- Solving equations by graphing
- Solving equations by graphing: intro
- Solving equations graphically: intro
- Solving equations by graphing: graphing calculator
- Solving equations graphically: graphing calculator
- Solving equations by graphing: word problems
- Solving equations graphically: word problems
- Equations: FAQ
Solving equations by graphing: graphing calculator
We can approximate the solutions to any equation by graphing both sides of the equation and looking for intersection points. If we have a graphing calculator, we can get a very good approximation of the solution.
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- Wouldn't the graph of -ln(2x)=2|x-4|-7 be a line with only one solution for x regardless of the y?(4 votes)
- Much like Sal did in the video, the graph has two functions, y=-ln(2x) and y=2|x-4|-7. If plotted on a graphing calculator, there are two intersections and solutions for x: 0.5 and 6.238.
Though I am unsure why the graph of -ln(2x)=2|x-4|-7 is a line, the line is approximately x=6.238, which is the other solution.(6 votes)
- is it possible to solve it by not using the calculator? Is it allowed to use calculator for tests?(4 votes)
- I suppose you could algebraically solve this particular problem by setting the functions equal to each other although this would probably be a harder process and you would still need to do calculations with e, which would probably require a calculator of some kind, if not a graphing one.
If having a physical calculator is the problem, there are programs you can use such as websites and apps that accomplish the same thing.
As for being allowed on tests, that is something to ask your teacher and they might even have a class set. For standardized tests like the SAT, there is a list of acceptable calculators (they won't let you use an app on your phone for obvious reasons) and while it is possible with a 4-function or scientific calculator, graphing is recommended. If the need be you could always borrow one from a friend or fellow student if it is just for one test (I did this myself when I took the SAT and borrowed a scientific calculator, but I was already familiar with it. Using a calculator for the first time on an important test could present a problem). Other tests might be possible without calculators in general, but become easier (and quicker) with them.(2 votes)
- how would you graph that graph?(2 votes)
- Use a graphing calculator! Sal is using desmos.com(1 vote)
- how do you solve the practice exercises if you dont have a Desmos graphing calculator
for example:-ln(2x)=2|x-4|-7(1 vote)
- You can also go online and search up Desmos Graphing Calc =)(1 vote)
- How does setting both sides of the equation equal to y a way to solve it and finding the intersections as solutions? I'm very confused on how this actually works.(1 vote)
- I use desmos.com(1 vote)
- Iam having trouble , iam not Abel to enter answers , like iam , solving equations when I have to put answer in square or cube , I can’t , like is that specific key I have to press on keyboard ⌨️ , and can I use calculator to solve the equations ,
- For some questions there won't be a "Show Calculator" button there, but if there is one, just click on the button. Answering your question about keyboard inputs, you could use the "^" key which you can access by pressing shift and 6 at the same time, and then enter in whatever number you want. But there is also another method. Which is just clicking the answer box and then selecting the small icon "e^x."
Keyboard Input Example
y = x^2 + 3 or y= x^3 + 3
(Note: You can also do this on your graphing calculator the same way, as the "^" symbol is also on there)
Hope that answers your question :)(1 vote)
- But how would you do something like x^2=3x-5=4x+3 using a table and graph to solve(1 vote)
- You have two equal signs which looses any meaning to what you are asking. Is it x^2 + (or -) 3x-5=4x+3?(1 vote)
- [Instructor] We are told we want to solve the following equation, the negative natural log of two x is equal to two times the absolute value of x minus four, all of that minus seven. One of the solutions is x is equal to 0.5. Find the other solution. They say hint, use a graphing calculator and round your answer to the nearest tenth. So pause this video and have a go at this if you like, and then we'll work on this together. And I encourage you to have a go on it, (laughing) a go at it. All right now let's work on this together. Now the key here is to realize that we might be able to solve this by graphing, or at least approximate the solutions to this by graphing. And the way we do that is, if we have an equation, especially a hairy equation like this, in one variable, we can set y equal to the left and then set y equal to the right and then graph each of those functions and then think about where they intersect. Because they'll intersect at an x-value that gives us the same y-value, and that means that the two sides are the same. So what do I mean by that? Well we could set y is equal to the negative natural log of two x. So we could have one equation or one function like that. And then we could have another equation or function that y is equal to two times the absolute value of x minus four minus seven. And let's see where they intersect. And the x-values where they intersect, those are going to be solutions to that. So I'm going to use Desmos as my graphing calculator. So let's type in the two sides. So first I'll do the left side. So if y is equal to the negative natural log of two x. And actually, let me make my color to be the same or as close as I can, so maybe closer to that bluer color. Okay and then the net one, I want y is equal to two times, two times the absolute value, actually I don't know whether Desmos prefers, I'll use that, actually that works, okay, x minus four, and then I will close my absolute value, and then I have minus seven. And I will do this in the red color so that we can keep track of things. Okay. So those are my two graphs. And now, I just need to think about where they intersect. And one of the solutions is x equal to 0.5. That's not the one they want. They want the other solution so to speak. So let's see. So let's see, we have one solution, actually let me zoom in a little bit. So when x is equal to 0.5, that's this solution, that's this solution right over here. It looks like y is equal to zero there. But then the other point of intersection seems to be right over here, and actually Desmos has a nice little feature where it'll tell us that point right over there. But you could even approximate it. You can see that x is over six and that each of these, let's see one, two, three, four, five, each of those is .2. So it's going to be 6.2 something is what I would do, and they want us to round to the nearest tenth anyway. So you don't even need to use that feature, but you can see very clearly that when x is equal to approximately 6.238, that we get a y is equal to negative 2.54. Or another way to think about it is, when x is approximately equal to, when x is approximately equal to 6.2, that the two sides of this equation are going to be equal to each other, are approximately equal to each other. And we're done, we've just solved using a, or at least approximated a solution using graphing in a graphing calculator.