1 \]

Algebraic Methods

Algebraic methods are techniques used to manipulate and solve equations. The elimination method is one such algebraic method for solving systems of equations.

Here’s a quick guide to using the elimination method:

• Step 1: Arrange the equations neatly, making sure similar terms align vertically.
• Step 2: Add or subtract the equations to eliminate one of the variables.
• Step 3: Solve the resulting equation for the remaining variable.
• Step 4: Substitute this value back into one of the original equations to find the other variable.
• Step 5: Write the solution as an ordered pair \((x, y)\).

Using our exercise, in Step 2, we added the equations:

\[ 7x + 2y - 15 = 0 \]
\[ + 3x - 2y + 5 = 0 \]
Adding these, we eliminated \(y\) to get:

\[ 10x - 10 = 0 \]

We solved for \(x\) and then substituted back to get \(y\). Thus, the solution to the system is \( (1, 4) \).

Understanding algebraic methods, like elimination, greatly enhances problem-solving abilities in algebra and beyond.