Problem 41 - 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. $$\left\\{\begin{array}{l}8 s-3 t=-3 \\\5 s-2 t=-1\end{array}\right.$$

Step-by-Step Solution

Answer

The system has a unique solution: \((s, t) = (-3, -7)\).

Step 1: Write the System of Equations

The given system of linear equations is: \(8s - 3t = -3\) and \(5s - 2t = -1\). We need to find the values of \(s\) and \(t\) that satisfy both equations simultaneously.

Step 2: Eliminate One Variable

We'll eliminate \(t\) by making the coefficients of \(t\) in both equations equal. Multiply the first equation by 2 and the second equation by 3 to make the coefficients of \(t\) equal: \[2(8s - 3t) = 2(-3)\] leading to \(16s - 6t = -6\), and \[3(5s - 2t) = 3(-1)\] leading to \(15s - 6t = -3\).

Step 3: Subtract the Modified Equations

Subtract the second modified equation from the first one: \((16s - 6t) - (15s - 6t) = -6 - (-3)\). This results in \(s = -3\).

Step 4: Substitute to Find the Other Variable

Substitute \(s = -3\) back into the second original equation \(5s - 2t = -1\): \(5(-3) - 2t = -1\). Simplifying this gives \(-15 - 2t = -1\), so \(-2t = 14\). Solving for \(t\), we get \(t = -7\).

Step 5: Check the Solution

Verify that \(s = -3\) and \(t = -7\) satisfy both original equations. For the first equation, \(8(-3) - 3(-7) = -24 + 21 = -3\) holds true. For the second equation, \(5(-3) - 2(-7) = -15 + 14 = -1\) also holds true. The solution is valid.

Key Concepts

Elimination Method

The elimination method is a technique used to solve systems of equations by removing one variable. This allows for a straightforward calculation of the remaining variable. In our exercise, we aim to eliminate the variable \(t\) from the equations:

• Equation 1: \(8s - 3t = -3\)
• Equation 2: \(5s - 2t = -1\)

To eliminate \(t\), we adjust the equations so their \(t\)-coefficients match. Multiply the first equation by 2, and the second by 3:

• First equation becomes \(16s - 6t = -6\)
• Second equation becomes \(15s - 6t = -3\)

Notice that both equations now have \(-6t\), allowing us to subtract one from the other and eliminate \(t\). This results in the simpler equation \(s = -3\).
By using elimination, we quickly find one of the variables, leading us directly into future steps of deducing the other variable.

Substitution Method

The substitution method involves solving one equation for one variable and plugging it into another equation. After using the elimination method, we determined \(s = -3\). Now, we can find \(t\) by substituting \(s = -3\) into one of the original equations.
Let’s choose the second equation, \(5s - 2t = -1\). Replace \(s\) with \(-3\):\(5(-3) - 2t = -1\), leading to \(-15 - 2t = -1\).
Solve for \(t\):

• Add 15 to each side, giving \(-2t = 14\)
• Divide by -2, resulting in \(t = -7\)

Using substitution, we systematically unravel the system from one known to another unknown. This method is especially effective after simplifying the system through elimination.

Solution Verification

Verifying a solution ensures that both values, \(s = -3\) and \(t = -7\), satisfy the original system of equations. Let's check each equation separately:

• Substitute into Equation 1: \(8s - 3t = -3\)
  • Calculating: \(8(-3) - 3(-7) = -24 + 21 = -3\)
  • Outcome: The first equation holds true.
• Substitute into Equation 2: \(5s - 2t = -1\)
  • Calculating: \(5(-3) - 2(-7) = -15 + 14 = -1\)
  • Outcome: The second equation also holds true.

Each original equation confirms that \(s = -3\) and \(t = -7\) are indeed the correct solutions. Verification is a vital step in solving systems to ensure no errors were made during calculations.