Problem 43 - A Graphical Approach to Precalculus with Limits

Question

Solve each system by elimination. $$\begin{aligned}&\frac{x}{2}+\frac{y}{3}=8\\\&\frac{2 x}{3}+\frac{3 y}{2}=17\end{aligned}$$

Step-by-Step Solution

Answer

x = 12, y = 6.

Step 1: Clear the Fractions

First, let's eliminate the fractions in both equations. Multiply the first equation by 6 (the Least Common Multiple of 2 and 3): \[ 3x + 2y = 48 \]Multiply the second equation by 6 (the Least Common Multiple of 3 and 2):\[ 4x + 9y = 102 \]

Step 2: Align for Elimination

To eliminate a variable, let's make the coefficients of one of the variables in both equations equal. We can eliminate \(y\) by obtaining equivalent coefficients. Multiply the first equation by 9:\[ 27x + 18y = 432 \]Multiply the second equation by 2:\[ 8x + 18y = 204 \]

Step 3: Subtract the Equations

Now, we'll subtract the second equation from the first:\[ (27x + 18y) - (8x + 18y) = 432 - 204 \] This simplifies to:\[ 19x = 228 \]

Step 4: Solve for x

Now solve for \(x\):\[ x = \frac{228}{19} = 12 \]

Step 5: Substitute to Find y

Substitute \(x = 12\) back into one of the original equations. We'll use the equation \(3x + 2y = 48\):\[ 3(12) + 2y = 48 \]This simplifies to:\[ 36 + 2y = 48 \]\[ 2y = 12 \]\[ y = 6 \]

Step 6: Check the Solution

Substitute \(x = 12\) and \(y = 6\) back into the original equations to verify the solution:For Equation 1: \[ \frac{12}{2} + \frac{6}{3} = 6 + 2 = 8 \text{ (True)} \]For Equation 2:\[ \frac{2 \cdot 12}{3} + \frac{3 \cdot 6}{2} = 8 + 9 = 17 \text{ (True)} \]Both equations are satisfied, confirming our solution.

Key Concepts

Elimination Method

The elimination method is a popular technique to solve systems of equations. It works by eliminating one variable, allowing you to solve for the remaining variable. This method can be particularly useful when equations contain fractions, as it simplifies the process by focusing on integer coefficients instead.

To apply the elimination method, follow these steps:

• First, multiply each equation by a suitable number to eliminate fractions. Often, this is achieved by multiplying by the least common multiple (LCM) of the denominators.
• Next, adjust the equations so that one variable can be eliminated when the equations are added or subtracted.
• Perform the addition or subtraction to eliminate one variable, resulting in an equation with a single variable.
• Solve the remaining equation for the single variable.

In the provided problem, the elimination method efficiently transforms a system with fractions into a solvable system, allowing you to find solutions with greater ease.

Fractions

Fractions in equations can make solving systems appear complex, but with the right approach, they can be managed easily. In this scenario, we begin by clearing the fractions to simplify our calculations. This approach involves multiplying each term by the least common multiple of the denominators.

Consider the equations:

• \(\frac{x}{2} + \frac{y}{3} = 8\),
• \(\frac{2x}{3} + \frac{3y}{2} = 17\).

To clear the fractions, calculate the LCM of the denominators for each equation. By multiplying all terms in the first equation by 6 (LCM of 2 and 3), the fractions disappear, transforming it into \(3x + 2y = 48\). Repeat the process for the second equation. This simplification allows you to perform subsequent steps without the added difficulty of fractions, streamlining the solution.

Substitution

Substitution is a critical complementary method in solving systems of equations. However, in problems involving elimination, substitution often plays a quieter role. It most often is used after one variable has been solved, allowing you to substitute back to find the other variable.

Once the elimination method simplifies the system to a single-variable equation, you solve for that variable. In our solved system, we found \( x = 12 \). Using substitution, we reintroduce this value into one of the original equations to find \( y \).

Plugging \( x = 12 \) into the equation \( 3x + 2y = 48 \), we substitute and solve for \( y \), simplifying our equation into \( 36 + 2y = 48 \) and eventually reaching \( y = 6 \). This step anchors the solution to both variables, providing a complete answer to the system using clear and simple calculations.

Verification

Verification is the final, yet crucial, part of solving systems of equations. This process reassures that the solution satisfies both of the original equations. It's like double-checking your work to ensure accuracy.

You can verify by substituting the solved values back into the original equations. For our problem, substituting \( x = 12 \) and \( y = 6 \) back into the equations:

• For the first equation: \(\frac{12}{2} + \frac{6}{3} = 8\). This checks out, as it simplifies to \(6 + 2 = 8\).
• For the second equation: \(\frac{2 \times 12}{3} + \frac{3 \times 6}{2} = 17\). Simplifying, \(8 + 9 = 17\), which also holds true.

Both equations are satisfied, confirming that our solutions \( x = 12 \) and \( y = 6 \) are correct. This verification provides confidence in the accuracy of your answers and emphasizes the reliability of your method.