Problem 33 - Elementary Algebra

Question

Solve each system by using either the substitution or the elimination-by- addition method, whichever seems more appropriate. $$\left(\begin{array}{l}\frac{x}{6}+\frac{y}{3}=3 \\ \frac{5 x}{2}-\frac{y}{6}=-17\end{array}\right)$$

Step-by-Step Solution

Answer

The solution is \((x, y) = (-6, 12)\).

Step 1: Rewrite the System of Equations

First, let's rewrite the given system of equations to make them easier to work with. The original equations are:\[ \frac{x}{6} + \frac{y}{3} = 3 \] and \[ \frac{5x}{2} - \frac{y}{6} = -17. \]To eliminate the fractions, multiply the first equation by 6 and the second equation by 6.

Step 2: Simplify the Equations

After multiplying by 6, the system of equations becomes:1) \[ x + 2y = 18 \]2) \[ 15x - y = -102 \].Now, we have a system of equations without fractions.

Step 3: Use Substitution Method

From equation 1, express \(x\) in terms of \(y\):\[ x + 2y = 18 \Rightarrow x = 18 - 2y. \]Substitute \(x = 18 - 2y\) into equation 2 to find \(y\).

Step 4: Solve for y

Substitute \(x = 18 - 2y\) into equation 2:\[ 15(18 - 2y) - y = -102 \].This simplifies to:\[ 270 - 30y - y = -102 \], which further simplifies to:\[ -31y = -372 \].Divide both sides by -31 to find \(y\):\[ y = 12. \]

Step 5: Substitute to find x

Now that we know \(y = 12\), substitute it back into equation \(x = 18 - 2y\):\[ x = 18 - 2(12) \].Calculating this gives:\[ x = 18 - 24 = -6 \].

Step 6: Verify the Solution

Substitute \(x = -6\) and \(y = 12\) back into the original equations to ensure they hold true:For the first equation:\[ \frac{-6}{6} + \frac{12}{3} = -1 + 4 = 3. \] This is correct.For the second equation:\[ \frac{5(-6)}{2} - \frac{12}{6} = -15 - 2 = -17. \] This is also correct.The solution \((x, y) = (-6, 12)\) satisfies both equations.

Key Concepts

Substitution Method

The substitution method is a technique often used to solve systems of equations, especially when one equation is easily reformulated to express one variable in terms of another.
It involves substituting this expression in the other equation to find the second variable.

In our example, once we have simplified our equations by removing fractions, we have:

• Equation 1: \( x + 2y = 18 \)
• Equation 2: \( 15x - y = -102 \)

To use substitution, we solve equation 1 for \( x \):

• \( x = 18 - 2y \)

We then substitute this expression for \( x \) into equation 2:

• \( 15 (18 - 2y) - y = -102 \)

This substitution transforms a system of two equations into a single equation with one variable, making it easier to solve.
The key is to always replace one variable, making sure to simplify step by step.
This approach is systematic and helps find the solution effectively.

Elimination Method

The elimination method, also known as the addition method, is another strategy used to solve systems of equations.
This method involves adding or subtracting equations to eliminate one of the variables, making it easier to solve for the remaining variable.

In our original problem, instead of substitution, we could have used elimination once we had our simplified equations:

• \( x + 2y = 18 \)
• \( 15x - y = -102 \)

To eliminate one of the variables, we would aim to make the coefficients of either \( x \) or \( y \) equal and opposite.
This could involve multiplying one or both equations by a number that strategically helps in canceling out a variable.
Once a variable is eliminated, a single equation in terms of one variable remains, which can then be solved easily.
After discovering the value of one variable, substitute back to find the other.
This method is useful when multiplication or division of the entire equations can simplify the problem.

Solving Linear Equations

Solving linear equations is a fundamental part of algebra that involves finding the value of the variables that make the equation true.
Linear equations are mathematical expressions that represent a straight line when graphed.

The process typically involves isolating the variable by performing operations that inverse those applied within the equation.
For example, if you need to solve the equation \( x + 2y = 18 \), you would do the following:

• Isolate \( x \) by subtracting \( 2y \) from both sides: \( x = 18 - 2y \).

This is the main mechanism behind solving equations, which can then be applied repeatedly within systems of equations to find an assignment that satisfies all equations.
Understanding how to manipulate and rearrange these equations is crucial for both isolation methods and substitution or elimination methods used in systems of equations.

Fractions in Equations

Dealing with fractions in equations may seem tricky, but it is completely manageable with the right approach.
Fractions often complicate calculations, so a common strategy is to eliminate them as early as possible.

In our original system of equations, the presence of fractions can be off-putting:

• \( \frac{x}{6} + \frac{y}{3} = 3 \)
• \( \frac{5x}{2} - \frac{y}{6} = -17 \)

To clear the fractions, multiply each equation by the least common multiple (LCM) of the denominators.
For these equations, multiplying by 6 clears the fractions, converting them into:

• \( x + 2y = 18 \)
• \( 15x - y = -102 \)

With fractions out of the way, the equations are now easier to handle and solve.
This technique simplifies the process and minimizes algebraic errors, making the path to solving the system much clearer.