Problem 73 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} x-2 y=-5 \\ 2 x-3 y=-4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = 7\), \(y = 6\).

Step 1: Solve one equation for one variable

Start with the first equation: \(x - 2y = -5\). Solve for \(x\): \(x = 2y - 5\).

Step 2: Substitute the solved expression into the second equation

Substitute \(x = 2y - 5\) into the second equation: \(2(2y - 5) - 3y = -4\).

Step 3: Solve for the remaining variable

Distribute and combine like terms: \(4y - 10 - 3y = -4\). Simplify to solve for \(y\): \(y - 10 = -4\), hence \(y = 6\).

Step 4: Substitute back to find the other variable

Use the value of \(y = 6\) in the expression from Step 1: \(x = 2(6) - 5\). Simplify to find \(x\): \(x = 12 - 5 = 7\).

Step 5: State the solution

The solution to the system of equations is \(x = 7\) and \(y = 6\).

Key Concepts

Systems of Equations

A system of equations consists of multiple equations that share common variables. In this exercise, we are working with two linear equations:
\ \[ \begin{array}{l} x - 2y = -5 \ 2x - 3y = -4 \ \end{array} \]
Our goal is to find values for the variables that satisfy both equations simultaneously. There are different methods to solve systems of equations, such as graphing, elimination, and substitution. For this task, we'll focus on the substitution method. This method involves substituting one variable's expression from one equation into the other equation.

Algebraic Substitution

The substitution method is a powerful tool for solving systems of equations. In this context, 'substitution' means replacing one variable with an algebraic expression derived from the other equation. Here are the steps we will follow:

• Step 1: Solve one of the equations for one variable.
• Step 2: Substitute this expression into the other equation.
• Step 3: Solve the resulting equation for the remaining variable.
• Step 4: Substitute back to find the initial variable.

Let's break these steps down with our given system.
First, solve the equation \( x - 2y = -5 \) for \( x \): \( x = 2y - 5 \).
Then substitute \( x = 2y - 5 \) into the second equation: \( 2(2y - 5) - 3y = -4 \).

Solving Linear Equations

After substitution, we need to solve the resulting linear equation. Let's continue with our current system:
Substitute \( x = 2y - 5 \) to get \( 2(2y - 5) - 3y = -4 \).
Distribute and simplify: \( 4y - 10 - 3y = -4 \).
Combine like terms: \( y - 10 = -4 \).
Solve for \( y \): \( y = 6 \).
Now we have the value of one variable.
Next, substitute \( y = 6 \) back into \( x = 2y - 5 \) to find \( x \): \( x = 2(6) - 5 \).
Simplify: \( x = 12 - 5 \), so \( x = 7 \).
We have now found the solution to our system of equations: \( x = 7 \), \( y = 6 \). Solving linear equations by substitution requires careful algebraic manipulation and attention to detail.
The key is to methodically follow each step and double-check your work for accuracy.