Problem 14 - College Algebra

Question

Solve each system by substitution. $$ \begin{aligned} 4 x-3 y+5 z &=31 \\ -x+2 y+4 z &=20 \\ x+5 y-2 z &=-29 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = -2\), \(y = -3\), \(z = 6\).

Step 1: Solve for One Variable in One Equation

Let's start by solving the third equation for variable \(x\):\[ x + 5y - 2z = -29 \]Rearrange it to express \(x\) in terms of \(y\) and \(z\):\[ x = -5y + 2z - 29 \]

Step 2: Substitute Expression into Another Equation

Substitute \(x = -5y + 2z - 29\) into the first equation:\[ 4(-5y + 2z - 29) - 3y + 5z = 31 \]Simplify and solve for \(y\) and \(z\):\[-20y + 8z - 116 - 3y + 5z = 31 \]Combine like terms:\[-23y + 13z = 147 \]

Step 3: Substitute Expression into Another Equation

Now substitute \(x = -5y + 2z - 29\) into the second equation:\[ -(-5y + 2z - 29) + 2y + 4z = 20 \]Simplify and solve for \(y\) and \(z\):\[ 5y - 2z + 29 + 2y + 4z = 20 \]Combine like terms:\[ 7y + 2z = -9 \]

Step 4: Solve the Two Equations

We now have two equations:1. \(-23y + 13z = 147\)2. \(7y + 2z = -9\)Let's solve this system using substitution. Solve the second equation for \(z\):\[ 2z = -9 - 7y \]\[ z = \frac{-9 - 7y}{2} \]

Step 5: Solve for Both \(y\) and \(z\)

Substitute \(z = \frac{-9 - 7y}{2}\) into the first equation:\[-23y + 13\left(\frac{-9 - 7y}{2}\right) = 147 \]Multiply through by 2 to clear the fraction:\[-46y + 13(-9 - 7y) = 294 \]Distribute:\[-46y - 117 - 91y = 294 \]Combine like terms:\[-137y - 117 = 294 \]Add 117:\[-137y = 411 \]Divide by -137:\[y = -3 \]

Step 6: Back-Substitute to Find \(z\)

Use \(y = -3\) in the equation for \(z\):\[ z = \frac{-9 - 7(-3)}{2} \]\[ z = \frac{-9 + 21}{2} \]\[ z = 6 \]

Step 7: Back-Substitute to Find \(x\)

Now use \(y = -3\) and \(z = 6\) in the expression for \(x\):\[ x = -5(-3) + 2(6) - 29 \]\[ x = 15 + 12 - 29 \]\[ x = -2 \]

Step 8: Verify the Solution

Substitute \(x = -2\), \(y = -3\), \(z = 6\) back into the original equations to ensure they all hold true:1. \(4(-2) - 3(-3) + 5(6) = 31\)2. \(-(-2) + 2(-3) + 4(6) = 20\)3. \(-2 + 5(-3) - 2(6) = -29\)All the equations are satisfied, confirming that the solution is correct: \(x = -2\), \(y = -3\), \(z = 6\).

Key Concepts

Substitution Method

The substitution method is a powerful strategy for solving systems of equations, especially when dealing with linear equations. This method involves solving one of the equations for one variable in terms of the others and then substituting this expression into the other equations. By doing so, you reduce the number of variables and solve the system step by step.
Generally, the substitution method involves the following steps:

• Choose an equation and solve for one variable. It's generally easier to start with an equation where a variable is already isolated or can easily be isolated.
• Substitute the expression obtained into the other equations. This incorporates the expression for the isolated variable into other equations.
• Solve these modified equations. You now work with simpler equations involving fewer variables.
• Back-substitute the solutions into the earlier expressions to find the values of the remaining variables.
• Verify the solution by substituting all variables back into the original equations to ensure they satisfy all the equations simultaneously.

Using the substitution method starts with simplifying the system as much as possible, leading to a clear and logical step-by-step process, making finding solutions approachable and less cumbersome.

Linear Equations

Linear equations are the building blocks of the system of equations in mathematics. They are algebraic expressions that represent straight lines when graphed on the coordinate plane. A linear equation typically looks like this:\[ ax + by + cz = d \] where \(x\), \(y\), and \(z\) are variables, and \(a\), \(b\), \(c\), and \(d\) are constants.
Key characteristics of linear equations include:

• The highest exponent of any variable is 1, which ensures the equation is linear.
• When graphed, they form straight lines on the plane.
• They are used to describe relationships with constant rates of change.

In a system, like the one presented in the exercise, multiple linear equations are combined. Solving such a system means finding the values of the variables that make all the equations true simultaneously. Often, methods like substitution are employed to manage this efficiently. Linear equations provide simplicity and utility in representing and solving real-world problems.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations and expressions to simplify or solve them. It forms the core of solving systems of equations, as seen in this exercise by the stepwise approach to substitution and simplification.
Here are some key components of algebraic manipulation:

• Transposition: Moving terms from one side of an equation to another to isolate a variable.
• Combining like terms: Adding or subtracting terms that have the same variables and powers to simplify equations.
• Distribution: Using the distributive property to expand or factor expressions, \(a(b + c) = ab + ac\).
• Clearing fractions: Multiplying through by the denominator to eliminate fractions in an equation.

Efficient algebraic manipulation relies on understanding these fundamental operations to reduce complexity step by step. This makes the solution path clearer and helps reach the answers more effectively. Proper application of these techniques ensures the problem-solving process is both systematic and accurate. Understanding and practicing algebraic manipulation helps in developing strong mathematical intuition and problem-solving skills.