Problem 15
Question
\(x=5, y=-4\)
Step-by-Step Solution
Verified Answer
The values given for the variables are x = 5, y = -4. Without further context or a specific question, this is all the information that can be extracted.
1Step 1: Identification of Variables
In the problem - we're given: x is equal to 5 and y equals -4.
2Step 2: Verify the Values
Given values are already provided. Here, there isn't anything to solve as we're merely given the values of the variables and there isn't a question requiring further problem solving. However, the given values for x and y can be substituted into any equations that might require these variable values.
Key Concepts
Algebraic VariablesSubstitution MethodSolving Algebraic Equations
Algebraic Variables
In algebra, algebraic variables are symbols that represent numbers. These variables can take on various values but, in a given context, they have specific ones. For instance, in the exercise \(x=5, y=-4\), \(x\) and \(y\) are algebraic variables with \(x\) representing the number 5 and \(y\) representing -4. Understanding variables is essential as they are foundational to algebraic expressions, equations, and functions.
When working with algebraic variables, it's crucial to differentiate between independent and dependent variables. In many equations, one variable depends on the value of another. For example, in the equation \(y = 3x + 2\), \(y\) is a dependent variable because its value depends on the value assigned to \(x\), the independent variable. Recognizing these relationships helps students to correctly form and solve equations.
When working with algebraic variables, it's crucial to differentiate between independent and dependent variables. In many equations, one variable depends on the value of another. For example, in the equation \(y = 3x + 2\), \(y\) is a dependent variable because its value depends on the value assigned to \(x\), the independent variable. Recognizing these relationships helps students to correctly form and solve equations.
Substitution Method
The substitution method is a technique used to find the value of one variable in terms of another or to solve a system of equations. This method involves replacing variables with their given or known values. Returning to our exercise example where \(x=5, y=-4\), if we have an equation such as \(2x + 3y = z\), we can substitute \(x\) and \(y\) with 5 and -4, respectively, to find \(z\).
Here's how it would work:
Here's how it would work:
- Replace \(x\) with 5: \(2(5) + 3y = z\)
- Replace \(y\) with -4: \(2(5) + 3(-4) = z\)
- Simplify the equation: \(10 - 12 = z\) or \(z = -2\)
Solving Algebraic Equations
The process of solving algebraic equations involves finding the value(s) of the variable(s) that make the equation true. Algebraic equations can range from simple, like \(2x = 10\), to complex, involving several variables and operations. The goal is to isolate the variable and determine its value.
To solve an algebraic equation, one typically performs a series of steps that may include distributing, combining like terms, and isolating the variable through addition, subtraction, multiplication, or division. For example, to solve \(2x = 10\), one would divide both sides of the equation by 2 to find \(x\), which is \(x = 5\).
It's important to understand that solving equations often requires checking your solution by substituting the value of the variable back into the original equation to ensure the left and right sides are equal. In the given exercise, there wasn't a need for this as we were provided the values for \(x\) and \(y\) directly. However, this step is crucial when you derive the variable's value as part of the solution process.
To solve an algebraic equation, one typically performs a series of steps that may include distributing, combining like terms, and isolating the variable through addition, subtraction, multiplication, or division. For example, to solve \(2x = 10\), one would divide both sides of the equation by 2 to find \(x\), which is \(x = 5\).
It's important to understand that solving equations often requires checking your solution by substituting the value of the variable back into the original equation to ensure the left and right sides are equal. In the given exercise, there wasn't a need for this as we were provided the values for \(x\) and \(y\) directly. However, this step is crucial when you derive the variable's value as part of the solution process.
Other exercises in this chapter
Problem 15
In Exercises 11–18, graph the function. State the domain and range. $$ h(x)=\frac{-3}{x+2} $$
View solution Problem 15
Find the least common multiple of the expressions. \(x^2+3 x-40, x-8\)
View solution Problem 15
Identify the least common denominator of the equation. $$\frac{x}{x+3}+\frac{1}{x}=\frac{3}{x}$$
View solution Problem 15
Find the product. $$ \frac{x^2-3 x}{x-2} \cdot \frac{x^2+x-6}{x} $$
View solution