Problem 44 - Precalculus Mathematics for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered pair form given in Example 6. $$\left\\{\begin{array}{l}\frac{3}{2} x-\frac{1}{3} y=\frac{1}{2} \\\2 x-\frac{1}{2} y=-\frac{1}{2}\end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((5, 21)\).

Step 1: Clear Fractions

Multiply each equation by the least common multiple of the denominators to eliminate fractions. For the first equation, the denominators are 2 and 3, so multiply by 6:\[ 6 \left( \frac{3}{2}x - \frac{1}{3}y \right) = 6 \times \frac{1}{2} \] Resulting in:\[ 9x - 2y = 3 \]For the second equation, the denominator is 2, so multiply by 2:\[ 2 \left( 2x - \frac{1}{2}y \right) = 2 \times -\frac{1}{2} \] Resulting in:\[ 4x - y = -1 \]

Step 2: Set Up for Elimination

Align the equations:\[ \begin{aligned} 9x - 2y &= 3 \ 4x - y &= -1 \end{aligned}\]

Step 3: Eliminate Variable y

Multiply the second equation by 2 to align with the coefficients of \(y\):\[ \begin{aligned} 9x - 2y &= 3 \ 8x - 2y &= -2 \end{aligned}\]Subtract the second equation from the first:\[ (9x - 2y) - (8x - 2y) = 3 - (-2)\]\[ x = 5 \]

Step 4: Substitute x Back

Substitute \(x = 5\) into one of the original equations, such as \(4x - y = -1\):\[ 4(5) - y = -1\]\[ 20 - y = -1\]Solve for \(y\):\[ -y = -21 \ y = 21\]

Step 5: Solution Verification

Now that \(x = 5\) and \(y = 21\), verify by plugging back into both original equations:For \( \frac{3}{2} x - \frac{1}{3} y = \frac{1}{2} \) :\[ \frac{3}{2} imes 5 - \frac{1}{3} imes 21 = 7.5 - 7 = 0.5\]Correct! For \( 2x - \frac{1}{2} y = -\frac{1}{2} \):\[ 2 imes 5 - \frac{1}{2} imes 21 = 10 - 10.5 = -0.5\]Correct! Both checks confirm solutions.

Key Concepts

Elimination Method

The elimination method is a powerful tool for solving systems of linear equations. It involves manipulating the equations to eliminate one of the variables, making it easier to solve for the remaining variables.
In the given problem, we start by aligning the equations:

• First equation: \(9x - 2y = 3\)
• Second equation: \(4x - y = -1\)

We then multiply the second equation by 2 to match the coefficient of \(y\) in the first equation. This gives us:

• First: \(9x - 2y = 3\)
• Second modified: \(8x - 2y = -2\)

By subtracting the second from the first, the \(y\) terms cancel out, simplifying the solution to find \(x\) as 5. Thus, eliminating the \(y\) variable makes it straightforward to solve for \(x\) using subtraction.

Substitution Method

The substitution method is another effective way to solve linear systems. Once you have one variable isolated, you can substitute it back into one of the equations to find the other variable.
In this problem, after using the elimination method, we discovered \(x = 5\). We then substitute \(x = 5\) back into the second simplified equation, \(4x - y = -1\):

• Substitute: \(4(5) - y = -1\)
• Solve for \(y: 20 - y = -1\)
• Therefore, \(y = 21\)

Substitution allows us to confirm and find the value of the second variable by knowing the first, simplifying the process even further after elimination.

Fractions in Equations

Fractions can make equations appear complex, but they are manageable when you clear them effectively. The key is to eliminate fractions at the start of the problem to work with whole numbers instead.
To clear the fractions in the given system:

• For the first equation, multiply by the least common multiple of 2 and 3, which is 6:
  \(6 \left( \frac{3}{2}x - \frac{1}{3}y \right) = 6 \times \frac{1}{2}\)
• This becomes \(9x - 2y = 3\).
• For the second equation, multiply by 2:
  \(2 \left( 2x - \frac{1}{2}y \right) = 2 \times -\frac{1}{2}\)
• This results in \(4x - y = -1\).

Managing and clearing fractions early simplifies the entire calculation process, setting up a clear path to solve the system using elimination or substitution.

Verifying Solutions

Verifying solutions is crucial to ensure that the obtained values satisfy the original equations. After calculating solutions with methods like elimination or substitution, always check your work.
For our problem, we found \(x = 5\) and \(y = 21\). To verify:

• First equation: \(\frac{3}{2} \times 5 - \frac{1}{3} \times 21 = 7.5 - 7 = 0.5\)
• This matches the right-hand side of the first equation: \(0.5\).
• Second equation: \(2 \times 5 - \frac{1}{2} \times 21 = 10 - 10.5 = -0.5\)
• This matches the right-hand side of the second equation: \(-0.5\).

Both values satisfy their respective original equations, confirming the solutions are correct. Always verify to prevent mistakes and ensure the accuracy of your solutions.