Solve the given linear equation. Check your solution. 4 ( x + 1 ) − 3 ( x − 2 ) = 2 x + 5 4\left(x+1\right)-3\left(x-2\right)=2x+5

we're asked to solve the given linear equation, then check our solution. The equation that we're given here is four times X plus one minus three times X minus two equals two X plus five. So we're going to jump right into our steps here. The first thing we wanna do is simplify both sides of our equation by distributing anything that we need to into parentheses and combining any like terms. So on this left hand side of my equation, I have this four that needs to get distributed to both x and one. I also have this negative three that I'll distribute into my x and this negative two. So four times x is gonna give me four x and then four times one is four. So that's four x plus four. Then make sure that you're paying attention to your signs here. This is a negative three. So negative three times x will be negative three x and then negative three times negative two will give me a positive six. Then that right hand side of my equation doesn't have anything that needs to get simplified, so I'm going to leave that alone. Two x plus five. Now we've distributed into parentheses, but we still need to combine like terms. Over here on this left hand side of my equation, I have this four x and I have negative three x. Those are like terms. They're the same variable raised to the same power. So I want to combine them. Four x minus three x is just going to give me x or one x, but we don't have to write that one. And then I have positive four and positive six, both constants, so I can combine those as well. Positive four, positive six is going to give me plus 10 here. So I have x plus 10 equals two x plus five. So now that everything is simplified, I can move on to step two, using my addition and subtraction properties of equality to collect all variable terms together on one side, all constant terms together on the other. So I have x and two x. I want to get those together on the same side. I also want to get ten and five together on the same side. I'm going to go ahead and just move this X over here with my two X. Since it's a positive X, I can move it over or cancel it out on this side by subtracting. Whatever I do on one side, I have to do to the other. So I have that x minus x cancel, just leaving me with 10. Then that's equal to two x minus x, which is just x. And then I have that plus five. Now I wanna go ahead and move that plus five over to the other side with that other constant 10. I can do that by subtracting. That will cancel out. 10 minus five gives me five. So I get that five equals X. That's all that's left on that side. So I've completed step two, and we can take a look at step three using our multiplication and division properties in order to isolate our variable. But our variable is already isolated. Its coefficient is already just one. So I already have my solution having just completed step two. No need to do any multiplication or division when solving here. We're already done. But, of course, we wanna check our solution by plugging that five back in for x to our original equation. So I'm gonna go ahead and do that down here since I need a little bit of extra space. So my original equation here, I have four and then plugging in five for x, that's gonna give me four times five plus one minus three times five minus two equals two times five plus five. So I have three instances of x in my original equation. I need to make sure I replace all of those x's with what I found to be my solution here, five equals x. So we're checking, is this going to give us a true statement? If it does, we know that we solved this correctly. So looking in my parentheses here, five plus one is six. So this becomes four times six minus three times five minus two. Five minus two is also three. So this is three times three. We're checking, is this equal to two times five plus five? I can go ahead and do that multiplication. Two times five gives me 10. So I have 10 plus five. Over here on this side doing our multiplication of these values, four times six is going to give me 24. So I have 24 minus nine, that's three times three. We're checking is this equal to 10 plus five? We can go ahead and do that addition. 10 plus five is 15. Finally, last step here, doing this subtraction, 24 minus nine is 15. So I get the statement that 15 equals 15. This is definitely a true statement. Those are the same exact value. So I know that I found my solution correctly, that five equals x. Let us know if you have any questions, and I'll see you in the next one.

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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

12. Quadratic Equations and Functions3h 1m

The Square Root Property
25m

Completing the Square
26m

The Quadratic Formula
40m

Graphing Quadratic Equations
1h 28m

13. Inverse, Exponential, & Logarithmic Functions2h 31m

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.
$4 (x + 1) - 3 (x - 2) = 2 x + 5$

$x=3$

$x=0$

$x=-5$

$x=5$

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Start by applying the distributive property to both terms on the left side of the equation: expand $4(x+1)$ to $4 \cdot x + 4 \cdot 1$ and expand $-3(x-2)$ to $-3 \cdot x + (-3) \cdot (-2)$.

Rewrite the equation after distribution: $4x + 4 - 3x + 6 = 2x + 5$.

Combine like terms on the left side: combine \$4x$ and $-3x$, and combine the constants $4$ and $6$.

After simplification, the equation will look like $x + 10 = 2x + 5$. Next, get all variable terms on one side by subtracting $x$ from both sides.

Then isolate $x$ by subtracting $5$ from both sides, resulting in an equation where $x$ is alone on one side. This will give you the solution for $x$.