Problem 46 - Intermediate Algebra

Question

For Problems 19-48, solve each system by using either the substitution or the elimination-by-addition method, whichever seems more appropriate. (Objective 2) $$ \left(\begin{array}{l} \frac{x}{2}+\frac{y}{3}=\frac{5}{72} \\ \frac{x}{4}+\frac{5 y}{2}=-\frac{17}{48} \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{4}\) and \(y = -\frac{1}{6}\).

Step 1: Simplify the Equations

First, let's eliminate the fractions by multiplying through by the common denominators. For the first equation, multiply every term by 72 to get:\[ 36x + 24y = 5 \]For the second equation, multiply every term by 48 to get:\[ 12x + 120y = -17 \]

Step 2: Rearrange the System

The simplified system of equations is now:\[ \begin{align*}36x + 24y &= 5 \12x + 120y &= -17 \\end{align*} \]

Step 3: Apply the Elimination Method

We aim to eliminate variable \(x\). Multiply the second equation by 3 to align \(x\) terms:\[ 36x + 360y = -51 \]Now, subtract the first equation from the newly scaled second equation:\[ (36x + 360y) - (36x + 24y) = -51 - 5 \]

Step 4: Solve for y

After subtracting, simplify:\[ 336y = -56 \]Solve for \(y\) by dividing both sides by 336:\[ y = -\frac{56}{336} = -\frac{1}{6} \]

Step 5: Substitute y back to find x

Use the value of \(y\) and substitute back into the simplified first equation:\[ 36x + 24\left(-\frac{1}{6}\right) = 5 \]This simplifies to:\[ 36x - 4 = 5 \]Add 4 to both sides:\[ 36x = 9 \]Divide by 36 to solve for \(x\):\[ x = \frac{9}{36} = \frac{1}{4} \]

Step 6: Verify the Solution

Verify by substituting \(x = \frac{1}{4}\) and \(y = -\frac{1}{6}\) back into both original equations:1. \(\frac{1}{2} * \frac{1}{4} + \frac{1}{3} * -\frac{1}{6} = \frac{5}{72} \) holds true.2. \(\frac{1}{4} * \frac{1}{4} + \frac{5}{2} * -\frac{1}{6} = -\frac{17}{48} \) holds true.The solution satisfies both equations.

Key Concepts

Substitution Method

The substitution method is a way to solve a system of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This technique is especially useful when one of the equations can be easily solved for one of the variables.
The steps for the substitution method are as follows:

• Solve one of the equations for one variable in terms of the other variable.
• Substitute the expression found into the other equation.
• Solve the resulting equation to find the value of the first variable.
• Substitute the value found back into one of the original equations to find the value of the second variable.

Using the substitution method can be effective in solving systems where one equation is straightforward to manipulate. However, if neither equation easily isolates a variable, the elimination method might be more straight-forward.

Elimination Method

The elimination method is an alternative method for solving systems of equations, particularly effective when the system includes equations that, with a bit of manipulation, can allow one variable to be eliminated when the equations are combined. This process involves aligning terms so that when the equations are added together, one variable cancels out completely.
Here’s how to use the elimination method:

• First, modify the equations (if necessary) to have opposite coefficients for one of the variables. This often involves multiplying one or both equations by a constant.
• Next, add or subtract the equations to eliminate one variable, which will leave you with a single equation to solve for the other variable.
• Solve this new equation for the remaining variable.
• Substitute the solution back into one of the original equations to find the eliminated variable.

This method is especially useful in our example, as it quickly allowed us to eliminate variable \(x\) by aligning its coefficients and then subtracting the equations.

Solving Linear Equations

Solving linear equations is foundational for algebra, requiring an understanding of operations and principles to isolate variables. Linear equations take the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.When solving linear equations:

• Identify and simplify the equations. Simplification often includes multiplying through by common denominators or combining like terms to streamline the problem.
• Use methods like substitution or elimination for systems of linear equations, choosing based on equation complexity and form.
• Verify your solutions by substituting them back into the original equations to ensure both original equations hold true.

Understanding these concepts not only helps in completing homework but also builds a foundation for solving more complex algebraic problems.