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How many solution sets do systems of linear inequalities have? Do solutions to systems of linear inequalities need to satisfy both inequalities? In what case might they not?
It can have none, one, or infinitely many. Here's a system with no solutions: x 5. Here's a system with one solution: x =3. [ ( = mean "less than or equal" and "greater than or equal") And here's a system with infinitely many solutions: x < 3 and x < 2. In order to satisfy both inequalities, a number must be in both solution sets. ]
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User: How many solution sets do systems of linear inequalities have? Do solutions to systems of linear inequalities need to satisfy both inequalities? In what case might they not?





Weegy: It can have none, one, or infinitely many. Here's a system with no solutions: x 5. Here's a system with one solution: x =3. [ ( = mean "less than or equal" and "greater than or equal") And here's a system with infinitely many solutions: x < 3 and x < 2. In order to satisfy both inequalities, a number must be in both solution sets. ]
Expert answered|bhebhekoh|Points 60|

User: Do the equations x = 4y + 1 and x = 4y – 1 have the same solution? How might you explain your answer to someone who has not learned algebra?

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Asked 10/25/2012 10:34:46 PM
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State one advantage that the method of substitution has over the intersection-of-graph method.
Weegy: In case the solutions are not perfect integers you make get more approximate value in substitution method then the intersection of graph method (More)
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When applying the method of substitution, how do you know that there are no solutions?
Weegy: If while solving a system of equations we get an inequality like 1=0 Since 1 is not equal to 0 the right side of the equation we can conclude the system of equations has no solution User: When applying the method of substitution, how do you know that there are infinitely many solutions? Weegy: We get to know that the system of equations has infinitely many solutions when the system deduces to an eqwuality like 0=0 Then we have infinite number of solutions (More)
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Asked 10/18/2012 8:48:02 PM
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can you provide an example that uses the substitution method to solve two linear equations.
Weegy: The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, [ "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works. (I'll use the same systems as were in a previous page.) Solve the following system by substitution. 2x – 3y = –2 4x + y = 24 The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But — some choices may be better than others. For instance, in this case, can you see that it would probably be simplest to solve the second equation for "y =", since there is already a y floating around loose in the middle there? I could solve the first equation for either variable, but I'd get fractions, and solving the second equation for x would also give me fractions. It wouldn't be "wrong" to make a different choice, but it would probably be more difficult. Being lazy, I'll solve the second equation for y: 4x + y = 24 y = –4x + 24 Now I'll plug this in ("substitute it") for "y" in the first equation, and solve for x: 2x – 3(–4x + 24) = –2 2x + 12x – 72 = –2 14x = 70 x = 5 Copyright © Elizabeth Sta ] (More)
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Asked 10/18/2012 9:20:41 PM
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How might you explain your answer toequations to someone who has not learned algebra?
Weegy: I would need to start to explain to him the basics of Algebra. I would teach it the way how I understood it and using simple terms that he could easily comprehend. I will provide lots of examples and make him try to answer them. [ [ Students learn best when examples are presented one by one, with increasing complexity. The first Algebra example problem that you work should be extremely simple so that every single student in the room will completely follow it. This builds their confident which will be very important for the remainder of the lesson. One by one, you increase the difficulty of the Algebra problems that you solve on the board. It is important not to choose problems that jump ahead in difficulty too far ahead of the previous problem. If the teacher did this, he or she would lose some students quickly. By choosing Algebra problems in a gradual progression of complexity, you can hold the student's attention all the way to the end of the lecture, building their confidence along the way. ] ] (More)
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Asked 10/25/2012 11:11:29 PM
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how do i use the product rule to simplify this expression 8^5·8^2
Weegy: When multiplying exponents with the same base, you simply add the exponents. 8^5*8^2 = 8^7 (More)
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Asked 10/26/2012 2:15:26 PM
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