Solve the following equations with fractions. x 4 − 2 3 = x 6 \frac{x}{4}-\frac23=\frac{x}{6}

Welcome to the practice. In this problem, we'd like to solve the equation x over four minus two thirds equals x over six. What makes this equation look a bit intimidating is that we have two fractions with an x in the numerator, x over four and x over six. So it doesn't really look like the one that we solved in the concept video. That means we should start by rewriting those two terms so that we can make the equation look a bit friendlier. So let's start with x over four. We know that x is equal to one x and that four is equal to four times one. So we can rewrite this as one x over four times one, which shows us how we can break up this fraction into one fourth times x over one. Now x over one is just x, so we can write this as one fourth x. And we can say that x over four is equal to one fourth x. We can do the same thing with x over six. Take a second, pause the video, follow the same steps, and you'll see that x over six is really just one sixth x. So we can rewrite this equation to be one fourth x minus two thirds is equal to one sixth x. And now we have it in the same form as the one that we solved in the concept video, so we're ready to follow our steps here to solve the equation, starting with step zero, which is clearing our fractions by multiplying by the least common denominator. So we wanna find the least common denominator of one fourth, minus two thirds, and one sixth. This means that we need to find the least common multiple of denominators four, three, and six, and that least common denominator is going to be 12. So let's multiply both sides of our equation by 12. That's gonna get us 12 times one fourth x minus two thirds is equal to 12 times one sixth x. And now we just need to distribute that 12. On the left, we end up getting 12 times one fourth x. Now, 12 times one fourth is three, so that's three x, and 12 times minus two thirds. 12 times two is 24, so this will get us minus 24 thirds. But because 24 divided by three is just eight, minus 24 thirds is really just minus eight. On the right side, we have 12 times one sixth x. Now 12 times one sixth is two, so this is just two x. Now we have no more fractions or decimals, so step zero is all set, and we would move on to step one. We don't have any parentheses or things we have to distribute, and we don't have any like terms on the left side or on the right side, so step one is all set, and we can move to step two, adding and subtracting to get variables on one side and constants on the other. We can see that our only constant is already on the left side, so let's keep constants on the left and move variables to the right. This means that we need to move this three x from the left, and we can get rid of it by subtracting three x from both sides of the equation. This is gonna cancel our three x on the left, leaving us with minus eight. And on the right, we get two x minus three x, which is minus x, or we can also think of it as minus one x. We're all set with step two, and now we can move on to step three using multiplication or division to isolate the variable. We wanna get rid of this minus one so we can divide both sides of the equation by minus one. And on the right, that's gonna cancel the minus one, just leaving us with x. And on the left, we get negative eight divided by negative one, which is just eight. So we end up getting that eight is equal to x. And that is step three all set. I encourage you to do step four and plug eight in for x back to the original equation to check your work. But once that's all set, you've solved the equation. Awesome job.

Indice

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 Functions5h 12m

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
53m

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 Functions3h 17m

The Product Rule
9m

The Quotient Rule
10m

Intro to the Power Rules
17m

The Power of a Quotient Rule
13m

Negative Exponents
24m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

Adding and Subtracting Polynomials
20m

Multiplying Polynomials
38m

Special Products
34m

7. Factoring2h 49m

Factoring by Greatest Common Factor & Grouping
38m

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 44m

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

Direct & Inverse Variation
27m

9. Roots, Radicals, and Complex Numbers3h 57m

Introduction to Radical Expressions
47m

Simplifying Radical Expressions
47m

Adding and Subtracting Radicals
19m

Multiplying, Dividing, and Rationalizing Radicals
53m

Introduction to Complex Numbers
18m

Multiplying and Dividing Complex Numbers
51m

10. Quadratic Equations and Functions3h 1m

The Square Root Property
25m

Completing the Square
26m

The Quadratic Formula
40m

Graphing Quadratic Equations
1h 28m

11. Inverse, Exponential, & Logarithmic Functions2h 30m

Introduction to Inverse Functions
25m

Exponential Functions
35m

Introduction to Logarithmic Functions
33m

Properties of Logarithms
54m

12. Conic Sections & Systems of Nonlinear Equations2h 24m

Graphing Circles
32m

Ellipses
35m

Parabolas
35m

Hyperbolas
41m

13. Sequences, Series, and the Binomial Theorem1h 50m

Introduction to Sequences
40m

Arithmetic Sequences
27m

Geometric Sequences
29m

Factorials
13m

2. Linear Equations and Inequalities

Solving Linear Equations

Intermediate Algebra

2. Linear Equations and Inequalities

Solving Linear Equations

Problema precedenteProssimo problema

Difficoltà con Intermediate Algebra?

Unisciti a migliaia di studenti che si affidano a noi per superare al meglio i loro esami!

Scelta multipla

Solve the following equations with fractions.
$\frac{x}{4} - \frac{2}{3} = \frac{x}{6}$

$x=$\frac$85$

$x=-8$

$x=8$

$x=-$\frac$85$

0 Commenti

Problema precedenteProssimo problema

Guida verificata passo dopo passo

Start with the given equation: $\frac{x}{4} - \frac{2}{3} = \frac{x}{6}$.

To eliminate the fractions, find the least common denominator (LCD) of 4, 3, and 6, which is 12. Multiply every term in the equation by 12 to clear the denominators.

After multiplying, rewrite the equation without fractions: $12 \times \frac{x}{4} - 12 \times \frac{2}{3} = 12 \times \frac{x}{6}$, which simplifies to $3x - 8 = 2x$.

Isolate the variable $x$ by subtracting \$2x$ from both sides: $3x - 2x - 8 = 0$, which simplifies to $x - 8 = 0$.

Finally, solve for $x$ by adding 8 to both sides: $x = 8$.