Problem 53 - College Algebra with Modeling and Visualization

Question

If possible, solve the system of linear equations and check your answer. $$ \begin{aligned} &0.2 x-0.1 y=0.5\\\ &0.4 x+0.3 y=2.5 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = 4\) and \(y = 3\).

Step 1: Simplify the coefficients

To make calculations easier, eliminate decimals by multiplying the entire first equation by 10 and the second equation by 10 as well. This gives us:\[\begin{aligned}&2x - y = 5 \&4x + 3y = 25\end{aligned}\]

Step 2: Eliminate one variable

We will eliminate the variable \(y\) by multiplying the entire first equation by 3 to align the \(y\) terms:\[\begin{aligned}&3(2x - y) = 3(5) \&4x + 3y = 25\end{aligned}\]This results in:\[\begin{aligned}&6x - 3y = 15 \&4x + 3y = 25\end{aligned}\]

Step 3: Add the equations

Now add the two equations:\[\begin{aligned}&(6x - 3y) + (4x + 3y) = 15 + 25\end{aligned}\]The \(y\) terms cancel out:\[10x = 40\]

Step 4: Solve for x

Divide both sides by 10 to isolate \(x\):\[x = 4\]

Step 5: Substitute x back to find y

Substitute \(x = 4\) into the first simplified equation:\[2(4) - y = 5\]This simplifies to:\[8 - y = 5\]Solve for \(y\):\[y = 8 - 5 = 3\]

Step 6: Check the solution

Substitute \(x = 4\) and \(y = 3\) back into the original equations to verify:- For the first equation: \(0.2(4) - 0.1(3) = 0.8 - 0.3 = 0.5\), which is correct.- For the second equation: \(0.4(4) + 0.3(3) = 1.6 + 0.9 = 2.5\), which is also correct. Both solutions satisfy the original system of equations.

Key Concepts

System of Equations

A system of equations consists of two or more equations with multiple variables. In our case, we are dealing with a system of linear equations involving two variables, \(x\) and \(y\). These equations are solved simultaneously, meaning we need to find values for \(x\) and \(y\) that satisfy both equations at the same time.

The given system of equations is:
\[\begin{aligned} &0.2x - 0.1y = 0.5\&0.4x + 0.3y = 2.5\end{aligned} \]

This system represents two lines on a graph, and the solution to the system is the point where the lines intersect. Solving the system means finding this intersection point by determining the values of \(x\) and \(y\). When solving such systems, it's important first to consider simplifying and manipulating the equations to make solving easier.

Variable Elimination

Variable elimination is a key strategy in solving systems of equations, especially when solving by algebraic methods such as substitution or elimination.

In the step-by-step solution, we choose to eliminate variable \(y\). The art of elimination requires aligning either the \(x\) or \(y\) coefficients, making them opposites so they cancel each other when added.

Here's how we proceed with our equations:

• We have the simplified equations: \(2x - y = 5\) and \(4x + 3y = 25\).
• Multiply the first equation by 3 to align the \(y\) coefficients: \(6x - 3y = 15\).
• Now add this to the second equation, \(4x + 3y = 25\).

The goal is to have the \(y\) terms cancel out, which allows further simplification. As a result, the \(y\) variable is eliminated, leaving us with a single equation in terms of \(x\): \(10x = 40\).

After finding \(x\), we substitute it back into one of the original equations to solve for \(y\). This two-step approach makes solving systems of equations more manageable and less error-prone.

Equation Simplification

Simplifying equations is a crucial early step in solving any system of equations, particularly when the coefficients are decimals, as they complicate calculations.

For our system, the first line of attack was to eliminate the decimals:

• Each equation was multiplied by 10. This moves the decimal point, transforming the equations into \(2x - y = 5\) and \(4x + 3y = 25\).
• This transformation produces integer coefficients, simplifying the math operations significantly.

By doing this, we also prevent potential calculation errors often introduced by handling decimals directly.

Equation simplification is about more than just making the numbers larger. It's about maintaining the relationships between variables while adjusting the form of the equations for easier manipulation. We want to work with simpler numbers while preserving the system's original properties. This is a vital skill and can greatly improve efficiency and accuracy when dealing with more complex systems of equations.