Problem 4 - Elementary Algebra Within Reach

Question

In Exercises 1-4, solve the system by the method of substitution. $$ \left\\{\begin{array}{r} x-y=0 \\ 5 x-3 y=10 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 5, y = 5\)

Step 1: Solve the First Equation

Isolate one variable in the first equation. Here, rearrange the first equation \(x - y = 0\) to solve for \(x\). This gives: \(x = y\)

Step 2: Substitute into the Second Equation

Substitute the expression of \(x = y\) from the first equation into the second equation. This would give us: \(5y - 3y = 10\)

Step 3: Solve for y

Solving the above equation gives: \(2y = 10\). Dividing both sides by 2, we then get \(y = 5\)

Step 4: Substitute y value back into the first equation

Substitute \(y = 5\) back into the first equation \(x = y\) and solve for \(x\). This gives: \(x = 5\)

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves expressing one variable in terms of another variable, then substituting this expression into another equation. This reduces the number of variables you need to solve for.

The first step in the substitution method is to choose an equation and solve for one variable. In some cases, one equation may already be conveniently isolated, like in the equation from our exercise: \(x - y = 0\). Here, it's easy to express \(x\) in terms of \(y\), so you can quickly write \(x = y\).

Once you have expressed one variable in terms of the other, substitute this expression into the other equation in the system. This creates a single-variable equation, which is much easier to solve. In our given problem, substituting \(x = y\) into \(5x - 3y = 10\) simplifies the equation to \(5y - 3y = 10\).

After substituting, solve the resulting single-variable equation to find the value of the variable. Using the substitution method effectively can greatly simplify the process of finding the solution to a system of equations.

Solving Linear Equations

Linear equations are foundational in algebra. They are equations that graph as straight lines and have variables raised to the power of one. Solving linear equations is a common skill needed not just in mathematics, but in many applied fields. Understanding how to solve them is critical for working with systems of equations.

In practice, solving a linear equation involves isolating the variable on one side of the equation. This can be done through a variety of operations such as adding, subtracting, multiplying, or dividing both sides by a constant.

For example, in the equation from our exercise \(2y = 10\), isolating \(y\) by dividing both sides of the equation by 2 gives \(y = 5\). Once the variable is isolated, it reveals its value, which can be substituted back into related equations within the system to find other variable values. This step is crucial for confirming the compatibility and accuracy of the solution.

Elementary Algebra

Elementary algebra is what you can think of as the basics of algebra. It includes the study of simple equations and relationships between numbers and variables.

Algebraic concepts such as solving for variables, substitution, and working with linear equations are all part of elementary algebra. These are the building blocks that students must understand in order to tackle more complex math problems.

In our exercise, the goal was to solve a system of linear equations using substitution. This process relied heavily on elementary algebra concepts like expressing one variable in terms of another, rearranging equations, and performing basic arithmetic operations.

• Identifying equations and understanding their structures.
• Applying correct operations to simplify and solve equations.
• Incorporating solutions back into original equations to validate results.

Mastering these basic tasks allows students to progress in algebra and develop their problem-solving skills in more advanced topics.