Problem 26 - A Graphical Approach to College Algebra

Question

Solve each system by substitution. $$\begin{aligned}0.6 x-0.2 y &=2 \\\\-1.2 x+0.4 y &=3\end{aligned}$$

Step-by-Step Solution

Answer

The system has no solution.

Step 1: Identify the More Suitable Equation for Substitution

Look for an equation that can be easily solved for one variable. In the given system: \[0.6x - 0.2y = 2\] and \[-1.2x + 0.4y = 3\], the first equation \(0.6x - 0.2y = 2\) is simpler to manipulate.

Step 2: Solve the First Equation for One Variable

Solve the first equation for \(x\):\[\begin{align*}0.6x - 0.2y &= 2 \0.6x &= 0.2y + 2 \x &= \frac{0.2y + 2}{0.6}\end{align*}\] After simplifying, we have:\[x = \frac{1}{3}y + \frac{10}{3}\]

Step 3: Substitute the Expression into the Second Equation

Substitute \(x = \frac{1}{3}y + \frac{10}{3}\) into the second equation:\[-1.2\left( \frac{1}{3}y + \frac{10}{3} \right) + 0.4y = 3\]

Step 4: Simplify the Equation and Solve for \(y\)

Expand and simplify:\[-0.4y - 4 + 0.4y = 3\]After simplifying further, \[-4 = 3\]This is a contradiction, indicating that the system of equations has no solution.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations. It's especially useful when one of the equations is easily solved for one variable. Here's how it works:

• Identify one of the equations that is easy to manipulate. Typically, look for equations where a variable has a coefficient of 1 or can be simplified easily.
• Solve this equation for one of the variables in terms of the other variable. This means you'll express one variable as a formula involving the other variable.
• Substitute the expression found into the other equation. This will replace the chosen variable with the expression, and you'll end up with an equation in one variable.

This method swaps one variable from the equation, making it easier to solve. In the exercise, we used the first equation to express \( x \) in terms of \( y \), allowing us to plug this back into the second equation.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate a particular variable. This step is sometimes necessary to simplify expressions or to enable substitution. In our exercise:

• Starting with \( 0.6x - 0.2y = 2 \), we rearranged terms to isolate \( x \).
• We brought terms involving \( y \) to one side and constants to the other side.
• Next, divide each term by the coefficient of \( x \) to solve for \( x \).

Our aim during manipulation is to make the equations easier to understand or simpler to solve. Correct manipulation helps prevent errors and ensures the correct steps are taken further in the problem-solving process. This approach led us to the expression \( x = \frac{1}{3}y + \frac{10}{3} \), which was crucial for substitution.

Solving Linear Equations

Solving linear equations involves finding the values of the variables that satisfy all equations in the system simultaneously. Here’s a general approach:

• Once we have a single equation in one variable after substitution, our goal is to simplify and solve it.
• In our example, after substituting \( x \) into the second equation, we ended up with an equation involving only \( y \).
• Attempt to combine like terms and reduce the equation to its simplest form.

Sometimes you may find a contradiction, such as \( -4 = 3 \). This indicates that there are no solutions, meaning the lines representing the equations are parallel and never intersect. Recognizing when a system is inconsistent is also a critical part of solving linear equations.