Solve the given linear equation. Check your solution. − 5 − y = 3 ( y + 9 ) -5-y=3\left(y+9\right)

Here we're asked to solve the given linear equation and then check our solution. The equation that we're given here is negative five minuteus y equals three times y plus nine. So let's get started with our steps here. The first thing we want to do is simplify both sides of our equation, which we can do by distributing anything into parenthesis and combining like terms. Here I can see this three that needs to get distributed to y and nine, so let's start there. I'm going to keep this left side of my equation the same because nothing needs to get simplified and just focus on this distribution of my three. So three times y will give me three y and then three times nine will give me 27. So I have positive 27 there. So everything is fully simplified on either side of my equation now. And I can move on to step two. Using my addition and subtraction properties of equality to collect all variable terms on one side and all constant terms on the other. Remember, it doesn't matter which side we put variables versus constants on so long as they're opposite sides and we put all those variables together, all the constants together. So we have this variable term at negative y and we have this three y. So I'm going to start with my variable terms. I want to move this negative y over to the other side, which I can do by adding y. If I add y on one side, I have to add it on the other. That's what my addition property of equality tells me. So I end up with just negative five because negative y and positive y canceled. And then I have three y plus y, which gives me four y. So I have four y plus 27 over here. Now, since I have my variable terms all together, I want to move this 27, my constant, to the other side of my equation, which I can do using subtraction. So subtracting 27 on both sides, that will cancel that positive 27 out. And I have negative five minus 27, which gives me negative 32. So I have negative 32 equals four y. Now from here, I can move on to step three using my multiplication and division properties of equality in order to fully isolate my variable. So I have this four y here. Four is multiplying y. So I can get rid of that multiplication by the four by dividing by four because that's that opposite operation. Whatever I do to ones that I have to do to the other, so making sure I divide that negative 32 by four as well. So these fours will cancel out. I have negative 32 divided by four, which gives me negative eight. So I have negative eight equals my variable here y. And this is my solution. Now I, of course, want to check my solution to make sure that I did this correctly by plugging that negative eight back in for y to my original equation. I have two instances of y here, so I want to make sure that I replace both of them with that negative eight when checking. So I have negative five minus negative eight equals three times negative eight plus nine. So we're checking, is this a true statement? So I have negative five minus negative eight. We know that two negatives make a positive, so I have negative five plus eight, which is positive three. And then I have three equals three times negative eight plus nine. Negative eight plus nine, just focusing inside of those parentheses, is going to be positive one. So this is three times one. Is this a true statement? Well, that three stays the same. And then three times one is just three. So I end up with this statement that three equals three. Definitely a true statement as those are the exact same value. So that tells me that I did solve this linear equation correctly. My solution is that negative eight equals y. Let us know if you have any questions and let's keep practicing.

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1. Review of Real Numbers2h 39m

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
40m

2. Linear Equations and Inequalities3h 38m

The Addition and Subtraction Properties of Equality
41m

The Multiplication and Division Properties of Equality
30m

Solving Linear Equations
1h 14m

Linear Inequalities in One Variable
1h 11m

3. Solving Word Problems2h 43m

Introduction to Problem Solving
37m

Formulas
21m

Percent Problem Solving
59m

Mixture Problem Solving
43m

4. Graphing Linear Equations in Two Variables3h 17m

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

5. Systems of Linear Equations1h 43m

Solving Systems of Linear Equations by Graphing
41m

Solving Systems of Linear Equations by Substitution
15m

Solving Systems of Linear Equations by Elimination
45m

6. Exponents and Polynomials3h 25m

The Product Rule
10m

The Quotient Rule
13m

Intro to the Power Rules
18m

The Power of a Quotient Rule
18m

Negative Exponents
28m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

Adding and Subtracting Polynomials
20m

Multiplying Polynomials
33m

Special Products
34m

7. Factoring2h 42m

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
33m

Quadratic Equations & Applications
41m

8. Rational Expressions and Equations3h 13m

Simplifying Rational Expressions
39m

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. Inequalities and Absolute Value2h 52m

Set Operations and Compound Inequalities
42m

Absolute Value Equations
38m

Absolute Value Inequalities
39m

Linear Inequalities in Two Variables
28m

Systems of Linear Inequalities
23m

10. Relations and Functions2h 9m

Introduction to Relations and Functions
57m

Function Notation
15m

Graphing Common Functions
5m

Composition of Functions
24m

Direct & Inverse Variation
27m

11. 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

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The Square Root Property
25m

Completing the Square
26m

The Quadratic Formula
40m

Graphing Quadratic Equations
1h 28m

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Introduction to Inverse Functions
25m

Exponential Functions
35m

Introduction to Logarithmic Functions
33m

Properties of Logarithms
55m

14. Conic Sections & Systems of Nonlinear Equations2h 24m

Graphing Circles
32m

Ellipses
35m

Parabolas
35m

Hyperbolas
41m

15. Sequences, Series, and the Binomial Theorem1h 45m

Introduction to Sequences
40m

Arithmetic Sequences
27m

Geometric Sequences
26m

Factorials
11m

2. Linear Equations and Inequalities

Solving Linear Equations

Beginning & Intermediate Algebra

2. Linear Equations and Inequalities

Solving Linear Equations

Problema precedenteProssimo problema

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Solve the given linear equation. Check your solution.
$- 5 - y = 3 (y + 9)$

$y=-8$

$y=8$

$y=-3$

$y=3$

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Start with the given equation: $-5 - y = 3\left(y + 9\right)$.

Distribute the 3 on the right side to both terms inside the parentheses: $3 \times y$ and $3 \times 9$, giving $-5 - y = 3y + 27$.

Next, get all the terms involving $y$ on one side and the constants on the other side. You can do this by adding $y$ to both sides and subtracting 27 from both sides: $-5 - y + y - 27 = 3y + y + 27 - 27$ which simplifies to $-5 - 27 = 4y$.

Combine the constants on the left side: $-5 - 27$ becomes $-32$, so the equation is now $-32 = 4y$.

Finally, solve for $y$ by dividing both sides of the equation by 4: $y = \frac{-32}{4}$.