Problem 36 - Algebra 2

Question

Solve each system of equations. $$ \begin{array}{l}{3 r-15 s+4 t=-57} \\ {9 r+45 s-t=26} \\ {-6 r+10 s+3 t=-19}\end{array} $$

Step-by-Step Solution

Answer

The solution is \( r = -\frac{4}{3} \), \( s = \frac{3}{5} \), \( t = -11 \).

Step 1: Label the Equations

Assign labels to each equation in the system for easy reference.\[\text{Equation 1: } 3r - 15s + 4t = -57 \\text{Equation 2: } 9r + 45s - t = 26 \\text{Equation 3: } -6r + 10s + 3t = -19\]

Step 2: Eliminate One Variable

To eliminate \(t\), we can multiply Equation 2 by 4 to align it with Equation 1 and Equation 3:\[4(9r + 45s - t) = 4(26) \36r + 180s - 4t = 104\]Add Equation 1 to the modified Equation 2:\[(3r - 15s + 4t) + (36r + 180s - 4t) = -57 + 104\]Simplifying gives Equation 4:\[39r + 165s = 47\]

Step 3: Eliminate Another Variable

Now, let's eliminate \(t\) again using Equation 2 multiplied by 3 and add it to Equation 3:\[3(9r + 45s - t) = 3(26) \27r + 135s - 3t = 78\]Add Equation 3:\[(-6r + 10s + 3t) + (27r + 135s - 3t) = -19 + 78\]This simplifies to Equation 5:\[21r + 145s = 59\]

Step 4: Solve for One Variable

Now, we have a system of two equations with two variables:\[39r + 165s = 47\]\[21r + 145s = 59\]Multiply Equation 5 by 13 and Equation 4 by 7 for elimination:\[13(21r + 145s) = 13(59) \273r + 1885s = 767 \]\[7(39r + 165s) = 7(47) \273r + 1155s = 329\]Subtract Equation 4 (scaled) from Equation 5 (scaled):\[273r + 1885s - (273r + 1155s) = 767 - 329\]Simplifies to:\[730s = 438\]Solve for \(s\):\[s = \frac{438}{730} = \frac{219}{365} = \frac{3}{5}\]

Step 5: Substitute to Find Another Variable

Substitute \(s = \frac{3}{5}\) back into Equation 4:\[39r + 165\left(\frac{3}{5}\right) = 47\]This simplifies to:\[39r + 99 = 47\]Subtract 99 from both sides:\[39r = -52\]Solve for \(r\):\[r = \frac{-52}{39} = -\frac{4}{3}\]

Step 6: Solve for the Last Variable

Substitute \(r = -\frac{4}{3}\) and \(s = \frac{3}{5}\) into Equation 1 to find \(t\):\[3\left(-\frac{4}{3}\right) - 15\left(\frac{3}{5}\right) + 4t = -57\]This simplifies to:\[-4 - 9 + 4t = -57\]Add 13 to both sides:\[4t = -44\]Solve for \(t\):\[t = -11\]

Step 7: Verify the Solution

Substitute \(r = -\frac{4}{3}\), \(s = \frac{3}{5}\), and \(t = -11\) back into the original equations to verify. They must satisfy all the equations. Carry out numerical checks to confirm that all equations hold true.

Key Concepts

Linear Equations

A linear equation is an equation that graphs as a straight line. These equations take the form \(ax + by + cz = d\), where \(a, b, c,\) and \(d\) are constants, and \(x, y,\) and \(z\) are variables. In the context of systems of equations, like the one given:

• 3r - 15s + 4t = -57
• 9r + 45s - t = 26
• -6r + 10s + 3t = -19

Each equation represents a plane in three-dimensional space. The solution to this system is the point where all three planes intersect, representing the values of \(r, s,\) and \(t\) that satisfy all equations simultaneously. It's important to align the coefficients effectively when solving them, as changing these coefficients affects the position and alignment of the planes they represent.

Variable Elimination

Variable elimination is a technique used to simplify systems of equations by removing one of the variables. In this exercise, we want to eliminate the variable \(t\) to make the system easier to solve. Here's how:

• Multiply Equation 2 by a factor that allows the \(t\) coefficients in Equations 1 and 2 to cancel each other out when added together.
• Add the adjusted Equation 2 to Equation 1, resulting in a new equation with only two variables, \(r\) and \(s\).
• Repeat this process with a different set of equations to eliminate \(t\) again, generating another two-variable equation.

The result is a simpler system where we only have to solve for \(r\) and \(s\). This step is crucial as it reduces the complexity, making the problem more manageable. It's like peeling layers of an onion, getting to the core where the simpler system lies.

Substitution Method

Substitution is another powerful tool in solving systems of equations, especially after variable elimination. Here's the process:

• Take one of the simplified equations and solve for one variable in terms of the other. For example, solve for \(s\) using one of the purely \(r\) and \(s\) equations.
• Once you find \(s\), substitute it back into the another remaining equation to find \(r\).
• Use the values of \(r\) and \(s\) to substitute back into the original equations to find the final variable, \(t\).

Through substitution, you gradually find all variable values. This step is like fitting the pieces together in a puzzle: once you have enough key pieces, the rest falls into place easily.

Verification of Solutions

Verification is a critical step to ensure that the solutions obtained are correct. After calculating \(r = -\frac{4}{3}\), \(s = \frac{3}{5}\), and \(t = -11\), substitute these values back into the original equations:

• Check each equation to see if the left-hand side equals the right-hand side.
• If all equations are satisfied, that confirms your solution is correct.
• If one or more equations do not hold true, re-check the calculations for errors in substitution or arithmetic.

Verification acts as a safety net, ensuring the solution found actually solves the system of equations. It provides confidence, confirming that all equations are balanced, much like checking the final math in a financial report.