Problem 23 - 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 sum of three numbers is \(16 .\) The sum of twice the first number, 3 times the second number, and 4 times the third number is \(46 .\) The difference between 5 times the first number and the second number is \(31 .\) Find the three numbers.

Step-by-Step Solution

Answer

The first number \(x\) is \(80/9\), the second number \(y\) is \(-49/9\), and the third number \(z\) is \(31/9\). Verification is left as an exercise for those following this solution.

Step 1: Formulate the Equations

First, you need to create a system of equations from the given conditions. These are the equations: \n 1) \(x + y + z = 16\) \n 2) \(2x + 3y + 4z = 46\) \n 3) \(5x - y = 31\)

Step 2: Solve the Equations

Next, you solve the system of equations. You might find it helpful to start by reordering them as follows: 1) \(x = 31 + y\) 2) \(2x + 3y + 4z = 46\) 3) \(x + y + z = 16\). Now, replace \(x\) in equations 2) and 3) with the value from 1): 4) \(63 + 5y + 4z = 46\) 5) \(16 = 31 + 2y + z\) After simplifying these equations, you have: 6) \(5y + 4z = -17\) 7) \(2y + z = -15\). Solve for \(y\) and \(z\) by multiplying equation 6) by 2 and adding it to 7), this gives \(9y = -49\), so \(y = -49/9\). Substitute \(y\) into 7) to solve for \(z\), this gives \(z = 31/9\). Finally, substitute \(y\) and \(z\) into 1) to get \(x = 80/9\).

Step 3: Verify the Solution

Lastly, substitute the values of \(x\), \(y\), and \(z\) into the original equations to verify that they are correct solutions.

Key Concepts

Algebraic Equations

Algebraic equations are fundamental to solving numerous problems in mathematics, providing a way to represent relationships between different quantities. These equations consist of variables and constants, and they involve operations like addition, subtraction, multiplication, and division. In the context of the provided exercise, we encounter a system of algebraic equations that describe relationships between three unknown numbers.

An example of an algebraic equation from the exercise is the sum of the three numbers, expressed as x + y + z = 16. In this equation, x, y, and z represent the variables, which are the unknown numbers we are trying to find, and 16 is the constant, the sum of these numbers.

Solving Systems of Linear Equations

A system of linear equations is a set of equations with multiple variables that you can solve simultaneously. Solving these systems is often about finding the values of the variables that satisfy all equations at once. There are several methods to do this, including graphing, substitution, elimination, and matrix methods. In our example, we seek values for x, y, and z that satisfy all three given equations.

To solve the system effectively, we usually look for ways to reduce the number of variables step by step. The step-by-step solution presented simplifies the system by expressing one variable in terms of another, then substituting that expression into the other equations. By doing so, you gradually decrease the number of unknowns and make the system more manageable. After repeated substitution and simplification, you eventually arrive at a point where each variable can be solved individually.

Substitution Method

The substitution method is a powerful technique for solving systems of linear equations. It involves solving one of the equations for one variable and then substituting the resulting expression into the other equations. This method transforms the system into a simpler one with fewer variables, eventually leading to a solution.

In the example exercise, we first solve the last equation for x, expressed as x = 31 + y. This expression is then substituted into the other equations in place of x. The process simplifies the system and allows us to solve for the other variables. With each substitution, the system becomes more straightforward until each variable is isolated and can be solved. The ease of the substitution method lies in its ability to reduce complexity step by step, making it a go-to approach, especially for systems that are well-suited to this kind of manipulation.