Problem 24 - Algebra and Trigonometry

Question

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

Therefore, the three numbers are \(x = 0\), \(y = 26/5\) and \(z = -1/5\).

Step 1: Writing down the system of equations

Let's form the three equations from the provided statements. The first statement gives us the equation, \(3x + y + 2z = 5\). From the second statement we get, \(x + 3z - 3y = 2\). And the third statement gives us the third equation, \(2x + 3y - z = 1\).

Step 2: Solving the first two equations simultaneously

To simplify, start by multiplying the second equation by 3 and the first by 1 and then add the resulting equations. \((3x + y + 2z) + 3(x + 3z - 3y) = 5 + 3*2\), which simplifies to \(6x + 11z = 11\).

Step 3: Solving the last two equations

Now, multiply the third equation by 3 and the second by 2 and then subtract the resulting equations. \((2x + 3y - z)*3 - 2*(x + 3z - 3y) = 3 - 4\), which simplifies to \(4x - 5z = -1\).

Step 4: Finding the value of x and z

We solve the two equations obtained above for \(x\) and \(z\). Multiply the first obtained equation by 5 and the second by 11, then add the resulting equations, it simplifies to \(76x = 0\), or \(x = 0\). Substitute \(x = 0\) into \(4x - 5z = -1\), we find \(z = -1/5\).

Step 5: Finding the value of y

Substitute the values of \(x\) and \(z\) into the first given original equation \(3x + y + 2z = 5\). This gives us \(y = 5 - 2*-1/5 = 5 + 2/5 = 26/5\).

Key Concepts

Solving Linear Equations

When dealing with linear equations, the aim is to find the values of unknown variables that satisfy all equations involved. In a system of equations with multiple linear equations, we seek solutions that make each equation true simultaneously. For instance, in our problem, we have three equations:

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

These equations are linked through the variables \(x\), \(y\), and \(z\). Solving them requires finding a single set of values for these variables that fulfill all three equations. Each equation represents a relationship among the variables, and solving them typically involves techniques like substitution or elimination. This process requires patience and a methodical approach to isolate and determine the values of each variable individually.

Substitution Method

The substitution method is a powerful tool for solving systems of equations. It involves solving one of the equations for one variable in terms of another, and then substituting this expression into the other equations. This reduces the number of equations and variables, making the system easier to solve. In our solution process, we used this method implicitly. Imagine if we focused initially on isolating one variable from an equation. After finding a simple expression for one, we substitute it back into the other equations, making them more manageable. This technique is especially useful when the system of equations is not readily solvable by inspection and needs simplification.

Algebraic Manipulation

Algebraic manipulation is at the heart of solving equations. It involves applying algebraic operations to simplify and solve equations. The step-by-step solution to our problem demonstrated several key manipulations:

• Multiplying equations by constants to align coefficients for easier elimination.
• Adding or subtracting equations to eliminate variables.
• Substituting known variable values back into equations to find unknowns.

For example, in Step 2, we manipulated the first and second equations to eliminate one of the variables, simplifying the problem. Such manipulations are vital as they transform complex equations into simpler forms that are easier to solve. Understanding and practicing these operations enhance your ability to tackle more intricate mathematical problems effectively.