Problem 30 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{rr} 2 x-3 y-z^{2}= & 0 \\ x-y-z^{2}= & -1 \\ x^{2}-x y= & 0 \end{array}\right. $$

Step-by-Step Solution

Answer

Solutions are \((-1, -1, 1)\) and \((-1, -1, -1)\).

Step 1: Solve for one variable in one of the equations

Let's start by solving the third equation for one of the variables. We have \(x^2 - x y = 0\). Factor out an \(x\) to get \(x(x-y) = 0\). This implies \(x = 0\) or \(x = y\).

Step 2: Substitute \(x = y\) into the first two equations

Substituting \(x = y\) into the first equation: \(2x - 3y - z^2 = 0\) becomes \(2y - 3y - z^2 = 0\), which simplifies to \(-y - z^2 = 0\), or \(y = -z^2\).Substituting \(x = y\) into the second equation: \(x - y - z^2 = -1\) becomes \(y - y - z^2 = -1\), which simplifies to \(-z^2 = -1\), or \(z^2 = 1\).

Step 3: Solve \(z^2 = 1\) to find \(z's\) values

From \(z^2 = 1\), we get \(z = 1\) or \(z = -1\).

Step 4: Find corresponding values for \(y\) and \(x\)

Using \(y = -z^2\) and the results from step 3:1. If \(z = 1\), then \(y = -1\).2. If \(z = -1\), then \(y = -1\).Since \(x = y\), for both \(z = 1\) and \(z = -1\), \(x = -1\).

Step 5: Verify solutions in all equations

Verify the solutions \((x, y, z) = (-1, -1, 1)\) and \((-1, -1, -1)\) in the original equations:1. For \((x, y, z) = (-1, -1, 1)\): - First equation: \(2(-1) - 3(-1) - 1^2 = 0\). - Second equation: \(-1 - (-1) - 1^2 = -1\). - Third equation: \((-1)^2 - (-1)(-1) = 0\). 2. For \((x, y, z) = (-1, -1, -1)\): - First equation: \(2(-1) - 3(-1) - (-1)^2 = 0\). - Second equation: \(-1 - (-1) - (-1)^2 = -1\). - Third equation: \((-1)^2 - (-1)(-1) = 0\).Both sets satisfy all equations.

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve systems of equations. It involves expressing one variable in terms of another and then substituting that expression into the other equations. This reduces the number of variables and equations we need to deal with at any given time, simplifying the problem.
In the original problem, the substitution method is applied by first manipulating the third equation: \(x^2 - xy = 0\). We factor this to get \(x(x-y) = 0\), revealing two potential solutions: \(x = 0\) or \(x = y\).
By choosing to substitute \(x = y\) into the first and second equations, it allows us to find simpler relationships between the remaining variables \(y\) and \(z\). This strategic choice makes the equations easier to solve step-by-step.

• First, solve for one variable in one equation.
• Substitute this variable in other equations.
• Solve the resulting simpler system.

Remember, correct substitution is key to simplifying complex equations!

Mathematics Problem Solving

Mathematics problem solving is a skill that involves understanding, planning, and carrying out a series of logical steps to arrive at a solution. For solving a system of equations, problem solving becomes a process of:

• Analyzing the system to identify which method (like substitution) is most suitable.
• Breaking down the problem into smaller, manageable steps.
• Executing each step carefully and validating the result each time.

In the context of the substitution method for this exercise, it's about understanding how one substitution can simplify the entire system. By analyzing each equation's potential for substitution and manipulation, we can create a logical path to the solution.
Moreover, mathematics problem solving requires verification. This is done by substituting the found solutions back into the original equations to ensure that they satisfy all conditions. It's a systematic approach where every step needs to be validated thoroughly to confirm the correctness of the solution.

Algebraic Solutions

An algebraic solution to a problem uses algebra, which involves symbols and the rules for manipulating these symbols to represent problems and solve them. When tackling a system of equations algebraically, like in this exercise:

• Begin by simplifying the equations as much as possible.
• Apply algebraic operations to manipulate the equations for easier solving.
• Look for opportunities to factor, combine terms, and substitute expressions.

In our example, algebra is used right from factoring \(x^2 - xy = 0\) to formulating substitutions \(x = y\) which reduce the system's complexity. The power of algebra comes through, as it helps in transforming the system of equations into something more manageable.
Finally, validate algebraic solutions by substituting back into the original equations. This is vital since it helps confirm that each potential solution indeed satisfies all original conditions, ensuring the correctness and reliability of the outcome.