Problem 28 - Algebra 2

Question

Write an example of a system of three equations in three variables that has (-3, 5, 2) as a solution. Show that the ordered triple satisfies all three equations.

Step-by-Step Solution

Answer

The system has the equations: \(2x - y + 3z = -5\), \(x + 4y - z = 15\), \(-x + 2y + 4z = 21\). The solution (-3, 5, 2) satisfies all equations.

Step 1: Choose Coefficients for Variables

To create a system of three equations, we need coefficients for each variable. Let's choose some random coefficients: For example, for the first equation, we could use coefficients 2, -1, and 3 for variables x, y, and z respectively.

Step 2: Construct the First Equation

Using the coefficients chosen in Step 1, substitute the coordinates of the solution (-3, 5, 2) into the equation: \(2x - y + 3z = d_1\). Calculate \(d_1\) by substituting: \(2(-3) - 5 + 3(2) = -6 - 5 + 6 = -5\). So, the first equation is \(2x - y + 3z = -5\).

Step 3: Choose Coefficients for Second Equation

Choose another set of coefficients for the second equation, such as 1, 4, and -1 for x, y, and z respectively.

Step 4: Construct the Second Equation

Substitute the solution (-3, 5, 2) into the second equation: \(x + 4y - z = d_2\). Calculate \(d_2\) as follows: \((-3) + 4(5) - 2 = -3 + 20 - 2 = 15\). Hence, the second equation is \(x + 4y - z = 15\).

Step 5: Choose Coefficients for Third Equation

Choose different coefficients for the third equation, such as -1, 2, and 4.

Step 6: Construct the Third Equation

Substitute the solution (-3, 5, 2) into the third equation: \(-x + 2y + 4z = d_3\). Calculate \(d_3\) as follows: \(3 + 2(5) + 4(2) = 3 + 10 + 8 = 21\). Thus, the third equation is \(-x + 2y + 4z = 21\).

Step 7: Verify the Solution

Check that the coordinates (-3, 5, 2) satisfy all three equations: 1. For \(2x - y + 3z = -5\), substitute to get \(-5 = -5\).2. For \(x + 4y - z = 15\), substitute to get \(15 = 15\).3. For \(-x + 2y + 4z = 21\), substitute to get \(21 = 21\).The ordered triple satisfies all three equations.

Key Concepts

Three Variables

A system of equations involving three variables is an extension of the simpler two-variable system. In our case, the variables are typically represented by letters like \(x\), \(y\), and \(z\). Each variable corresponds to a coordinate in a three-dimensional space.

When dealing with three variables, you're essentially working within a 3D environment. This gives us more complexity compared with the two-dimensional systems, such as those found on a flat graph.

Each equation in a system of three variables represents a plane in this 3D space. The solution to this system is the point or set of points where these planes intersect.

In simple terms, a solution is an ordered triple \((x, y, z)\) that makes each equation in the system true when you substitute \(x\), \(y\), and \(z\) into them. Our example has a solution \((-3, 5, 2)\), meaning substituting these values into all three equations validates them, confirming the intersection at this point.

Solutions of Equations

Finding solutions to a system of equations with three variables involves determining the values of the variables that satisfy every equation simultaneously.

Sometimes these systems have:

• A unique solution: One single triplet \((x, y, z)\) that satisfies all equations.
• Infinitely many solutions: A line or plane of triplets that satisfy all equations.
• No solution at all: The planes do not intersect at a common point.

To confirm a solution exists, like our given \((-3, 5, 2)\), we substitute these numbers into each equation to check if they hold true. In our exercise:

• Equation 1: \(2x - y + 3z = -5\) checks out.
• Equation 2: \(x + 4y - z = 15\) is satisfied.
• Equation 3: \(-x + 2y + 4z = 21\) also holds true.

This demonstrates \((-3, 5, 2)\) is indeed the solution to the system.

Substitution Method

The substitution method is a powerful strategy for solving systems of equations, including those with three variables.

This method involves solving one of the equations for a single variable, then substituting this expression into the other equations.

By doing so, you reduce the complexity of the system through systematic elimination.

For three variables, you'll typically start by expressing \(x\), \(y\), or \(z\) from one equation in terms of the others. Then, substitute these expressions into the remaining equations to solve for the remaining variables. This method requires careful organization but can simplify finding the solution:

• Choose an equation and solve for one variable.
• Substitute the result into the other equations.
• Solve for the next variable using these simplified equations.

In our step-by-step solution, while the example didn't explicitly use substitution from scratch, verifying \((-3, 5, 2)\) as a solution demonstrates how substitution and checking are integral parts of solving systems of equations.