Problem 39 - Elementary Algebra Within Reach

Question

In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{array}{l} 5 x+3 y=15 \\ 2 x-3 y=6 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=3, y=0\).

Step 1: Choose and Resolve One of the Equations

Choose one of the equations to resolve for one of the variables. In this case, it seems easier to choose the second equation and resolve for \(x\), so that \(x = \frac {6+3y} {2}\).

Step 2: Substitute in the Other Equation

Substitute the expression found in Step 1 into the other equation. Thus, the first equation becomes: \(5(\frac {6+3y} {2}) + 3y = 15\), which simplifies to \(15+y=15\).

Step 3: Solve for the Variable

Solving the equation \(15+y=15\) for \(y\), we find that \(y=0\).

Step 4: Substitute \(y\) in the First Equation

Plug \(y=0\) into the first equation to find \(x\); therefore, the first equation becomes: \(5x+3(0)=15\), which simplifies to \(5x=15\).

Step 5: Solve for the Variable \(x\)

Solving the equation \(5x=15\) for \(x\), we find that \(x=3\).

Key Concepts

Substitution Method

The substitution method is one technique used to solve systems of linear equations. It involves expressing one variable in terms of the other from one equation and substituting this expression into the other equation. This method simplifies the system by reducing two equations to one, making it easier to solve.

• Start by choosing one of the equations that seems simplest for isolating a variable.
• Solve this equation for the chosen variable.
• Substitute the expression obtained into the second equation.
• Solve the resulting equation for the other variable.
• Once you have one variable, substitute it back into the expression obtained in the first step to find the value of the second variable.

In our example, we solved the second equation for \(x\), which was convenient because it allowed us to easily substitute it into the first equation. This approach is efficient in ensuring the calculations remain straightforward and manageable.

Solving Algebraic Equations

Solving algebraic equations is a fundamental skill in algebra. It involves finding the values of the unknown variables that satisfy the given equation. The process typically requires patience and a systematic approach.When solving algebraic equations within the substitution method:

• Substitute an expression involving one variable into the other equation, forming a single equation in one variable.
• Combine like terms and simplify the equation, if necessary.
• Use inverse operations—such as addition, subtraction, multiplication, and division—to isolate the variable.

In the example exercise, solving the equation \(15 + y = 15\) for \(y\) gives us \(y = 0\). This value is a critical stepping-stone toward finding the value of \(x\) in the context of a system of equations.

Linear Equations

Linear equations are the backbone of algebra. They describe a straight line when graphed and take the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. The solutions of linear equations are key in many real-world applications ranging from business to engineering.

• Linear equations can have one, infinitely many, or no solutions.
• A system of linear equations is a set of two or more equations with the same variables.
• The aim is to find common values for these variables that satisfy all the equations simultaneously.

In our exercise, the equations \(5x + 3y = 15\) and \(2x - 3y = 6\) form a system where substitution helps find the point \((x, y)\) where the two lines intersect, providing the solution to the system.

Elementary Algebra

Elementary algebra is the foundation of all higher-level math courses. It introduces key concepts such as variables, constants, coefficients, expressions, and equations. Mastery of these basics is essential to tackle more complex topics. Key aspects of elementary algebra include:

• Understanding variables as symbols that represent unknown quantities.
• Learning how to manipulate algebraic expressions through the use of operations and properties of equality.
• Developing strategies to solve different types of equations, including linear equations, as seen in our exercise.

In the exercise, foundational algebraic techniques such as combining like terms and applying inverse operations to solve for variables are demonstrated. This example highlights the step-by-step logic and procedural approach that is vital in elementary algebra.