Problem 23 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{rr} 4 x-5 y= & 3 \\ 8 x+15 y= & -24 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{3}{4}\) and \(y = -\frac{6}{5}\).

Step 1: Isolate Variable for Substitution

Start with the first equation: \(4x - 5y = 3\). We will solve for \(x\) by isolating it. This can be done by adding \(5y\) to both sides and then dividing by 4. Hence, \(x = \frac{5y + 3}{4}\).

Step 2: Substitute Back into Second Equation

Take the expression for \(x\) from Step 1 and substitute it into the second equation: \(8x + 15y = -24\). Substituting for \(x\), we get: \(8(\frac{5y + 3}{4}) + 15y = -24\).

Step 3: Simplify Substituted Equation

Simplify the equation from Step 2. Multiply through by 8 to clear the fraction: \(10y + 6 + 15y = -24\). Combine like terms to get: \(25y + 6 = -24\).

Step 4: Solve for \(y\)

Isolate \(y\) in the simplified equation. Subtract 6 from both sides to find \(25y = -30\). Divide by 25 to find \(y = -\frac{6}{5}\).

Step 5: Substitute \(y\) Value Back to Find \(x\)

Using the expression for \(x\) from Step 1, substitute \(y = -\frac{6}{5}\) into it: \(x = \frac{5(-\frac{6}{5}) + 3}{4}\). Simplify: \(x = \frac{-6 + 3}{4}\), which gives \(x = -\frac{3}{4}\).

Step 6: Solution to the System

The solution to the system of equations is \(x = -\frac{3}{4}\) and \(y = -\frac{6}{5}\).

Key Concepts

System of Equations

When faced with a system of equations, we're dealing with two or more equations that have several unknowns. The goal is to find a solution that satisfies all equations simultaneously. In our exercise, we have a system of two linear equations in two variables:

• Equation 1: \(4x - 5y = 3\)
• Equation 2: \(8x + 15y = -24\)

These equations represent straight lines on a coordinate plane. Solving the system means finding the point where these lines intersect. In this case, we use the substitution method to solve it effectively. This method involves isolating one variable in one of the equations and substituting this expression into the other equation, simplifying the process of finding both variables and gaining an intersection point in the system.

Linear Equations

Linear equations, like those in our example, are equations of the first degree. This means that the variables are raised only to the power of one. They appear in the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.

• These equations often represent straight lines when graphed.
• The slope and intercept can be used to determine their representation on a coordinate plane.

In our problem, substituting and simplifying them leads us to find specific numerical solutions for \(x\) and \(y\). These solutions give the exact point of intersection for the two lines represented by the equations. Therefore, linear equations are fundamental in forming the framework for solving systems of equations and are used in various real-life applications, from physics to economics.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. It is a unifying thread in almost all of mathematics. Learning algebra allows us to solve equations, which is exactly what we did in this system of equations.

In our exercise, we used algebraic techniques like substitution and simplification:

• We isolated variables, allowing us to express one variable in terms of others.
• Algebra lent us the 'power' to solve for unknowns sequentially and systematically.

These fundamental skills learned in algebra are the bedrock upon which more complex mathematical concepts are built. They allow us to not only understand equations but also to apply these concepts to different problems and situations, thereby solving real-world issues effectively.