Given the linear inequality 4 x + 2 3 y ≥ 6 4x+\frac23y\ge6 , determine if the following ordered pair is a solution to the inequality. ( 0 , 9 ) (0,9)

So in this problem, we need to determine if our ordered pair zero nine is a solution to the given inequality four x plus two thirds y is greater than or equal to six. Now I know that the zero is my x, and then the nine is my y, so let's go ahead and try this in our inequality. So we're going to have four times zero plus two thirds times nine, and nine is my y, that's gonna be greater than or equal to six. Now, four times zero, or anything times zero, is just going to give you zero. Now nine, I know is the same thing as three times three. Now I can cancel one of the three's, giving me two times three is greater than or equal to six. So we have zero plus two times three, which all just comes out to six on the left side, is greater than or equal to six on the right side. Is this a true statement? Well, yes, this actually is a true statement. Because six has to be greater than or equal to six. It's not greater than six, but it is in fact equal to six. Now something I want to mention here is, yes, this would be true. This is a solution to the inequality, but I also want to mention what the graph would look like real quick. So if we were to deal with a graph for this inequality, right, let's say that we had our line that we had, and then we had our shaded region as well. When it comes to a situation where you have the solution for your inequality, it has to be in the region. This ordered pair would actually end up somewhere on the line. So zero nine, this would actually end up somewhere on this line exactly. So this is actually one of those cases where you're not gonna end up in the region, you're going to end up on the line. And that can be a little bit of of a a suspicious territory when it comes to inequalities, but it's actually fine when it's greater than or equal to. If it was just greater than, like six is greater than six, then actually, we would not want it to end up on the line. And in this case, we would see that this is not a true statement. So that's just something to keep in mind when dealing with graphs and when dealing with inequalities in general. But in this case, you can see it is a solution. And if we were dealing with the graph, it would end up on the line. So the answer to this problem is yes. The ordered pair zero nine is a solution to this inequality.

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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

Slope-Intercept Form
38m

Point Slope Form
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

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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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Factoring by Greatest Common Factor & Grouping
38m

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11m

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

Factoring Special Products
41m

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41m

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Simplifying Rational Expressions
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

Introduction to Radical Expressions
47m

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47m

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19m

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53m

Introduction to Complex Numbers
18m

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51m

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25m

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26m

The Quadratic Formula
40m

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1h 28m

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Introduction to Inverse Functions
25m

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35m

Introduction to Logarithmic Functions
33m

Properties of Logarithms
54m

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32m

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35m

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35m

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41m

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Introduction to Sequences
40m

Arithmetic Sequences
27m

Geometric Sequences
29m

Factorials
13m

4. Graphs and Functions

Linear Inequalities in Two Variables

Intermediate Algebra

4. Graphs and Functions

Linear Inequalities in Two Variables

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Given the linear inequality $4 x plus two thirds y is greater than or equal to 6$ , determine if the following ordered pair is a solution to the inequality.
$open paren 0 comma 9 close paren$

Yes

Cannot be determined

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Start with the given inequality: $4x + \frac{2}{3}y \ge 6$.

Substitute the values of the ordered pair $(0, 9)$ into the inequality, replacing $x$ with $0$ and $y$ with $9$.

This gives the expression: $4(0) + \frac{2}{3}(9)$.

Simplify the expression by calculating $4 \times 0$ and $\frac{2}{3} \times 9$ separately.

Compare the simplified result to $6$ to check if the inequality $\ge 6$ holds true; if it does, the ordered pair is a solution, otherwise it is not.

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Padroneggia Intro to Linear Inequalities in Two Variables con una breve spiegazione video di Patrick Ford

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