Problem 21 - Intermediate Algebra

Question

Solve each of the following problems by setting up and solving a system of three linear equations in three variables. The sum of the digits of a three-digit number is 14 . The number is 14 larger than 20 times the tens digit. The sum of the tens digit and the units digit is 12 larger than the hundreds digit. Find the number.

Step-by-Step Solution

Answer

The number is 194.

Step 1: Define the variables

Let's assign variables to represent the digits of the number. Let \( x \) be the hundreds digit, \( y \) be the tens digit, and \( z \) be the units digit. The number itself can be expressed as \( 100x + 10y + z \).

Step 2: Write the first equation

According to the problem, the sum of the digits of the number is 14. This can be expressed by the equation: \( x + y + z = 14 \).

Step 3: Write the second equation

The problem states that the number is 14 larger than 20 times the tens digit. Thus, the equation representing this statement is \( 100x + 10y + z = 20y + 14 \).

Step 4: Write the third equation

The problem also states that the sum of the tens digit and the units digit is 12 larger than the hundreds digit. This can be expressed by \( y + z = x + 12 \).

Step 5: Simplify second equation

Rearrange the second equation \( 100x + 10y + z = 20y + 14 \) to get \( 100x + z = 10y + 14 \).

Step 6: Develop a system of equations

Now we have three equations:1. \( x + y + z = 14 \)2. \( 100x + z = 10y + 14 \)3. \( y + z = x + 12 \).We'll solve this system using substitution or elimination.

Step 7: Substitution for elimination

From equation 3, express \( z = x + 12 - y \). Substitute into equations 1 and 2:- Equation 1 becomes \( x + y + (x + 12 - y) = 14 \) simplifies to \( 2x + 12 = 14 \), hence \( 2x = 2 \), so \( x = 1 \).- Substitute \( x = 1 \) into \( z = x + 12 - y \) gives \( z = 1 + 12 - y = 13 - y \).

Step 8: Solve for variable y

Using the values substituted in equation 2: \( 100(1) + (13-y) = 10y + 14 \) simplifies to \( 113 - y = 10y + 14 \), results in \( 113 - 14 = 11y \), thus \( 99 = 11y \), so \( y = 9 \).

Step 9: Solve for variable z

Substitute \( y = 9 \) in \( z = 13 - y \), gives \( z = 13 - 9 = 4 \).

Step 10: Verify and conclude

With \( x = 1 \), \( y = 9 \), and \( z = 4 \), verify with the original three conditions:- The sum of the digits: \( 1 + 9 + 4 = 14 \).- The number \( 194 = 20(9) + 14 \).- \( 9 + 4 = 1 + 12 \).All are satisfied, hence the three-digit number is correct.

Key Concepts

Three-digit Number Problems

Three-digit number problems are exercises where you break down a number into its hundreds, tens, and units (ones) digits. Understanding these helps solve problems where each digit holds a specific mathematical relationship with the other digits. For example, if you are given that the sum of the digits of a number has a particular value, you set up an equation that reflects this. In our exercise, it was:

• The sum of the digits: \( x + y + z = 14 \)

This exercise requires you to assign variables to each digit, often choosing:

• \( x \) for the hundreds digit,
• \( y \) for the tens digit,
• \( z \) for the units digit.

By expressing the entire number as \( 100x + 10y + z \), you can plug this expression into other equations based on the problem statement. This makes you apt at identifying patterns and creating equations that represent real-world scenarios.

Solving by Substitution

Solving systems of equations using substitution involves choosing one variable from one equation and expressing it in terms of other variables. This substitution is then used in other equations, diminishing the number of variables, thereby simplifying the equation.

Here's what the substitution method looked like in the exercise:

• From \( y + z = x + 12 \), solve for \( z \) yielding \( z = x + 12 - y \).

Once substituted into another equation, like \( x + y + z = 14 \), the equation simplifies to fewer variables:

• The rearranged equation becomes \( 2x + 12 = 14 \), leading us to determine that \( x = 1 \).

Substitution is a strategic choice because it allows one to isolate variables easily and can be more intuitive for certain systems, like when one of the variables seems naturally easier to express in terms of others.

Solving by Elimination

Elimination is a classic method used to solve a system of equations by removing variables, step by step, until one equation is left with one variable. Once this variable is found, you backtrack to find the others.

In the exercise, the elimination concept was closely tied with substitution. After substituting to reduce variables, we still used elimination by:

• Arriving at equations with fewer variables, such as \( 113 - y = 10y + 14 \).
• Combining and simplifying equations: Solving \( 113 - 14 = 11y \) to find \( y \), and then using \( y \) to find \( z \).

This method is favorable when equations are numerous or when they easily allow for straightforward addition or subtraction to eliminate terms. Elimination remains a powerful tool particularly in linear systems with more than two equations.