Problem 49 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} \frac{3}{4} x+\frac{2}{3} y=7 \\ \frac{3}{5} x-\frac{1}{2} y=18 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 20 \) and \( y = -12 \).

Step 1: Clear the Fractions

To eliminate the fractions from the equations, let's find a common denominator for each equation and multiply through. For the first equation, the least common multiple of 4 and 3 is 12. Multiply every term in the equation \( \frac{3}{4}x+\frac{2}{3}y=7 \) by 12 to get: \[ 12 \times \left( \frac{3}{4}x \right) + 12 \times \left( \frac{2}{3}y \right) = 12 \times 7 \] Simplifying gives: \[ 9x + 8y = 84 \] For the second equation, the least common multiple of 5 and 2 is 10. Multiply every term in the equation \( \frac{3}{5}x-\frac{1}{2}y=18 \) by 10 to get: \[ 10 \times \left( \frac{3}{5}x \right) - 10 \times \left( \frac{1}{2}y \right) = 10 \times 18 \] Simplifying gives:\[ 6x - 5y = 180 \]

Step 2: Solve using Elimination

Now that the fractions are removed, our system is:\[ \begin{align*} 9x + 8y &= 84 \ 6x - 5y &= 180 \end{align*} \]We will use the elimination method. Multiply the first equation by 5 and the second equation by 8 to make the coefficients of \( y \) in both equations the same:\[ \begin{align*} 5(9x + 8y) &= 5(84) \ 8(6x - 5y) &= 8(180) \end{align*} \]Solving gives us:\[ \begin{align*} 45x + 40y &= 420 \ 48x - 40y &= 1440 \end{align*} \]

Step 3: Add the Equations

Add the equations from Step 2 to eliminate \( y \):\[ (45x + 40y) + (48x - 40y) = 420 + 1440 \]This simplifies to:\[ 93x = 1860 \]

Step 4: Solve for \( x \)

Divide both sides of the equation by 93 to find \( x \):\[ x = \frac{1860}{93} = 20 \]

Step 5: Substitute to Find \( y \)

Substitute \( x = 20 \) back into one of the original equations to find \( y \). Using \( 9x + 8y = 84 \):\[ 9(20) + 8y = 84 \]This simplifies to:\[ 180 + 8y = 84 \ 8y = 84 - 180 \]\[ 8y = -96 \]Divide by 8:\[ y = -12 \]

Step 6: Verify Solution

Substitute \( x = 20 \) and \( y = -12 \) into the other original equation to ensure it holds:\[ \frac{3}{5}(20) - \frac{1}{2}(-12) = 18 \]Calculate term by term:\[ 12 + 6 = 18 \]This confirms our solution is correct.

Key Concepts

Elimination Method

The elimination method is a popular technique for solving systems of linear equations. It strategically removes one of the variables by adding or subtracting equations.
To effectively deploy this method, follow these steps:

• Firstly, identify a variable to eliminate. It’s often easiest when one of the coefficients is 1 or -1, but otherwise, you may need to perform some multiplication.
• Then, modify each equation so that adding or subtracting them will remove one variable. You achieve this by multiplying each equation by necessary coefficients to make the variable’s coefficients equal but opposite.
• Lastly, add or subtract the equations. This should give you a single-variable equation which you can then solve.

In our specific exercise, by multiplying the first equation by 5 and the second equation by 8, both coefficients of the variable \( y \) became equal in magnitude but opposite in sign. This allowed us to add the equations together, thus eliminating the \( y \)-variable and solving for \( x \). After finding \( x \), we substituted it into one of the original equations to find \( y \). The elimination method effectively reduces a system of equations to a single unknown, simplifying the solution process.

Fractions in Equations

Working with fractions in equations can often complicate the process. To simplify solving, it’s common practice to eliminate fractions by using the least common multiple (LCM). Here's how:

• Identify the denominators in the equation. For an effective strategy, calculate the LCM of these denominators to eliminate them.
• Multiply every term in the equation by this LCM. This operation transforms the entire equation into one without fractions, making arithmetic operations straightforward.

In our given system, both equations contained fractions. The first equation used denominators 4 and 3, while the second used 5 and 2. Calculating the LCM, we found they were 12 and 10 respectively. We then multiplied through each equation by these LCMs, which transformed the system into a straightforward linear equation format, ready for solving with methods like elimination.
By clearing fractions first, the equations transform to a simpler form, making calculations clearer and reducing chances for error.

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra, involving various techniques. Once an equation is simplified or fractions are eliminated, here’s how to proceed:

• Simplification: Combine like terms and arrange the equation to isolate variables. This results in more manageable terms.
• Substitution: If solving a system, upon finding one variable using methods like elimination, substitute it back into another equation to find the second variable.
• Verification: Once solutions are found for variables, always substitute back into the original equations to verify correctness.

In the exercise, after using the elimination method to find \( x = 20 \), we substituted \( x \) into one of the purged equations to solve for \( y = -12 \). Finally, verifying by plugging these values into the other equation ensured the solution was correct.
Solving linear systems accurately often involves multiple approaches and confirms understanding by verifying solutions against original equations.