Problem 8 - Elementary and Intermediate Algebra

Question

Determine whether \((-1,-3,2)\) is a solution of the system $$\begin{array}{r} {x-y+z=4} \\ {x-2 y-z=3} \\ {3 x+2 y-z=1} \end{array}$$

Step-by-Step Solution

Answer

No, (-1, -3, 2) is not a solution.

Step 1 - Substitute the point into the first equation

Substitute \(x = -1\), \(y = -3\), and \(z = 2\) into the first equation \(x - y + z = 4\).\[-1 - (-3) + 2 = 4\]Simplify:\[-1 + 3 + 2 = 4\]\[4 = 4\]This is true.

Step 2 - Substitute the point into the second equation

Substitute \(x = -1\), \(y = -3\), and \(z = 2\) into the second equation \(x - 2y - z = 3\).\[-1 - 2(-3) - 2 = 3\]Simplify:\[-1 + 6 - 2 = 3\]\[3 = 3\]This is true.

Step 3 - Substitute the point into the third equation

Substitute \(x = -1\), \(y = -3\), and \(z = 2\) into the third equation \(3x + 2y - z = 1\).\[3(-1) + 2(-3) - 2 = 1\]Simplify:\[-3 - 6 - 2 = 1\]\[-11 eq 1\]This is not true.

Key Concepts

solution verification

Verifying a solution involves checking if a given set of values satisfies all equations in the system. For this exercise, our goal was to determine if \((-1, -3, 2)\) is a solution to the system of equations. Simply put, we substitute these values into each equation one by one to see if the left-hand side equals the right-hand side.

For the first equation \(x - y + z = 4\):

\((-1) - (-3) + 2 = 4\):

Simplifying gives us: \-1 + 3 + 2 = 4\. This is true.

So, the point satisfies the first equation.

For the second equation \(x - 2y - z = 3\):

\((-1) - 2(-3) - 2 = 3\):

Simplifying gives us: \-1 + 6 - 2 = 3\. This is true.

The point satisfies the second equation, too.

For the third equation \(3x + 2y - z = 1\):

\3(-1) + 2(-3) - 2 = 1\:

Simplifying gives us: \-3 - 6 - 2 = 1\. This is false because \-11 ≠ 1\.

This means \((-1, -3, 2)\) doesn't satisfy the system of equations because it fails the third equation.

linear equations

Linear equations are equations where each term is either a constant or the product of a constant and a single variable. They can be written in the form \(ax + by + cz = d\), where \(a, b, c,\) and \(d\) are constants.

Linear equations are central to solving systems of equations. In this exercise, we dealt with three linear equations:

• \x - y + z = 4\
• \x - 2y - z = 3\
• \3x + 2y - z = 1\

Each of these equations represents a plane in 3-dimensional space.

The intersection point of these planes is the solution to the system, if it exists.

In our original exercise, we checked whether the point \((-1, -3, 2)\) lay on all three planes. It did on the first two, but not on the third. Hence, it is not a solution to the system.

substitution method

The substitution method is a way to solve a system of equations by replacing one variable with an expression involving the other variables. This method simplifies the system step by step.

Here is how to use the substitution method:

• Solve one equation for one variable
• Substitute that expression into the other equations, reducing the number of variables
• Repeat until you have a single linear equation
• Solve for the unknown variable, then backtrack to find other variables

In our exercise, although we didn’t explicitly use the substitution method, understanding how to substitute values correctly was crucial.

Substituting \((-1, -3, 2)\) into each equation helped us check if it was a valid solution.

In more complex problems, using the substitution method can simplify the process of finding correct solutions for systems of equations.