Problem 68 - College Algebra

Question

will help you prepare for the material covered in the next section. In each exercise, refer to the following system: $$ \left\\{\begin{array}{c} 3 x-4 y+4 z=7 \\ x-y-2 z=2 \\ 2 x-3 y+6 z=5 \end{array}\right. $$ Show that \((12 z+1,10 z-1, z)\) satisfies the system for \(z=0\)

Step-by-Step Solution

Answer

By substituting \(z=0\) into \((12 z+1,10 z-1, z)\), we get the point \((1,-1,0)\). Checking this point against each of the three equations verifies that it does indeed satisfy the entire system of equations. Therefore it can be concluded that \((12 z+1,10 z-1, z)\) satisfies the system when \(z=0\).

Step 1: Insert z into the variables

Replace \(z\) in \(12 z+1,10 z-1, z\) by \(0\) to get the point \((1,-1,0)\).

Step 2: Substitute into First Equation

Check the first equation in the system, \(3x - 4y + 4z = 7\), substituting the values of \(x=1, y=-1, z=0\). Upon calculation, it becomes \(3(1)-4(-1)+4(0) = 7\), which simplifies to 7=7, and is true.

Step 3: Substitute into Second Equation

Now check the second equation in the system, \(x - y - 2z = 2\) using \(x=1, y=-1, z=0\). It simplifies to \(1-(-1)-2(0)=2\), which is true as 2=2.

Step 4: Substitute into Third Equation

Finally, validate the last equation \(2x-3y+6z=5\) using \(x=1, y=-1, z=0\). This results in \(2(1)-3(-1)+6(0)=5\), which simplifies to 5=5, and is also true.

Key Concepts

Solution Verification

Solution verification is the process of ensuring the solution satisfies all given equations in a system. This is especially important to confirm correctness after solving a system of equations. To verify a solution, we insert the solution values into each equation and check if they satisfy the equality.

• Start by substituting the given solution into the first equation.
• Check if the left-hand side equals the right-hand side.
• Repeat the process for each subsequent equation in the system.

When all equations hold true, you can be confident that the solution is correct. In our exercise, substituting \((x, y, z) = (1, -1, 0)\) into each equation confirmed that each holds true, verifying the solution.

Substitution Method

The substitution method is a fundamental technique for solving systems of linear equations. It involves solving one equation for one variable and substituting that expression into another equation.
Here's a breakdown of the substitution method:

• Firstly, solve one of the equations for one of its variables. In our exercise, the equations are already solved for expressions involving \(z\).
• Substitute the expression from the first step into the other equations. This reduces the system to fewer variables.
• Solve the resulting equations one by one for the variables.

This method is especially helpful when dealing with systems that are easily manipulated, like when one variable is isolated, as it simplifies the overall process.

Linear Equations

Linear equations form the backbone of systems of equations often encountered in algebra. They appear in one or more variables and are graphed as straight lines.
Characterizing linear equations:

• They possess a constant rate of change, described by a slope in two dimensions.
• They can be expressed in a standard form such as \(ax + by + cz = d\).
• Linear systems with three variables can be visualized as intersecting planes.

In a system like the one in our exercise, each equation represents a plane. The solution \((12z+1, 10z-1, z)\) is verified to satisfy all planes at once when \(z = 0\), corresponding to a point of intersection.