Problem 98 - Introductory and Intermediate Algebra for College Students 4th

Question

Determine whether the given number is a solution of the equation. $$\frac{2}{3} y+\frac{5}{6} y=2 ; 1 \frac{1}{3}$$

Step-by-Step Solution

Answer

Yes, \(1\frac{1}{3}\) is a solution of the equation. The left side of the equation simplifies to 2 when y = \(1\frac{1}{3}\), exactly matching the right side of the equation.

Step 1: Understand the Problem

We are asked to determine if \(1\frac{1}{3}\) is a solution to the equation \(\frac{2}{3}y + \frac{5}{6}y = 2\). This means, when we substitute \(y\) with \(1\frac{1}{3}\), the equation should hold.

Step 2: Substitute the Value of y

Substitute \(y = 1\frac{1}{3} = \frac{4}{3}\) into the equation. The equation becomes \(\frac{2}{3}(\frac{4}{3}) + \frac{5}{6}(\frac{4}{3})\).

Step 3: Simplify the Equation

By performing the multiplications, we have \(\frac{8}{9} + \frac{10}{9} = 2.\) Probability theory says that if two fractions have the same denominator they can be added together. So, we add \(\frac{8}{9} + \frac{10}{9} = \frac{18}{9}\). The result \(\frac{18}{9}\) simplifies to \(2\).

Step 4: Conclude

Since both sides of the equation match, \(1\frac{1}{3}\) is a solution of the given equation.

Key Concepts

Equation Solving

Solving algebraic equations is a fundamental skill in algebra. An equation represents a balance between two expressions, and solving it means finding the value or values of the variable that make the two sides equal. When you're given an equation, like \(\frac{2}{3} y + \frac{5}{6} y = 2\), the goal is to isolate the variable \(y\) and determine its value.

Equation solving often involves a series of steps, starting with simplifying expressions, combining like terms, and then using operations to isolate the variable. When faced with a question about whether a certain number is a solution, you plug in that number in place of the variable and simplify to see if the two sides of the equation are equal. If they are, then that number is indeed a solution to the equation.

Substitution Method

The substitution method is a technique used in algebra to find the value of variables. It involves replacing a variable with a given number or expression. In the context of our exercise, to verify if \(1\frac{1}{3}\) is a solution, we substitute \(y\) with \(\frac{4}{3}\), converting the original equation into one that can be evaluated with arithmetic.

When you perform the substitution correctly and the expressions on both sides of the equation simplify to the same value, then the substituted value is a correct solution. Substitution is a critical concept because it can be applied to more complex equations, systems of equations, and various algebraic contexts.

Simplifying Fractions

Simplifying fractions is a process of reducing the numerator and denominator to their smallest possible values while keeping the same value of the fraction. This is done by dividing both by their greatest common factor. The equation \(\frac{2}{3} y + \frac{5}{6} y = 2\) eventually leads to the fraction \(\frac{18}{9}\), which can be simplified since 18 and 9 are both divisible by 9.

To simplify \(\frac{18}{9}\), divide both the numerator (18) and the denominator (9) by 9 to get \(2\), which is a whole number. Understanding how to simplify fractions is crucial not only to solve equations but also to perform operations with fractions, compare them, and apply them in various mathematical contexts.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. They do not have an equal sign as equations do. In our problem, \(\frac{2}{3} y + \frac{5}{6} y\) is an example of an algebraic expression. Working with algebraic expressions involves understanding how to combine like terms and perform operations with variables.

In an expression like \(\frac{2}{3} y + \frac{5}{6} y\), we identify 'y' as a common variable and then use the coefficients \(\frac{2}{3}\) and \(\frac{5}{6}\) to combine terms, leading to a simplified expression. Mastery in manipulating algebraic expressions is essential for higher level math, where expressions become more complex, and you often need to rearrange them to solve for variables.