Problem 2 - Elementary and Intermediate Algebra

Question

Solve. The sum of three numbers is \(5 .\) The first number minus the second plus the third is \(1 .\) The first minus the third is 3 more than the second. Find the numbers.

Step-by-Step Solution

Answer

The numbers are 4, 2, and -1.

Step 1: Set Up Equations

Let's define the three numbers as follows: let the first number be denoted by \(x\), the second by \(y\), and the third by \(z\). Based on the problem, we can write the following system of equations: 1. \(x + y + z = 5\)2. \(x - y + z = 1\)3. \(x - z = y + 3\)

Step 2: Simplify Equations

We can simplify the third equation to isolate \(y\). Rearrange the equation to get: \(x - z = y + 3\) thus, \(y = x - z - 3\)

Step 3: Substitute into First and Second Equations

Substitute \(y = x - z - 3\) into the first two equations:For Equation 1: \(x + (x - z - 3) + z = 5\) Simplify: \(2x - 3 = 5\) thus, \(2x = 8\) so, \(x = 4\)For Equation 2: \(x - (x - z - 3) + z = 1\)Simplify: \(x - x + z + 3 + z = 1\) to get: \(2z + 3 = 1\) thus, \(2z = -2\) so, \(z = -1\)

Step 4: Find the Value of y

Now use the value of \(x\) and \(z\) to find \(y\): \(y = x - z - 3\)\(y = 4 - (-1) - 3\)\(y = 4 + 1 - 3\)\(y = 2\)

Step 5: Verify the Solution

Verify the values by substituting \(x = 4\), \(y = 2\), and \(z = -1\) back into the original equations:1. \(x + y + z = 4 + 2 - 1 = 5\) (True)2. \(x - y + z = 4 - 2 - 1 = 1\) (True)3. \(x - z = y + 3\) which simplifies to \(4 - (-1) = 2 + 3 = 5\) (True)All equations are satisfied.

Key Concepts

Algebraic Equations

Algebraic equations involve variables, numbers, and mathematical operations. In the given exercise, we deal with three algebraic equations:

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

These equations describe relationships between three unknown variables (\(x\), \(y\), \(z\)). To solve such systems, we need strategies to find the values of these variables.

Substitution Method

The substitution method is used to solve equations by expressing one variable in terms of the others and substituting this into the remaining equations. In the exercise, we isolated \( y \) in terms of \( x \) and \( z \) from Equation 3: \( y = x - z - 3 \)Next, we substituted \( y \) in Equations 1 and 2:

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

This substitution step allows us to reduce the number of variables and simplify the solving process.

Verification of Solutions

Verification ensures the solution satisfies all original equations. After solving, we obtained \( x = 4 \), \( y = 2 \), and \( z = -1 \). We substitute these values back into the original equations:

• Equation 1: \( 4 + 2 - 1 = 5 \) (Ttrue)
• Equation 2: \(4 - 2 - 1 = 1 \) (Ttrue)
• Equation 3: \( 4 - (-1) = 2 + 3 \) (Ttrue)

Each equation holds true, confirming the solution is correct.

Linear Equations

Linear equations are equations of the first order, meaning each term is either a constant or the product of a constant and a single variable. The solved exercise consisted of three linear equations. Linear equations such as: \( x + y + z = 5 \) have solutions that can be found using methods like substitution or elimination. By understanding the properties of linear equations, students can apply systematic methods to find the values of unknowns effectively.