Solving Algebra 2 Problems

Solving Algebra 2 Problems-60
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.So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4.

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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.

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I was homeschooled (that's not the confession part), and in 8th grade my algebra textbook had the answers to half the problems in the back. That seems to be the premise behind app called Socratic. The app lets you take a picture of a problem (you can also type it in, but that's a little laborious), and it'll not only give you an answer, but the steps necessary to to arrive at that answer — and even detailed explanations of the steps and concepts if you need them.

Of course, cheating at math is a terrible way to learn, because the whole point isn't to know the answer to 2x 2 = 7x - 5, it's to understand the learn?is the third math course in high school and will guide you through among other things linear equations, inequalities, graphs, matrices, polynomials and radical expressions, quadratic equations, functions, exponential and logarithmic expressions, sequences and series, probability and trigonometry.This is divided into 13 chapters and each chapter is divided into several lessons.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.But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation.If this is equal to that, anything we do to that, we also have to do to this. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that.To do this, I simply inverted each value for "hours to complete job": My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together.In this case, I know the "together" time, but not the individual times.So in order to do that, let's get rid of this 2.And the best way to get rid of that 2 is to subtract it.


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