Problem 47 - College Algebra with Modeling and Visualization

Question

If possible, solve the system of linear equations and check your answer. $$ \begin{aligned} \frac{1}{2} x-y &=-5 \\ x+\frac{1}{2} y &=10 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = 6\) and \(y = 8\).

Step 1: Eliminate Fractions

First, we want to eliminate the fractions in the equations. Multiply the first equation by 2 to get rid of the fraction: \[ 2 \left(\frac{1}{2} x - y \right) = 2(-5) \]This simplifies to:\[ x - 2y = -10 \]Next, multiply the second equation by 2:\[ 2 \left(x + \frac{1}{2} y \right) = 2(10) \]This simplifies to:\[ 2x + y = 20 \]

Step 2: Solve for One Variable

We now have the system:\[\begin{aligned}x - 2y &= -10 \ 2x + y &= 20 \end{aligned}\]Let's solve the second equation for \(y\):\[ y = 20 - 2x \]

Step 3: Substitute and Solve

Substitute \(y = 20 - 2x\) from Step 2 into the first equation:\[ x - 2(20 - 2x) = -10 \]Simplify and solve for \(x\):\[ x - 40 + 4x = -10 \]Combining like terms:\[ 5x - 40 = -10 \]Add 40 to both sides:\[ 5x = 30 \]Divide by 5:\[ x = 6 \]

Step 4: Back-substitute to Solve for y

Using \(x = 6\), substitute back into the equation \(y = 20 - 2x\) to find \(y\):\[ y = 20 - 2(6) \]Simplify:\[ y = 20 - 12 \]\[ y = 8 \]

Step 5: Verify the Solution

Check that \(x = 6\) and \(y = 8\) satisfy the original equations:For the first equation:\[ \frac{1}{2}(6) - 8 = -5 \]\[ 3 - 8 = -5 \]This is correct.For the second equation:\[ 6 + \frac{1}{2}(8) = 10 \]\[ 6 + 4 = 10 \]This is also correct. Hence, \(x = 6\) and \(y = 8\) is a solution.

Key Concepts

Eliminating Fractions

When solving systems of linear equations, having fractions can make calculations difficult. To simplify the process, it's helpful to eliminate fractions early in the solution. This is done by multiplying every term in the equation by a common denominator. In our example, we start with two equations:\[ \frac{1}{2} x - y = -5 \]\[ x + \frac{1}{2} y = 10 \]Both of these equations have terms with a denominator of 2. Thus, we multiply the entire equation by 2 to eliminate fractions.- For the first equation: Multiply each term by 2 to get \( x - 2y = -10 \).- For the second equation: Multiply each term by 2 to get \( 2x + y = 20 \).Now, both equations no longer have fractions, making them easier to work with in subsequent steps.

Substitution Method

The substitution method is a common technique used to solve systems of equations. After simplifying the equations, choose one equation to express one variable in terms of the other. In our system:\[ x - 2y = -10 \]\[ 2x + y = 20 \]We choose to express \( y \) from the second equation: \( y = 20 - 2x \).This expression for \( y \) is then substituted back into the first equation. By doing this, we reduce the system of two equations with two variables to a single equation with one variable:- Substitute \( y = 20 - 2x \) into \( x - 2y = -10 \) which then becomes \( x - 2(20 - 2x) = -10 \).This method allows us to solve for one variable at a time, simplifying the problem further.

Verifying the Solution

It's essential to verify that your solutions satisfy the original equations, especially when variables are involved. After finding the values of \( x = 6 \) and \( y = 8 \), substitute these values back into the original equations to check:- For the first equation: \( \frac{1}{2}(6) - 8 = -5 \), simplify and confirm.- For the second equation: \( 6 + \frac{1}{2}(8) = 10 \), simplify and confirm.Both equations hold true with the solutions found, confirming that \( x = 6 \) and \( y = 8 \) is indeed correct. Verifying your solution gives confidence that the answer is correct and complete.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate variables and find solutions. This concept is crucial for solving equations effectively. In the given problem, once fractions are removed and the substitution method is applied, you end up with:\[ x - 40 + 4x = -10 \]Here, algebraic manipulation involves:- Combining like terms: \( 5x - 40 = -10 \).- Isolating \( x \) by adding 40 to both sides:\[ 5x = 30 \].- Solving for \( x \) by dividing both sides by 5:\[ x = 6 \].This logical and systematic approach helps us solve for one variable at a time, making the process of finding solutions clearer and more manageable.