Problem 14 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the system. $$ \left\\{\begin{array}{l} 0.11 x-0.03 y=0.25 \\ 0.12 x+0.05 y=0.70 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x \approx 3.68 \) and \( y \approx 5.16 \).

Step 1: Simplify the equations

Firstly, let's simplify and rewrite the given system of equations for clarity: 1. Equation 1: \( 0.11x - 0.03y = 0.25 \)2. Equation 2: \( 0.12x + 0.05y = 0.70 \).

Step 2: Eliminate one variable

We will eliminate \( y \) by making the coefficients of \( y \) equal in both equations. Multiply Equation 1 by 5 and Equation 2 by 3:- New Equation 1: \( (0.11 \times 5)x - (0.03 \times 5)y = 0.25 \times 5 \Rightarrow 0.55x - 0.15y = 1.25 \)- New Equation 2: \( (0.12 \times 3)x + (0.05 \times 3)y = 0.70 \times 3 \Rightarrow 0.36x + 0.15y = 2.10 \).

Step 3: Add equations to solve for x

Add the simplified equations together to eliminate \( y \): \((0.55x - 0.15y) + (0.36x + 0.15y) = 1.25 + 2.10 \)\(0.55x + 0.36x = 3.35 \)\(0.91x = 3.35 \)Divide both sides by 0.91:\( x = \frac{3.35}{0.91} \approx 3.68 \).

Step 4: Substitute x back to find y

Substitute the value of \( x \) back into either original equation to find \( y \). Using Equation 1: \(0.11 \times 3.68 - 0.03y = 0.25 \)\( 0.4048 - 0.03y = 0.25 \)\(-0.03y = 0.25 - 0.4048 \Rightarrow -0.03y = -0.1548 \)Divide both sides by \(-0.03\):\( y = \frac{-0.1548}{-0.03} \approx 5.16 \).

Step 5: Verify the solution

Substitute \( x = 3.68 \) and \( y = 5.16 \) back into the second equation to verify:\(0.12 \times 3.68 + 0.05 \times 5.16 \approx 0.70 \)\(0.4416 + 0.258 \approx 0.70 \)\(0.6996 \approx 0.70 \)The values satisfy the equation, confirming our solution.

Key Concepts

Elimination Method

The Elimination Method is used to solve systems of linear equations by eliminating one of the variables. This involves manipulating the equations so that when they're added or subtracted, one variable is cancelled out. This allows you to solve for the remaining variable easily. In this case, we eliminated the variable \(y\) from our system.

Here's how it's done:

• First, choose which variable you want to eliminate; we chose \(y\) in this example.
• Next, adjust the equations by multiplying each by a suitable number to make the coefficients of \(y\) the same (but opposite in sign) in both equations. For instance, multiplying the first equation by 5 and the second by 3 accomplished this.
• Then, add or subtract the two equations to eliminate \(y\). In our solution, adding the equations cancelled out \(y\), letting us solve for \(x\).
• Finally, substitute the value of \(x\) back into one of the original equations to find \(y\).

This method is intuitive in that it focuses on simplifying what can sometimes feel like a complex problem. It's particularly useful when the equations are set up nicely, allowing straightforward manipulation.

Substitution Method

The Substitution Method provides another elegant approach to solving systems of linear equations. Instead of trying to eliminate a variable through adding or subtracting equations, we express one variable in terms of the other and substitute it into the second equation.

Here's a breakdown of how it works:

• Begin with one of the given equations, ideally one that is easy to manipulate, and solve it for one of the variables in terms of the other. This transforms the first equation into an expression with one variable.
• Next, substitute this expression into the other equation. This way, the equation now contains only one variable, making it solvable.
• Solve this new equation to find the value of the isolated variable.
• Finally, use this value back in the expression from the first step to solve for the second variable.

Although the original step-by-step solution utilized the Elimination Method, knowing the Substitution Method is beneficial. It provides an alternative strategy and is especially useful in cases where solving or rearranging equations seems more feasible than setting up coefficients for elimination.

Linear Equations

Linear Equations form the backbone of systems solving methods. A linear equation is any equation that can be written in the form \( ax + by = c \), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. These equations graph as straight lines, which is why they are called "linear."

What to know about Linear Equations:

• Each equation represents a line on a Cartesian coordinate system. The solution to a system of linear equations is where those lines intersect, providing a visual way to understand solutions.
• Systems can have a single solution, infinite solutions, or no solution at all. A single intersection between two lines corresponds to a unique solution like in this exercise. Parallel lines never intersect, meaning no solution exists. Identical lines overlap completely, leading to infinite solutions.
• Understanding their algebraic representation is key; they can tell us how changing coefficients alter the slope and position of their graphical line representation.

Solving linear equations requires mastering basic algebraic manipulations and understanding of graphing principles. This understanding is crucial for techniques like the Elimination and Substitution Methods, bringing a strategic approach to interpreting and solving linear systems.