Problem 11 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{c} y=\frac{2}{5} x-1 \\ 3 x+5 y=4 \end{array}\right) $$

Step-by-Step Solution

Answer

\( x = \frac{9}{5} \), \( y = \frac{-7}{25} \)

Step 1: Identify the Equations

The system of equations is given as: \( y = \frac{2}{5}x - 1 \) and \( 3x + 5y = 4 \). We will use the substitution method to solve this system of equations.

Step 2: Substitute for y in the Second Equation

Substitute the expression for \( y \) from the first equation into the second equation: \( 3x + 5\left(\frac{2}{5}x - 1\right) = 4 \).

Step 3: Simplify the Substituted Equation

Simplify the equation: \( 3x + 2x - 5 = 4 \).

Step 4: Solve for x

Combine like terms and solve for \( x \): \( 5x - 5 = 4 \) Add 5 to both sides: \( 5x = 9 \) Divide by 5: \( x = \frac{9}{5} \).

Step 5: Substitute Back to Find y

Substitute \( x = \frac{9}{5} \) back into the expression for \( y \): \( y = \frac{2}{5}\left(\frac{9}{5}\right) - 1 \) \( y = \frac{18}{25} - 1 \) Convert 1 to \( \frac{25}{25} \): \( y = \frac{18}{25} - \frac{25}{25} \).

Step 6: Simplify to Find y

Simplify the fraction: \( y = \frac{18 - 25}{25} = \frac{-7}{25} \).

Step 7: State the Solution

The solution to the system of equations is \( x = \frac{9}{5} \) and \( y = \frac{-7}{25} \).

Key Concepts

Systems of Equations

Think of a system of equations as a set of mathematical speaking partners. These partners are equations that relate with each other by sharing variables. Each equation gives us information about the variables and helps us find the solution when they work together. For instance, in this exercise, we have two equations:

• First Equation: \( y = \frac{2}{5}x - 1 \)
• Second Equation: \( 3x + 5y = 4 \)

Both these equations have the same variables, namely \( x \) and \( y \). The goal is to find values of \( x \) and \( y \) that make both equations true at the same time. Solving systems of equations is a fundamental part of algebra, and it is applied to real-world problems where two or more conditions must be satisfied together.

Solving Linear Equations

Solving linear equations involves finding the value of the variable that makes the equation true. When linear equations are part of a system, it can be a bit more complex because we're dealing with more than one equation to solve for the unknowns. In the substitution method, we solve one equation for one variable and use this expression to replace the variable in another equation. In our example, we first solved the equation \( y = \frac{2}{5}x - 1 \) to express \( y \) in terms of \( x \). Then, substituting this expression in the second equation simplifies the problem to one equation in terms of just one variable. This step is crucial because it reduces the system into something easier to manage: a single linear equation. Let's see how this works:

• By substituting, we get: \( 3x + 5\left(\frac{2}{5}x - 1\right) = 4 \)
• Simplifying gives us: \( 3x + 2x - 5 = 4 \)
• This leads to finding \( x \) easily.

Once \( x \) is found, we plug it back into our expression for \( y \) to find its value, solving the entire system.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to solve for variables. It involves operations such as addition, subtraction, multiplication, and division applied systematically to isolate the variables.In our example, after substituting \( y \) in the second equation, the main focus is to combine like terms and simplify step by step:

• With \( 3x + 2x - 5 = 4 \), first combine like terms to simplify this to \( 5x - 5 = 4 \).
• Next, add 5 to both sides to isolate terms involving \( x \): \( 5x = 9 \).
• Finally, divide both sides by 5 to solve for \( x \): \( x = \frac{9}{5} \).

Once \( x \) is identified, we substitute it back into the equation for \( y \) to find the exact value of \( y \). Converting fractions, as shown when finding \( y \), is also part of algebraic manipulation. Each step in algebraic manipulation draws us closer to finding the solution by making the expression simpler and more manageable.