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 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

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Introduction to Problem Solving
37m

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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

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38m

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22m

Linear Inequalities in Two Variables
28m

Introduction to Relations and Functions
53m

Function Notation
15m

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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

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20m

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38m

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34m

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37m

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11m

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

Factoring Special Products
41m

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41m

8. Rational Expressions and Functions3h 44m

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42m

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25m

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19m

Least Common Denominators
32m

Adding and Subtracting Rational Expressions with Different Denominators
32m

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44m

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27m

9. Roots, Radicals, and Complex Numbers3h 57m

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47m

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47m

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19m

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53m

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18m

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51m

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25m

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26m

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40m

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1h 28m

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25m

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35m

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33m

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54m

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32m

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41m

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40m

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27m

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29m

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13m

2. Linear Equations and Inequalities

Solving Linear Equations

Intermediate Algebra

2. Linear Equations and Inequalities

Solving Linear Equations

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Multiple Choice

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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Verified step by step guidance

Start with the given equation: $4\left(x+1\right) - 3\left(x-2\right) = 2x + 5$.

Apply the distributive property to remove the parentheses: multiply 4 by each term inside the first parentheses and -3 by each term inside the second parentheses, resulting in $4x + 4 - 3x + 6 = 2x + 5$.

Combine like terms on the left side: $4x - 3x$ and $4 + 6$ to simplify the equation to $x + 10 = 2x + 5$.

Isolate the variable $x$ by subtracting $x$ from both sides and subtracting 5 from both sides, which gives $10 - 5 = 2x - x$ or $5 = x$.

Check the solution by substituting $x = 5$ back into the original equation to verify both sides are equal.

5:27m

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