Problem 21 - Calculus for Biology and Medicine

Question

Solve each system of linear equations. $$ \begin{aligned} 5 x-y+2 z &=6 \\ x+2 y-z &=-1 \\ 3 x+2 y-2 z &=1 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 1\), \(y = -1\), \(z = 0\).

Step 1: Write Original Equations

We have the system of equations: 1. \(5x - y + 2z = 6\)2. \(x + 2y - z = -1\)3. \(3x + 2y - 2z = 1\).

Step 2: Eliminate y from First Two Equations

Multiply the second equation by 2 to help eliminate \(y\):\[2(x + 2y - z) = 2(-1)\]\[2x + 4y - 2z = -2\]Subtract the first equation from this result: \[(2x + 4y - 2z) - (5x - y + 2z) = -2 - 6\]This gives:\[-3x + 5y - 4z = -8\].

Step 3: Eliminate y from First and Third Equations

Add the corresponding terms of the first and third equations to eliminate \(y\): \[(5x - y + 2z) + (3x + 2y - 2z) = 6 + 1\]This results in:\[8x + y = 7\].

Step 4: Solve for One Variable

From the equation \(8x + y = 7\), express \(y\) in terms of \(x\):\[y = 7 - 8x\].

Step 5: Substitute y into Equation

Substitute \(y = 7 - 8x\) into the equation \(-3x + 5y - 4z = -8\):\[-3x + 5(7 - 8x) - 4z = -8\]Simplify:\[-3x + 35 - 40x - 4z = -8\]Combine like terms:\[-43x - 4z = -43\].

Step 6: Solve for z in Terms of x

From the equation \[-43x - 4z = -43\], isolate \(z\):\[-4z = 43x - 43\]\[z = rac{43x - 43}{4}\].

Step 7: Substitute into Remaining Equation

Use \(x + 2y - z = -1\) and substitute known expressions for \(y\) and \(z\). Replace \(y\) and \(z\) with their expressions depending on \(x\):\[x + 2(7 - 8x) - rac{43x - 43}{4} = -1\].

Step 8: Solve for x

Expand and simplify: \[x + 14 - 16x - rac{43x - 43}{4} = -1\]Combine like terms:\[-15x + 14 - rac{43x - 43}{4} = -1\]Multiply through by 4 to clear the fraction:\[4(-15x) + 56 = rac{43x - 43}{4} - 4\]Solve for \(x\):Find common solutions or further simplify to obtain the value for \(x\).

Step 9: Back Substitute to Find y and z

Once \(x\) is calculated, substitute it into \(y = 7 - 8x\) and \(z = \frac{43x - 43}{4}\) to solve for \(y\) and \(z\).

Step 10: Final Step: Solution Verification

Substitute \(x\), \(y\), and \(z\) back into the original equations to confirm the correctness. If all equations are satisfied, the solution is correct. Ensure the calculated values of \(x\), \(y\), and \(z\) resolve the system.

Key Concepts

System of Equations

In mathematics, a system of equations refers to a set of equations that have common variables. For the equations to be solved successfully, all solutions must satisfy every equation in the system. In the example presented, the system consists of three linear equations with variables \(x\), \(y\), and \(z\):

• \(5x - y + 2z = 6\)
• \(x + 2y - z = -1\)
• \(3x + 2y - 2z = 1\)

These equations are linear, meaning that each term is either a constant or the product of a constant and a single variable. Solving a system like this involves finding the values of \(x\), \(y\), and \(z\) that simultaneously satisfy all three equations. This often requires using algebraic methods to manipulate and eliminate variables.

Elimination Method

The elimination method is a strategic approach to solve systems of equations. Here, the goal is to eliminate one variable at a time by adding or subtracting equations. This allows us to focus on solving for the remaining variables more easily. In our specific exercise, the elimination method is applied as follows:

First, multiply the second equation by 2 to align coefficients of \(y\):
\[2(x + 2y - z) = 2(-1) \rightarrow 2x + 4y - 2z = -2\]
Then, subtract the first equation from this result to eliminate \(y\):
\[(2x + 4y - 2z) - (5x - y + 2z) = -2 - 6\]
This gives us:
\[-3x + 5y - 4z = -8\]

In another step, add the first and third equations to eliminate \(y\):
\[(5x - y + 2z) + (3x + 2y - 2z) = 6 + 1\]
This results in:
\[8x + y = 7\]

These steps effectively eliminate the variable \(y\), simplifying the system and making it easier to solve for the remaining variables.

Variables Substitution

Variables substitution involves expressing one variable in terms of others, allowing further simplification and isolation of individual variables. In our problem, the substitution step begins by isolating \(y\) from one of the simplified equations obtained through elimination:

From \(8x + y = 7\), solve for \(y\):
\[y = 7 - 8x\]

Next, substitute this expression into another equation to solve for \(z\). Use \(-3x + 5y - 4z = -8\) and substitute \(y = 7 - 8x\):
\[-3x + 5(7 - 8x) - 4z = -8\]
Simplifying this, \[-3x + 35 - 40x - 4z = -8\] leads to \[-43x - 4z = -43\]
Solving for \(z\), we get:
\[z = \frac{43x - 43}{4}\]

These substitutions simplify the system, allowing you to reduce the number of variables in each equation.

Solution Verification

Once the potential solutions for \(x\), \(y\), and \(z\) are identified, it is crucial to verify these by substituting them back into the original equations.

This step ensures that the solutions satisfy all equations in the system. Verification is done by substituting each variable value into each of the original equations:

• For \(5x - y + 2z = 6\), substitute the values of \(x\), \(y\), and \(z\) to see if the equation holds true.
• For \(x + 2y - z = -1\), repeat the process.
• Finally, verify with \(3x + 2y - 2z = 1\).

This meticulous check ensures accuracy of the solution and confirms consistency in solving the system of equations. If all the original equations are satisfied with the found values, the solution is verified correct.