Problem 89 - College Algebra

Question

For Exercises \(86-89\) use your graphing utility to graph. each side of the equations in the same viewing rectangle. Based on the resulting graph, label each equation as conditional, inconsistent, or an identity. If the equation is conditional, use the \(x\) -coordinate of the intersection point to find the solution set. Verify this value by direct substitution into the equation. $$ \frac{2 x-1}{3}-\frac{x-5}{6}=\frac{x-3}{4} $$

Step-by-Step Solution

Answer

The equation is conditional with the solution set {3}.

Step 1: Express the equation in standard form

Rewrite the equation like this: \\(2x-1)\/3=(x-5)\/6 + (x-3)\/4\\. Then, adjust the fractions to common denominators and rewrite the equation again: 4(2x-1) = 2(x-5) + 3(x-3), which after simplifying becomes 8x - 4 = 2x - 10 + 3x - 9

Step 2: Solve for x

Simplify the above equation further: 8x - 4 = 5x - 19. Combine like terms to the one side of the equation, that gives: 8x - 5x = -19 + 4. Therefore, x = 3.

Step 3: Check the solution

Substitute x = 3 back into the original equation. This will allow you to confirm that the solution is correct. That gives: \\(2(3)-1)\/3=(3-5)\/6 + (3-3)\/4\\. This simplifies to 2=2, which is correct.

Step 4: Interpret the result

As the graphs of both sides of the equation will intersect at x=3, it means this equation is conditional. So, the solution set for this given equation is {3}.

Key Concepts

Conditional Equations

When graphing linear equations, encountering a conditional equation is quite common. A conditional equation is true for specific values of its variables. It sets up a condition that must be met for the equation to hold true.

For instance, consider the equation presented in the exercise: \(\frac{2x-1}{3}-\frac{x-5}{6}=\frac{x-3}{4}\). Upon graphing each side of the equation and finding the intersection point, we discover that the graphs meet at a specific point. This means the equation is satisfied only at that particular intersection point, which corresponds to a certain value of x.

The essence of a conditional equation is this specificity––a unique solution exists. In the exercise, after manipulating the equation and solving for x, we obtain x = 3. This value is the only solution that satisfies the given conditional equation, and by substituting x with 3 into the original equation, we validate its truthfulness, as the equation simplifies to a true statement.

Inconsistent Equations

In contrast to conditional equations, inconsistent equations are types of equations that have no solution. When graphed, the lines representing each side of an inconsistent equation will never intersect—they are parallel lines.

In practical terms, if you attempt to solve an inconsistent equation, you will eventually arrive at a statement that is always false, such as 0 = 5. No value for the variable can make such a statement true. It's essential to identify these equations because they indicate a problem has no solution, and spending further time trying to solve them is not productive.

Understanding the nature of inconsistent equations frees students from the frustration of endless calculation loops and directs their efforts toward more fruitful mathematical tasks.

Identities in Algebra

Diving deeper into the world of algebraic equations, we encounter identities in algebra. An identity is an equation that is always true, no matter what value is substituted for the variable. In graphical terms, if you were to graph both sides of an identity, you would end up with exactly the same graph; they are superimposed on each other.

An example of an identity would be \(2x = x + x\), where, irrespective of the value you choose for x, both sides of the equation will always equate. When dealing with identities, it's important to recognize them early on, as they don't present a challenge to solve for a specific value. Instead, they confirm a universal truth within the scope of their variables.

Solving Algebraic Equations

At the heart of graphing linear equations lies the fundamental process of solving algebraic equations. Solving an equation involves finding the value(s) of the variable that make the equation true. When graphing, this translates to finding the point(s) at which the graphs of two expressions intersect.

The exercise we look at zeroes in on precisely this aspect. The step-by-step solution guides you through the process: from expressing the equation in standard form to manipulating it to isolate the variable. The outcome is a clear-cut value for x that can be verified by substitution into the original equation.

Mastery of solving algebraic equations is not just about performing rote steps; it's about understanding the relationships between the elements of the equation and effectively using algebraic techniques to uncover these hidden values. It's a journey from a jumbled set of terms to a singular solution that unveils the equation's true essence.