How many solutions does each of the following equations have? 1 2 ( x + 4 ) − 1 3 = 1 6 x − 2 \frac12\left(x+4\right)-\frac13=\frac16x-2

Welcome back. In this problem, we've been given the equation one half times x plus four minus one third equals one sixth x minus two. And we want to see if this equation has one solution, infinite solutions, or no solution. We're going to have to solve the equation to answer this question, so let's get into it. We can see that the equation has fractions. We have one half, minus one third, and one sixth. So we need to find the LCD of those fractions so we can multiply by that LCD on both sides of the equation. Our denominators are two, three, and six, so the LCD is going to be six. So when we multiply by six on both sides of the equation, we get six times one half times x plus four minus one third, on the left, equals six times one sixth x minus two, on the right. Let's distribute that six on both sides of the equation. Starting with on the left, six times one half times x plus four. Six times one half is three, so we get three times x plus four. And then we have six times negative one third, and six times one third is two, so we get minus two. On the right, we have six times one sixth x, and six times one sixth is one, so we get one x, which we can also just write as x. And then we have six times minus two, which is minus 12. Next, we have to simplify our equation. So let's distribute that three on the left side. We have three times x, which is three x, and then three times four, which is 12, and we still have that minus two on the left. And then on the right, we still have x minus 12. Next, we have to combine our like terms on the left side of the equation. We have plus 12 and minus two, which combine to 10, so we get three x plus 10 on the left side of the equation equals x minus 12 on the right. Now we have to get variable terms on one side of the equation and constants on the other, and we currently have both variable terms and constant terms on both sides of the equation. So let's try to get the variables to the left and the constants to the right, starting with the constants. We have to get rid of the plus 10 on the left side of the equation, which we can do by subtracting 10 on both sides. This is gonna cancel the plus 10 on the left, leaving us with three x equals x minus 12 minus 10, which is minus 22. Next, we want to get rid of the x on the right, which we can do by subtracting x from both sides of the equation. It gets rid of the x on the right, so we get minus 22 on the right, and we have three x minus x, which is two x on the left. Next, we wanna isolate our x, which we can do by dividing by two on both sides of the equation. This is gonna cancel our coefficient of two on the left, leaving us with just x. And then we have negative 22 divided by two on the right, which is negative 11. So we end up getting the solution x equals negative 11, which means if we plug in that one solution, negative 11, to our equation, we get a true statement. But if we tried to plug in anything else, we wouldn't get a true statement. That means that we only have the one solution, x equals negative 11. Great work.

Table of contents

Skip topic navigation

1. Review of Real Numbers2h 43m

Simplifying Fractions
30m

Evaluating Exponents
29m

Evaluating Expressions
15m

Translating Phrases to Expressions
16m

Translating Sentences to Equations
9m

The Distributive Property
17m

Simplifying Expressions
44m

2. Linear Equations and Inequalities5h 35m

Solving Linear Equations
2h 23m

Linear Inequalities in One Variable
46m

Set Operations and Compound Inequalities
1h 7m

Absolute Value Equations
38m

Absolute Value Inequalities
39m

3. Solving Word Problems2h 46m

Introduction to Problem Solving
37m

Formulas
25m

Percent Problem Solving
59m

Mixture Problem Solving
43m

4. Graphs and Functions4h 44m

The Rectangular Coordinate System
44m

Graph Linear Equations in Two Variables
24m

Graph Linear Equations Using Intercepts
23m

Slope of a Line
44m

Slope-Intercept Form
38m

Point Slope Form
22m

Linear Inequalities in Two Variables
28m

Introduction to Relations and Functions
25m

Function Notation
15m

Composition of Functions
17m

5. Systems of Linear Equations1h 53m

Solving Systems of Linear Equations by Graphing
29m

Solving Systems of Linear Equations by Substitution
15m

Solving Systems of Linear Equations by Elimination
45m

Systems of Linear Inequalities
23m

6. Exponents, Polynomials, and Polynomial Functions1h 27m

The Product Rule
9m

The Quotient Rule
10m

Intro to the Power Rules
17m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
5m

Multiplying Polynomials
38m

7. Factoring2h 49m

Factoring by Greatest Common Factor & Grouping
37m

Factoring Trinomials of the Form x² + bx + c
11m

Factoring Trinomials of the Form ax² + bx + c
37m

Factoring Special Products
41m

Quadratic Equations & Applications
41m

8. Rational Expressions and Functions3h 16m

Simplifying Rational Expressions
42m

Multiplying and Dividing Rational Expressions
25m

Adding and Subtracting Rational Expressions with Common Denominators
19m

Least Common Denominators
32m

Adding and Subtracting Rational Expressions with Different Denominators
32m

Rational Equations
44m

9. Roots, Radicals, and Complex Numbers2h 33m

Introduction to Radical Expressions
11m

Simplifying Radical Expressions
27m

Adding and Subtracting Radicals
19m

Multiplying, Dividing, and Rationalizing Radicals
23m

Introduction to Complex Numbers
18m

Multiplying and Dividing Complex Numbers
51m

10. Quadratic Equations and Functions1h 23m

The Square Root Property
25m

Completing the Square
26m

The Quadratic Formula
31m

11. Inverse, Exponential, & Logarithmic Functions1h 5m

Introduction to Inverse Functions
25m

Exponential Functions
13m

Introduction to Logarithmic Functions
26m

12. Conic Sections & Systems of Nonlinear Equations58m

Graphing Circles
32m

Ellipses
15m

Parabolas
10m

13. Sequences, Series, and the Binomial Theorem1h 21m

Introduction to Sequences
40m

Arithmetic Sequences
27m

Factorials
13m

2. Linear Equations and Inequalities

Solving Linear Equations

Intermediate Algebra

2. Linear Equations and Inequalities

Solving Linear Equations

Previous problemNext problem

Struggling with Intermediate Algebra?

Join thousands of students who trust us to help them ace their exams!

Multiple Choice

How many solutions does each of the following equations have?
$\frac{1}{2} (x + 4) - \frac{1}{3} = \frac{1}{6} x - 2$

One solution

Infinite solutions

No solution

0 Comments

Previous problemNext problem

Verified step by step guidance

Start by writing down the given equation: $\frac{1}{2}(x + 4) - \frac{1}{3} = \frac{1}{6}x - 2$.

Distribute the $\frac{1}{2}$ on the left side to both terms inside the parentheses: $\frac{1}{2} \times x$ and $\frac{1}{2} \times 4$.

Simplify the left side after distribution: this gives $\frac{1}{2}x + 2 - \frac{1}{3}$.

Combine the constant terms on the left side: $2 - \frac{1}{3}$, by finding a common denominator.

Rewrite the equation with simplified terms and then isolate the variable $x$ by moving all $x$ terms to one side and constants to the other, then analyze the resulting equation to determine the number of solutions.

6:06m

Watch next

Master Introduction to Linear Equations with a bite sized video explanation from Patrick Ford

Struggling with Intermediate Algebra?

How many solutions does each of the following equations have?12(x+4)−13=16x−2\(\frac\)12\(\left\)(x+4\(\right\))-\(\frac\)13=\(\frac\)16x-2

Watch next

How many solutions does each of the following equations have?
$\frac{1}{2} (x + 4) - \frac{1}{3} = \frac{1}{6} x - 2$