Problem 24 - College Algebra

Question

In Exercises \(23-24\), let \(x\) represent the first number, \(y\) the second number, and \(z\) the third number. Use the given conditions to write a system of equations. Solve the system and find the numbers. The following is known about three numbers: Three times the first number plus the second number plus twice the third number is \(5 .\) If 3 times the second number is subtracted from the sum of the first number and 3 times the third number, the result is \(2 .\) If the third number is subtracted from 2 times the first number and 3 times the second number, the result is \(1 .\) Find the numbers.

Step-by-Step Solution

Answer

The values of the numbers \(x, y, z\) are 1, 2 and 3 respectively.

Step 1: Translating Words into Equations

Based on the given conditions, we can create the following system of equations: \n\n1. \(3x + y + 2z = 5\)\n2. \(x + 3z - 3y = 2\)\n3. \(2x + 3y - z = 1\)

Step 2: Solving the Equation

Use substitution or elimination to find values for \(x, y,\) and \(z\). Let's employ substitution: \n\nFrom equation (2), express \(x\) in terms of \(y\) and \(z\): \(x = 3y + 2 - 3z\). \n\nSubstitute \(x\) into equation (3):\n \(2(3y + 2 - 3z) + 3y - z = 1 \)\n Simplify: \(6y + 4 - 6z + 3y - z = 1 \)\n Put like terms together and simplify: \(9y - 7z = -3 \)\n\n Amend equation (1) to look like equation (4) by multiplying it by 3: \(9x + 3y + 6z = 15\)\n\nSubstitute equation (4) into improved equation (1):\n \(9(3y + 2 - 3z) + 3y + 6z = 15\)\n Simplify: \(27y + 18 - 27z + 3y + 6z = 15\)\n Put like terms together and simplify: \(30y - 21z = -3\)\n\nNow we have two new equations: \(9y - 7z = -3\) and \(30y - 21z = -3\). Solve them to find \(y\), and \(z\). Afterwards, we will find the value of \(x\).

Step 3: Finding the Values

Multiply the first equation by 30 and the second by 9 \n\n270y - 210z = -90 (5)\n270y - 189z = -27 (6)\n\nSubtract equation (6) from (5) to get z:\n \n-21z = -63\nSo, z = 3\n\nSubstitute z = 3 into the equation 9y - 7z = -3 to get y:\n\n9y - 7(3) = -3\n9y - 21 = -3\nSo, y = 2\n\nSubstitute y = 2 and z = 3 into the equation x = 3y + 2 - 3z to get x:\n\nx = 3(2) + 2 - 3(3) \nSo, x = 1

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve systems of equations. It involves expressing one variable in terms of others, then substituting this expression into other equations in the system. This method helps you find the values of the unknowns step-by-step.
To use substitution, begin by isolating one of the variables in one of the equations. For example, from equation (2) in the original problem, we expressed \(x\) in terms of \(y\) and \(z\): \(x = 3y + 2 - 3z\).
Next, substitute this expression into another equation. We substituted \(x = 3y + 2 - 3z\) into equation (3), which led to:

• \(2(3y + 2 - 3z) + 3y - z = 1\)
• This further simplifies to \(9y - 7z = -3\)

By this substitution, we created a new equation with only two variables. This new simplified system can then be solved using the elimination method or repetition of substitution.

Elimination Method

The elimination method is another effective approach for solving systems of equations. It involves adding or subtracting equations to eliminate a variable, making it possible to solve for the remaining variables.
In the context of the original exercise, after using substitution, two new equations were formed:

• \(9y - 7z = -3\)
• \(30y - 21z = -3\)

To eliminate variables using the elimination method, align the equations so that adding them cancels out one variable. This may require multiplying one or both equations by a coefficient to align them.
In this exercise, equations were multiplied to make the coefficients of \(y\) equal, resulting in:

• \(270y - 210z = -90\) (from multiplying the first equation by 30)
• \(270y - 189z = -27\) (from multiplying the second equation by 9)

Subtracting these equations eliminates \(y\) and solves for \(z\). Afterwards, substitute back to find the other unknowns. This step-by-step elimination leads to finding the solution for all variables efficiently.

Solving Linear Equations

Solving linear equations is an essential algebraic skill which involves finding the values of unknown variables that make the equation true. It is at the heart of solving systems of equations.
Linear equations can typically be written in the form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are coefficients, and \(x\), \(y\), and \(z\) are variables. In the given exercise, each equation relates these concepts:

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

The goal is to find a common solution for these equations. By converting the given information into solvable linear equations and applying methods like substitution or elimination, variable values can be systematically deduced.
Whether using substitution or elimination, the key to solving linear equations is breaking them down into simpler components. This simplification can then be used strategically to isolate and solve for each variable individually, building to the complete solution step by step.