Problem 29 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{aligned} x^{2}+z^{2} &=5 \\ 2 x+y &=1 \\ y+z &=1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are (1, -1, 2) and (-1, 3, -2).

Step 1: Express y in terms of z

From the equation \(y + z = 1\), subtract \(z\) from both sides to express \(y\) in terms of \(z\):\[ y = 1 - z \]

Step 2: Substitute y in the second equation

Substitute \(y = 1 - z\) into the second equation \(2x + y = 1\) to get:\[ 2x + (1 - z) = 1 \]Simplify:\[ 2x + 1 - z = 1 \]Subtract 1 from both sides:\[ 2x - z = 0 \]Add \(z\) to both sides to express \(z\) in terms of \(x\):\[ z = 2x \]

Step 3: Substitute z in the first equation

Substitute \(z = 2x\) into the first equation \(x^2 + z^2 = 5\) to get:\[ x^2 + (2x)^2 = 5 \]Simplify:\[ x^2 + 4x^2 = 5 \]Combine like terms:\[ 5x^2 = 5 \]Divide both sides by 5:\[ x^2 = 1 \]Taking the square root of both sides, we find:\[ x = 1 \text{ or } x = -1 \]

Step 4: Solve for z using x values

For \(x = 1\), substitute into \(z = 2x\):\[ z = 2(1) = 2 \]For \(x = -1\), substitute into \(z = 2x\):\[ z = 2(-1) = -2 \]

Step 5: Solve for y using z values

For \(z = 2\) (when \(x = 1\)), use \(y = 1 - z\):\[ y = 1 - 2 = -1 \]For \(z = -2\) (when \(x = -1\)), use \(y = 1 - z\):\[ y = 1 - (-2) = 3 \]

Step 6: Conclusion: Solutions to the System

The solutions to the system are \((x, y, z) = (1, -1, 2)\) and \((x, y, z) = (-1, 3, -2)\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations involving the same set of variables. In our case, we have three equations with three variables: \(x\), \(y\), and \(z\). This system can be solved to find the values of these variables that satisfy all three equations simultaneously.
Solving a system of equations can involve methods like graphing, substitution, or elimination. Here, we focus on substitution, a powerful method that can simplify complex systems.

• Identify one equation to express a variable in terms of another.
• Substitute this expression into other equations to reduce the number of variables.
• Continue substituting until each equation has only one variable.

Once we find the values of each variable, we can verify the solution by plugging them back into the original equations.

Solving Quadratic Equations

Quadratic equations are polynomial equations of degree two, generally in the form \(ax^2 + bx + c = 0\). They are often solved using the quadratic formula or factoring.
In this exercise, after substituting for \(z\), we arrive at the equation \(5x^2 = 5\), which simplifies to the quadratic equation \(x^2 = 1\).
Solving \(x^2 = 1\), we find the values of \(x\) by taking the square roots:

• \(x = 1\)
• \(x = -1\)

These solutions represent the potential values for \(x\), and each leads to a different value for \(z\) and subsequently \(y\) in the original system.

Algebraic Manipulation

Algebraic manipulation involves using operations like addition, subtraction, multiplication, and division to rearrange and simplify equations. This is a critical skill when solving systems of equations, especially with substitution.
In our system, algebraic manipulation allowed us to:

• Express \(y\) in terms of \(z\) with \(y = 1 - z\).
• Substitute \(y\) and rearrange the second equation to find \(z = 2x\).
• Simplify equations to reduce them to simpler forms, such as \(5x^2 = 5\).

This step-by-step manipulation is key to isolating variables and solving the system efficiently.

Coordinate Geometry

Coordinate geometry helps us understand solutions to systems of equations by visualizing them as intersections of geometric figures. For example, each equation in our system can be thought of as a surface or line in a three-dimensional coordinate space.
In particular:

• The equation \(x^2 + z^2 = 5\) represents a circular cylinder aligned along the \(y\)-axis.
• The equation \(2x + y = 1\) is a plane in 3D space.
• The equation \(y + z = 1\) is another plane.

The solutions \((1, -1, 2)\) and \((-1, 3, -2)\) are the points in space where these geometric shapes intersect, illustrating the power of algebra in solving spatial problems.