Problem 51 - Algebra A Combined Function

Question

The fraction \(\frac{1}{24}\) can be written as the following sum: \(\frac{1}{24}=\frac{x}{8}+\frac{y}{4}+\frac{z}{3}\) where the numbers \(x, y,\) and \(z\) are solutions of \(\left\\{\begin{aligned} x+y+z=& 1 \\ 2 x-y+z=& 0 \\\\-x+2 y+2 z=&-1 \end{aligned}\right.\) Solve the system and see that the sum of the fractions is \(\frac{1}{24}\).

Step-by-Step Solution

Answer

\(x = 5\), \(y = 3\), \(z = -1\); the fraction sum is \(\frac{1}{24}\).

Step 1: Write Down the System of Equations

We have the following system of equations based on the problem: \[\begin{aligned}x+y+z &= 1 \2x - y + z &= 0 \-x + 2y + 2z &= -1 \\end{aligned}\] Our task is to solve these equations to find the values of \(x\), \(y\), and \(z\).

Step 2: Eliminate Variables

We will start by eliminating one of the variables to simplify the system. Let's eliminate \(z\) by subtracting the first equation from the second: \( (2x - y + z) - (x + y + z) = 0 - 1 \), which simplifies to \(x - 2y = -1\).

Step 3: Substitute and Simplify

Using the equation \(x - 2y = -1\) and the third original equation \(-x + 2y + 2z = -1\), substitute \(x = 2y - 1\) from the first equation into the third one. So \(- (2y - 1) + 2y + 2z = -1\), which simplifies to \(2z = -2\) or \(z = -1\).

Step 4: Find Remaining Variables

Substitute \(z = -1\) back into equation \(x + y + z = 1\): \(x + y - 1 = 1\) gives \(x + y = 2\). Additionally, using the reduced equation \(x - 2y = -1\), substitute \(x = 2 + y\): \(2 + y - 2y = -1\), or \(2 - y = -1\), which gives \(y = 3\). Now, substitute \(y = 3\) back to find \(x\): \(x = 2 + 3 = 5\).

Step 5: Write Fraction Substitute Results

We substitute back into the fraction sum: \[\frac{5}{8} + \frac{3}{4} + \frac{-1}{3}\]. Calculate: \(\frac{5}{8} = 0.625\), \(\frac{3}{4} = 0.75\), and \(\frac{-1}{3} = -0.3333\). Adding those: \(0.625 + 0.75 - 0.3333 = 0.0417\) which is \(\frac{1}{24}\). This confirms our solution is correct.

Key Concepts

Fractions

Fractions are parts of a whole number. They consist of a numerator, the top part of the fraction, and a denominator, the bottom part.
In our problem, we deal with fractions like \(\frac{1}{24}, \frac{x}{8}, \frac{y}{4},\) and \(\frac{z}{3}\).
These fractions help express the equation:\[\frac{1}{24} = \frac{x}{8} + \frac{y}{4} + \frac{z}{3}\]By manipulating fractions, we can find the values of unknowns like \(x, y, z\).
When adding fractions, it's important to have a common denominator to combine them properly. This eventually verifies if the sum equals to \(\frac{1}{24}\).
Remember, fractions can represent very small or very large numbers, making them versatile in solving systems of linear equations.

Elimination Method

The elimination method is a technique used to solve systems of linear equations. By canceling out one variable, you can simplify the equations.
In our exercise, we first eliminated \(z\) by subtracting one equation from another:
\[(2x - y + z) - (x + y + z) = 0 - 1\]
This simplifies to \(x - 2y = -1\).

By doing this, we reduced the number of variables, making it easier to find the specific values for each variable.
The goal of elimination is to reduce the system of equations to a simpler form, which can easily be solved by back-substitution or further elimination.
Elimination method requires careful arithmetic but once mastered, it can speed up solving linear systems.

Substitution Method

The substitution method involves replacing one variable with another expression derived from one of the equations.
In this problem, after using the elimination method, we substitute \(x = 2y - 1\) into another equation in place of \(x\).
This simplifies the equation, allowing us to find specific values for the variables.
For instance, substituting \(x = 2y - 1\) into the equation \(-x + 2y + 2z = -1\) results in:\[-(2y - 1) + 2y + 2z = -1\]
This solves to \(z = -1\).

By substituting back and forth between the derived expressions and original equations, all variable values can be determined efficiently.
This method perfectly complements elimination when solving complex systems.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations such as addition and subtraction.
In our system of equations, equations are algebraic expressions where unknowns \(x, y,\) and \(z\) are combined in a meaningful pattern.
Each equation like \(x+y+z=1\) represents a relationship between these variables.
Understanding how to manipulate algebraic expressions using methods like distribution, combining like terms, and simplification is crucial in solving equations.
When you rewrite these expressions in different forms, either for substitution or elimination, it often reveals the solutions systematically.
Mastery of algebraic expressions lays the foundation for efficiently using methods like elimination and substitution.