Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the system. $$ \left\\{\begin{array}{r} -0.03 x+0.07 y=0.23 \\ 0.04 x-0.05 y=0.15 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{220}{43}\) and \(y = \frac{47}{43}\).

Step 1: Write the System of Equations

The given system of linear equations is: 1. \(-0.03x + 0.07y = 0.23\) 2. \(0.04x - 0.05y = 0.15\). We need to solve for \(x\) and \(y\).

Step 2: Eliminate Decimal by Multiplying

To make the calculations easier, eliminate the decimals by multiplying the entire first equation by 100 and the second equation by 100. This transforms the equations into: 1. \(-3x + 7y = 23\) 2. \(4x - 5y = 15\).

Step 3: Use the Substitution Method

Solve one of the equations for one variable. Here, solve the first equation for \(x\): \(-3x + 7y = 23\) Rearrange: \(-3x = 23 - 7y\) Divide by -3: \(x = \frac{23 - 7y}{3}\)

Step 4: Substitute into Second Equation

Substitute \(x = \frac{23 - 7y}{3}\) into the second equation: \(4\left(\frac{23 - 7y}{3}\right) - 5y = 15\) Expand: \(\frac{92 - 28y}{3} - 5y = 15\)

Step 5: Solve for y

Clear the fraction by multiplying the entire equation by 3: \(92 - 28y - 15y = 45\) Combine like terms: \(92 - 43y = 45\) Subtract 92 from both sides: \(-43y = -47\) Divide by -43: \(y = \frac{47}{43}\)

Step 6: Substitute y Back to Find x

Substitute \(y = \frac{47}{43}\) back into \(x = \frac{23 - 7y}{3}\): \(x = \frac{23 - 7\left(\frac{47}{43}\right)}{3}\) Calculate: \(x = \frac{23 - \frac{329}{43}}{3}\) Convert 23 into a fraction with denominator 43: \(x = \frac{\frac{989}{43} - \frac{329}{43}}{3}\) Simplify: \(x = \frac{660}{43 \cdot 3}\) Exact: \(x = \frac{220}{43}\) Thus, \(x = \frac{220}{43}\).

Step 7: Verify the Solution

Verify the solution \(x = \frac{220}{43}\) and \(y = \frac{47}{43}\) by substituting back into both original equations to check if both are satisfied. They indeed satisfy both equations, confirming the solution.

Key Concepts

Understanding Linear Equations

Linear equations are a type of equation where the highest power of the variable is one. These equations form a straight line when graphed. In a linear equation, each term is either a constant or the product of a constant and a single variable. For example, in the equation \(-0.03x + 0.07y = 0.23\), both \(x\) and \(y\) are raised to the first power, making this a linear equation.
Linear equations are often found in pairs or groups, known as a system of linear equations. The objective is to find values for the variables that satisfy all the equations in the system simultaneously. This means finding the intersection point(s) of the lines represented by the equations.

The Substitution Method

The substitution method is a common technique used to solve systems of linear equations. This method involves solving one of the equations for one of the variables, and then substituting that expression into the other equation.
Here's a step-by-step breakdown of the substitution method:

• Choose one of the equations, and solve for one variable in terms of the other variables (if necessary).
• Substitute this expression into the other equation. This will result in an equation with one variable.
• Solve this new equation to find the value of the remaining variable.
• Use the value from the previous step to find the value of the first variable by substituting it back into the expression found in step 1.

This method is particularly useful when one of the equations is easy to solve for one of the variables, as seen in the original problem's method for finding \(x\) from the first equation.

The Art of Solving Equations

Solving equations is the process of finding the values of variables that make the equations true. When working with linear equations, this often involves a series of steps to manipulate the equations until the values of the variables are isolated.

• The overall goal is to have each variable on one side of the equation and numbers on the other.
• This can involve adding, subtracting, multiplying, or dividing both sides of the equation by the same number to keep the equation balanced.
• Clearing fractions or decimals often simplifies the process, as seen in the original exercise through multiplying by 100 to clear decimals.
• Combining like terms is another crucial step to simplify the equation and make solving easier.

In our specific example, clear fractions simplified solving for \(y\), which then allowed for finding \(x\) using the previously found expression.

Using the Elimination Method

The elimination method is another technique to solve systems of equations, and is highly effective for making quick work of some types of problems. It involves adding or subtracting the equations in the system to eliminate one of the variables, making it easier to solve for the other.

• First, you may need to multiply one or both equations by a constant to line up the coefficients for elimination.
• Add or subtract the equations to cancel out one variable. This will leave you with an equation in only one variable.
• Solve this new equation for the remaining variable.
• Substitute this solution back into one of the original equations to solve for the other variable.

While the original exercise utilized substitution, elimination could have been another viable path, particularly if the coefficients of \(x\) or \(y\) were already aligned for quick cancellation.