Problem 31 - Understanding Elementary Algebra with Geometry

Question

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{l} 6 x+y=6 \\ 4 x+1=y \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{2} \) and \( y = 3 \).

Step 1: Rewrite the second equation

Start by identifying the equations given. The system of equations is: 1) 6x + y = 6 2) 4x + 1 = y Rewrite the second equation to make y the subject: \( y = 4x + 1 \).

Step 2: Substitute y in the first equation

Now substitute \( y = 4x + 1 \) into the first equation: \( 6x + (4x + 1) = 6 \).

Step 3: Simplify and solve for x

Combine like terms and solve for x: \( 6x + 4x + 1 = 6 \) \( 10x + 1 = 6 \) Subtract 1 from both sides: \( 10x = 5 \) Divide both sides by 10: \( x = \frac{1}{2} \).

Step 4: Substitute x back into the expression for y

Substitute \( x = \frac{1}{2} \) back into the expression \( y = 4x + 1 \): \( y = 4 \times \frac{1}{2} + 1 \) \( y = 2 + 1 \) \( y = 3 \).

Step 5: Write the solution

The solution to the system of equations is \( x = \frac{1}{2} \) and \( y = 3 \).

Key Concepts

solving linear equations

Solving linear equations is a fundamental concept in algebra. Here, we work with equations where each variable is raised to the power of one.

To solve a system of linear equations, we need to find the values of the variables that make both equations true at the same time. In our exercise, we can use different methods like graphing, substitution, and elimination. For simplicity, we'll focus on the substitution method in this context. The ultimate goal: isolate the variables and find their values.

Consider the system:
\[ 6x + y = 6 \]
\[ 4x + 1 = y \]
Following a systematic approach helps you avoid mistakes and better understand the underlying concepts.

substitution method

The substitution method involves substituting one equation into another. This helps reduce the system to a single equation with only one variable.

### Breaking It Down
1. **Solve one equation for one variable**: Start by rearranging one of the equations to solve for one of the variables. In our example, we rearrange the second equation:
\[ y = 4x + 1 \]
2. **Substitute the expression**: Next, substitute this expression into the other equation. This creates an equation with only one variable:
\[ 6x + (4x + 1) = 6 \]
3. **Solve for the variable**: Simplify the expression to solve for x:
\[ 10x + 1 = 6 \] \[ 10x = 5 \] \[ x = \frac{1}{2} \]
4. **Find the other variable**: Substitute the found value back into the first solved equation:
\[ y = 4 \times \frac{1}{2} + 1 = 2 + 1 = 3 \]
Following these steps ensures you accurately find the variables' values.

algebraic simplification

Algebraic simplification is all about making an equation as straightforward as possible. This includes combining like terms and simplifying both sides of an equation.

### Steps for Simplification
* **Combine Like Terms:** Bring together terms containing the same variable or constants. From our problem:
\[ 6x + 4x + 1 = 6 \] simplifies to:
\[ 10x + 1 = 6 \]
* **Isolate the Variable:** Perform operations to isolate the variable on one side:
\[ 10x = 5 \] by subtracting 1 from both sides
\[ x = \frac{1}{2} \] by dividing both sides by 10

Practicing these steps boosts understanding and confidence in tackling various algebraic problems.

Remember, the key is to perform identical operations on both sides to keep the equation balanced. This helps in maintaining the equality and eventually finding the correct values of variables.