Problem 38 - A Graphical Approach to Precalculus with Limits

Question

Solve each system analytically. If the equations are dependent, write the solution set in terms of the variable \(z\). $$\begin{aligned} 10 x+2 y-3 z &=0 \\ 5 x+4 y+6 z &=-1 \\ 6 y+3 z &=2 \end{aligned}$$

Step-by-Step Solution

Answer

Unique solution: \((x, y, z) = \left(-\frac{1}{5}, \frac{1}{2}, -\frac{1}{3}\right)\).

Step 1: Solve for y and z from the third equation

Start by solving the third equation, which is simpler as it involves only variables \(y\) and \(z\). The equation is \(6y + 3z = 2\). Isolate \(y\) to express it in terms of \(z\):\[6y = 2 - 3z\] \[y = \frac{2 - 3z}{6} = \frac{1}{3} - \frac{1}{2}z\].

Step 2: Substitute y into the remaining equations

Now that \(y\) is expressed in terms of \(z\), substitute \(y = \frac{1}{3} - \frac{1}{2}z\) into the first and second equations:1. For the first equation: \(10x + 2\left(\frac{1}{3} - \frac{1}{2}z\right) - 3z = 0\) Simplify: \(10x + \frac{2}{3} - z - 3z = 0\) \[10x - 4z = -\frac{2}{3}\]2. For the second equation: \(5x + 4\left(\frac{1}{3} - \frac{1}{2}z\right) + 6z = -1\) Simplify: \(5x + \frac{4}{3} - 2z + 6z = -1\) \[5x + 4z = -1 - \frac{4}{3}\].

Step 3: Simplify second set of equations

Simplifying the adjusted versions for both equations, we have:- First new equation: \(10x - 4z = -\frac{2}{3}\)- Second new equation: \(5x + 4z = -\frac{7}{3}\), from simplifying \(-1 - \frac{4}{3}\).The next step involves solving this smaller system of equations.

Step 4: Add the two equations

To eliminate \(z\), add the two equations:\[ (10x - 4z) + (5x + 4z) = -\frac{2}{3} + -\frac{7}{3} \]\[ 15x = -\frac{9}{3} = -3 \]Solving for \(x\): \[x = -\frac{3}{15} = -\frac{1}{5}\].

Step 5: Substitute x back to find z

Now that \(x\) is known, substitute \(x = -\frac{1}{5}\) back into either of the simplified equations to find \(z\). Using the first new equation:\[10\left(-\frac{1}{5}\right) - 4z = -\frac{2}{3}\]\[-2 - 4z = -\frac{2}{3}\]Add 2 to both sides:\[-4z = -\frac{2}{3} + 2\]\[-4z = \frac{4}{3}\]\[z = -\frac{1}{3}\].

Step 6: Check for dependent solutions

From Step 4 and Step 5, we have particular values for \(x\) and \(z\). Substitute \(z = -\frac{1}{3}\) back to find \(y\):\[y = \frac{1}{3} - \frac{1}{2}\left(-\frac{1}{3}\right) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\].Given distinct values for all variables, the system is not dependent in terms of \(z\). The unique solution to the system is \((x, y, z) = \left(-\frac{1}{5}, \frac{1}{2}, -\frac{1}{3}\right)\).

Key Concepts

Analytical Methods

In the realm of solving systems of equations, analytical methods often stand out due to their precision and logical approach. These methods involve manipulating algebraic expressions to find exact solutions of variables involved.

Analytical methods commonly used include substitution, elimination, and matrix techniques. These are used to convert complex systems into simpler forms that are easier to solve.

• Substitution: Replace one variable with an expression involving another variable.
• Elimination: Add or subtract equations to eliminate a variable, simplifying the system.
• Matrix Methods: Utilize matrices and vectors to systematically solve the equations.

In our specific example, we used substitution - expressing one variable in terms of another, and then simplifying the problem step by step. This stepwise method allows handling one piece of the puzzle at a time, leading to a complete solution.

Dependent Equations

Equations are termed dependent when they are essentially multiples of each other, or when combining them doesn't provide new information about the system. Dependent equations do not lead to a unique solution. Instead, they produce infinitely many solutions that satisfy the given relations.

The key indicators of dependent equations can include:

• Identical Equations: After simplification, two or more equations appear to be the same.
• Constant Ratio: The coefficients of the equations are proportional across the entire system.

In our scenario, we initially needed to verify if the system was dependent. However, since we obtained a unique solution with distinct values for all variables, this indicates that the system was independent. This verification process ensures no surprise mishaps in assumptions about the system.

Substitution

Substitution is a straightforward method used to solve systems of equations whereby one equation is solved for one variable, and this expression is substituted into the other equation(s). It reduces the complexity by decreasing the number of variables in the remaining equations.

Here's a brief look at how substitution works:

• Isolate: Solve one of the equations for one of the variables.
• Substitute: Plug this expression into the other equation(s) to reduce the number of unknowns.
• Solve: Simplify the new equation and solve for the remaining variables.

In the example provided, we began by isolating \(y\) in terms of \(z\) using the simplest equation. This substituted value was then used in the other two equations to form a new, simpler system of two equations with two unknowns, making the solution accessible. The solution proceeds by simplifying and solving this reduced system step by step.