Problem 16 - College Algebra Essentials

Question

Find all values of \(x\) satisfying the given conditions. \(y_{1}=2.5, y_{2}=2 x+1, y_{3}=x,\) and the difference between 2 times \(y_{1}\) and 3 times \(y_{2}\) is 8 less than 4 times \(y_{3}.\)

Step-by-Step Solution

Answer

The value of \(x\) that satisfies the given conditions is \(x = 3\).

Step 1: Set up equation

From the given problem, we know that the difference between 2 times \(y_{1}\) and 3 times \(y_{2}\) is 8 less than 4 times \(y_{3}\). This can be written as: \(2y_{1} - 3y_{2} = 4y_{3} - 8\).

Step 2: Substitute the given values of \(y_{1}\), \(y_{2}\) and \(y_{3}\)

Now we substitute the given values into the derived equation in Step 1. Substituting the value of \(y_{1}\) as 2.5, \(y_{2}\) as \(2x+1\), and \(y_{3}\) as \(x\) into the equation gives us: \(2(2.5) - 3(2x+1) = 4(x) - 8\).

Step 3: Solve the equation

Solving the equation from Step 2 gives: \(5 - (6x+3) = 4x - 8\). Simplifying both sides gives: \(-2x = -6\). Dividing by -2 gives the value of \(x\).

Step 4: Find the value of \(x\)

Finally, dividing by -2 gives us the value of \(x = 3\).

Key Concepts

Equation Solving

Solving an equation is like cracking a code. It involves finding the value of the variable that makes the equation true. In our original exercise, the goal is to find the value of \(x\) that satisfies the given condition. Here’s how we did it:

• The problem gives us a relationship between different expressions involving \(x\).
• First, we set up an equation based on the problem statement: \(2y_{1} - 3y_{2} = 4y_{3} - 8\).
• Next, we substitute the known values of \(y_{1}, y_{2},\) and \(y_{3}\) into this equation.
• This substitution transformed our equation into a solvable format.

Once the equation was properly set up and values substituted, we used algebraic methods to solve for \(x\). It's important to carefully perform each operation to maintain the balance of the equation, ensuring the solution is correct.

Substitution Method

The substitution method is a powerful technique used to simplify and solve equations. It involves replacing variables in an equation with known values or expressions. Here’s how it was applied in our problem:

• We identified the values of \(y_{1}, y_{2},\) and \(y_{3}\).
• Then, we substituted \(y_{1} = 2.5\), \(y_{2} = 2x + 1\), and \(y_{3} = x\) into the main equation.
• This substitution reduced the number of variables, making the equation easier to solve.

The main benefit of substitution is that it turns complex expressions into simpler ones, ultimately leading to the value of the variable we’re trying to find. This method is especially useful in systems of equations or when functions are given in terms of other variables.

Linear Equations

Linear equations are equations of the first degree, meaning they involve no exponents or powers higher than one. The standard form of a linear equation in one variable is \(ax + b = c\). In the original problem:

• We ended up with a linear equation: \(-2x = -6\).
• This was derived by combining like terms and moving all terms involving \(x\) to one side of the equation.

Solving linear equations involves straightforward arithmetic:

• We need to isolate the variable \(x\).
• This is achieved by performing the same operation on both sides of the equation.
• By dividing both sides by \(-2\), we found \(x = 3\).

Linear equations are one of the simplest types of equations to solve, providing a straightforward pathway to finding the value of a variable.