Problem 22 - Calculus for Biology and Medicine

Question

In Problems 19-24, solve each system of linear equations. $$ \begin{array}{r} -2 x+4 y-z=-1 \\ x+7 y+2 z=-4 \\ 3 x-2 y+3 z=-3 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{2}{7}\), \(y = -\frac{12}{35}\), \(z = -\frac{33}{35}\).

Step 1: Identify the System of Equations

We start with the system of linear equations:\[\begin{align*}-2x + 4y - z &= -1, \x + 7y + 2z &= -4, \3x - 2y + 3z &= -3.\end{align*}\]

Step 2: Choose a Variable to Eliminate

We will eliminate the variable \(x\). This can be done by making the coefficients of \(x\) the same in two different equations and then subtracting them.

Step 3: Eliminate \(x\) from Equations 1 and 2

Multiply Equation 2 by 2, then add it to Equation 1 to eliminate \(x\):\[\begin{align*}2(x + 7y + 2z) &= 2(-4) \2x + 14y + 4z &= -8.\end{align*}\]Add this result to Equation 1:\[\begin{align*}(-2x + 4y - z) + (2x + 14y + 4z) &= -1 - 8 \18y + 3z &= -9. \\end{align*}\]Simplifying, we get:\[ 6y + z = -3. \]

Step 4: Eliminate \(x\) from Equations 1 and 3

Multiply Equation 3 by 2/3, then add it to Equation 1 to eliminate \(x\):\[\begin{align*}2/3(3x - 2y + 3z) &= -2 \2x - \frac{4}{3}y + 2z &= -2. \end{align*}\]Now subtract this from Equation 1:\[\begin{align*}(-2x + 4y - z) - (2x - \frac{4}{3}y + 2z) &= -1 - (-2) \\frac{16}{3}y - 3z &= 1.\end{align*}\]Multiply everything by 3 to clear the fractions:\[16y - 9z = 3.\]

Step 5: Solve the New System in Terms of \(y\) and \(z\)

We have two new equations:\[\begin{align*}6y + z &= -3, \16y - 9z &= 3.\end{align*}\]We'll solve this system by elimination or substitution. Let's use substitution. Solve the first for \(z\):\[z = -3 - 6y.\]

Step 6: Substitute for \(z\) and Solve for \(y\)

Substitute \(z = -3 - 6y\) into the second equation:\[16y - 9(-3 - 6y) = 3.\]Distribute:\[16y + 27 + 54y = 3.\]Combine like terms:\[70y = 3 - 27.\]\[70y = -24.\]Solve for \(y\):\[y = -\frac{24}{70} = -\frac{12}{35}.\]

Step 7: Solve for \(z\)

Substitute \(y = -\frac{12}{35}\) into the equation for \(z\):\[z = -3 - 6(-\frac{12}{35})\]\[z = -3 + \frac{72}{35}.\]Convert \(-3\) to a fraction:\[-3 = -\frac{105}{35}.\]\[z = -\frac{105}{35} + \frac{72}{35} = -\frac{33}{35}.\]

Step 8: Solve for \(x\)

Using \(x + 7y + 2z = -4\), substitute \(y = -\frac{12}{35}\) and \(z = -\frac{33}{35}\).\[ x + 7(-\frac{12}{35}) + 2(-\frac{33}{35}) = -4\]\[ x - \frac{84}{35} - \frac{66}{35} = -4\]Combine terms:\[ x - \frac{150}{35} = -4 \]Solve for \(x\):\[ x = -4 + \frac{150}{35} \]Convert \(-4\) to a fraction:\[-4 = -\frac{140}{35}\]. Solve: \[\Rightarrow x = \frac{10}{35} = \frac{2}{7}.\]

Step 9: Solution Verification

Check the solutions \(x = \frac{2}{7}, y = -\frac{12}{35}, z = -\frac{33}{35}\) in all three original equations to verify correctness.

Key Concepts

Elimination Method

The elimination method is an efficient technique for solving systems of linear equations. It simplifies the problem by removing, or "eliminating," one variable at a time, making it possible to solve the system using the remaining variables.

First, we identify pairs of equations and decide which variable to eliminate. This involves manipulating the equations to make the coefficients of one of the variables equal in two different equations. Once the coefficients are equal, we add or subtract the equations to eliminate the chosen variable.

For example, in our exercise, we start with the goal of eliminating the variable \(x\) from the system. By carefully selecting operations (like multiplying an equation by a constant and then adding or subtracting from another), we can transform the system into a simpler one. In our case, this gave us new equations involving just \(y\) and \(z\), which are easier to handle.

This method is particularly useful in systems where one can easily manipulate equations through multiplication or addition to eliminate a variable. It works excellently in linear systems with two or three variables.

Substitution Method

The substitution method is another powerful approach to solving systems of linear equations. This method involves solving one of the equations for one variable explicitly and then substituting that expression into the other equations.

In practice, we begin by solving one of the equations in the system for one variable. We then replace that variable in the other equation(s) with the expression we found.

For instance, our example involved transforming the first of our new simplified equations for \(z\): we expressed \(z\) in terms of \(y\) as \(z = -3 - 6y\). This formula was used to substitute \(z\) in the other equation (the one without \(x\) after elimination), making it possible to solve it for \(y\).

Once \(y\) is determined, we can find the value of \(z\) by substituting \(y\) back into the expression for \(z\). Substitution is particularly useful when one of the equations in a system easily allows solving for one of the variables.

Solving Linear Systems

Solving linear systems refers to finding values for variables that satisfy all equations in the system simultaneously. A successful solution involves getting expressions where each variable can be independently determined.

These systems can be solved using various methods such as substitution, elimination, or even graphing.

• Substitution: This method involves expressing one variable in terms of others and substituting it across the equations.
• Elimination: This technique focuses on eliminating terms systematically until the system is reduced to a single-variable equation.
• Graphing: When dealing with two-variable systems, graphing each equation and finding their intersection can provide the solution.

For complex systems involving three or more equations, like in our exercise, often a combination of substitution and elimination is employed. In this specific exercise, after applying the elimination method to simplify variables \(x\) and \(z\), substitution helped reach the solution effectively.

Efficiently solving linear systems requires understanding the advantages and limitations of each method, allowing you to strategically choose the best approach for the given system.