*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.

*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.*

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.

I like the Socratic interface and explanations a bit better, but I'm glad to see this is a vibrant market.

## Solving Algebra 2 Problems

If you're seeing this message, it means we're having trouble loading external resources on our website.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.If you're behind a web filter, please make sure that the domains *.and *.are unblocked.We have the equation negative 16 is equal to x over 4, plus 2. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms.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.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.Under each lesson you will find theory, examples and video lessons.Mathplanet hopes that you will enjoy studying Algebra 2 online with us!

## Comments Solving Algebra 2 Problems

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