Problem 40 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 2(2 x+3 y)=5 \\ 8 x=3(1+3 y) \end{array}\right. $$

Step-by-Step Solution

Answer

Solution: \(x = \frac{3}{4}, y = \frac{1}{3}\). The system is consistent and independent.

Step 1: Simplify the Equations

Begin by simplifying the given equations. For the first equation, distribute the 2 to get: \(2(2x + 3y) = 5\) which simplifies to \(4x + 6y = 5\). For the second equation, distribute the 3 to get: \(8x = 3(1 + 3y)\) which simplifies to \(8x = 3 + 9y\).

Step 2: Rearrange the Second Equation

Rearrange the second equation to the standard form by bringing all terms involving variables to one side: \(8x - 9y = 3\). Now, the system of equations is:1. \(4x + 6y = 5\)2. \(8x - 9y = 3\).

Step 3: Use Elimination Method

Choose a method to solve the system. In this case, we'll use the elimination method. Adjust the first equation to match the x-term coefficients of both equations. Multiply the first equation by 2 to get \(8x + 12y = 10\). Then subtract the second equation \(8x - 9y = 3\) from this equation.

Step 4: Subtract the Equations

Perform the subtraction from Step 3:\((8x + 12y) - (8x - 9y) = 10 - 3\)Simplifying yields:\(21y = 7\)Thus, \(y = \frac{7}{21} = \frac{1}{3}\).

Step 5: Substitute y Solution into an Equation

Substitute \(y = \frac{1}{3}\) back into one of the original equations to solve for \(x\). Use \(4x + 6\left(\frac{1}{3}\right) = 5\):\(4x + 2 = 5\)Solving for \(x\), we subtract 2 from both sides:\(4x = 3\)Divide by 4:\(x = \frac{3}{4}\).

Step 6: Verify the Solution

Verify the solution \(x = \frac{3}{4}\), \(y = \frac{1}{3}\) by substituting into both original equations to ensure they are satisfied:1. First equation: \(4\left(\frac{3}{4}\right) + 6\left(\frac{1}{3}\right) = 3 + 2 = 5\), true.2. Second equation: \(8\left(\frac{3}{4}\right) - 9\left(\frac{1}{3}\right) = 6 - 3 = 3\), true.

Key Concepts

Elimination Method

The elimination method is a powerful technique used to solve systems of equations. It involves combining equations in such a way that one of the variables is eliminated, making it easier to solve the system. The goal is to create an equation with just one variable, which can then be solved using simple algebra.

• Start by aligning the equations such that similar terms are in columns.
• Choose a variable to eliminate. This often depends on the coefficients that make elimination straightforward with minimal adjustments.
• Modify one or both equations so that adding or subtracting them will cancel one of the variables. This is usually done by multiplying one or both equations by constants.
• Once one of the variables is eliminated, solve the remaining equation for the other variable.
• Finally, back-substitute the found value into one of the original equations to solve for the other variable.

This method is especially useful when the system has whole numbers and small coefficients, making calculations simpler.

Solving Linear Equations

Solving linear equations is a fundamental skill that applies across many areas of mathematics and real-world problem-solving. A linear equation represents a straight line when graphed on a coordinate plane, and solving these equations typically involves finding the values of the variables that make the equation true.
There are many methods to solve a linear equation:

• Isolation of Variables: Rearranging the equation to isolate the variable on one side can often solve simple linear equations.
• Substitution Method: This involves substituting one variable in terms of another into an equation to simplify it. It's particularly useful when dealing with more than one equation.
• Elimination Method: As explained, this technique uses the addition or subtraction of equations to remove one variable entirely, simplifying the system.

For accuracy, be sure to verify your solutions by substituting them back into the original equations ensuring they satisfy all given conditions. Mistakes often happen in calculations, so this step is crucial.

Consistent and Dependent Systems

In the realm of solving systems of equations, understanding the nature of the solutions is key. A system can be classified as consistent, inconsistent, or dependent based on its solutions.

• Consistent Systems: These systems have at least one solution. If you end up with distinct solutions like specific values for all variables, the system is consistent.
• Inconsistent Systems: These have no solutions. This outcome typically occurs when you try to solve and meet a contradiction, like deriving an equation that states something false, such as 0 = 3.
• Dependent Systems: These systems have infinitely many solutions. This happens when the equations are essentially the same (or multiple equations represent the same line), leading to solutions that are one in terms of the other variable.

Identifying the type of system is critical as it guides you in predicting the nature of the solutions. When solving, returning consistent results validates that the system is indeed consistent, as seen in our worked-out example, which provided explicit solutions for both variables.