Problem 11 - Elementary Algebra Within Reach

Question

In Exercises 5-14, solve the system by the method of substitution. $$ \left\\{\begin{array}{l} 2 x-y=-2 \\ 4 x+y=5 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution of the system of equations is x = 0.5, y = 3

Step 1: Solve one equation for one variable

From the first equation, we can isolate \(y\). \[2x - y = -2\] => \[y = 2x + 2\]

Step 2: Substitute the expression of \(y\) into the second equation

Now we substitute \(y = 2x + 2\) into the second equation:\[4x + (2x + 2) = 5\], which simplifies to:\[6x + 2 = 5\]

Step 3: Solve the resulting equation for \(x\)

Subtract 2 from both sides:\[6x = 3\] Divide by 6 to solve for \(x\):\[x = 0.5\]

Step 4: Substitute \(x\) into the equation for \(y\)

Finally, substitute \(x = 0.5\) into the equation \(y = 2x + 2\):\[y = 2 * 0.5 + 2 = 3\]

Key Concepts

Substitution Method

The substitution method is a common technique used to solve systems of equations. It involves solving one of the equations for one variable first. Once you have expressed one variable in terms of the other, substitute the resulting expression into the other equation.
This method is particularly useful when one of the equations is easy to manipulate, making it simpler to isolate one variable.Here's a quick overview of the steps:

• First, solve one of the equations for one variable. In our example, we solved the first equation to isolate \(y\), giving us \(y = 2x + 2\).
• Next, substitute this expression into the other equation. This reduces the system of equations to a single equation with one variable, which is easier to solve.
• Finally, solve the resulting equation for the remaining variable, and substitute it back into the equation used earlier to find the value of the other variable.

This method is efficient when equations are simple to re-arrange and substitute, streamlining the path to find the solution.

Elementary Algebra

Elementary algebra is the most fundamental part of algebra, teaching basic principles that students need to understand more complex concepts later on. It provides the groundwork for solving equations like those in our exercise.
In elementary algebra, you'll engage with:

• Variables, which are symbols used to represent unknown numbers.
• Expressions, which are combinations of variables and numbers formed using addition, subtraction, multiplication, and division.
• Equations, where two expressions are set equal, allowing us to find the value of unknown variables.

Solving equations in elementary algebra typically includes combining like terms, using arithmetic operations to isolate variables, and applying the distributive property to simplify expressions. For instance, in our exercise, you'd begin by isolating one of the variables to make substitution easier, which is a standard skill developed through practice with this branch of algebra.

Linear Equations

Linear equations are mathematical statements that show equality between expressions, and they involve constant rates of change. They form a straight line when graphed on a coordinate plane.
The general format for a linear equation in two variables is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants:

• If you have two linear equations, they create what is known as a 'system of equations', like in our example system.
• These equations depict lines that can intersect, be parallel, or be the same line, and solving them involves finding points (if they exist) where the two equations are true simultaneously, which we refer to as their intersection point.
• In the given exercise, solving using the substitution method revealed the intersection at \((0.5, 3)\), which is the solution to the system.

Understanding linear equations and their properties is crucial because they frequently appear in various real-world applications, making them essential concepts to master in algebra.