Solving Algebra 2 Problems

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One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times. Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time: Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate: Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable.

If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him.

"Work" problems usually involve situations such as two people working together to paint a house.

You are usually told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together.

There are a lot of computer-based algebra solvers out there, but for Socratic they had to do some extra engineering to get at the steps a human would need to solve the same problem.

Also, I'd be remiss not to mention Photomath, which has been doing this since 2014, and actually has step-by-step explanations in the recently released Photomath paid version (there's a free trial).But for algebra this thing is I pointed it at 2x 2 = 7x - 5, which I wrote down at random, and it gave me a 10 step process that results in x = 7/5.It has trouble with word problems, but if you can write down a word problem in math notation it shouldn't be an issue.I could write just a plus 0, but I think that's a little unnecessary. And our whole goal here is to isolate the x, to solve for the x.And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. And then the right-hand side, negative 18 plus 2, that's negative 16. This right-hand side, when x is equal to negative 72, does indeed equal negative 16.The method of solution for "work" problems is not obvious, so don't feel bad if you're totally lost at the moment.There is a "trick" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does Since the unit for completion was "hours", I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour.Under each lesson you will find theory, examples and video lessons.Mathplanet hopes that you will enjoy studying Algebra 2 online with us!It's technology that augments a human brain, not just a distraction.The creator of Socratic just open sourced its step-by-step solver, called mathsteps.


Comments Solving Algebra 2 Problems

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