- Number of solutions to equations
- Worked example: number of solutions to equations
- Number of solutions to equations
- Creating an equation with no solutions
- Creating an equation with infinitely many solutions
- Number of solutions to equations challenge
Creating an equation with infinitely many solutions
Sal shows how to complete the equation 4(x - 2) + x = 5x + __ so that it has infinitely many solutions. Created by Sal Khan.
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- is it possible for an equation to have more than one solution but not infinite?(17 votes)
- It is possible for other equations (e.g., quadratic equations) to have more than one solution, but in terms of linear, no.(16 votes)
- Is there a faster way to find out what math problems are infinite solutions? I do the problems and then see if they are infinite, but it takes a long time...(5 votes)
- I simplify each side of the equation. If they are the same, then you've got infinitely many solutions.
Here is an example:
23x-5-12x=11(x-1)-6 is equal to 11x-5=11x-5
Hope this helps. God bless!(8 votes)
- is anyone else finding this too be incredibly difficult? like I have watched this video numerous times but the subject is still extremely difficult and foreign for my brain to wrap around. am I stupid or is it a difficult subject?(7 votes)
- Creating problems is a much hinger level thinking process than just solving. In this case, the idea is that you have to create something that makes both the right side of the equation and the left side to be equal to each other which gives you an infinite number of solutions. so if you have 5x-8 on the left, you need 5x-8 on the right for everything to cancel and end up with 0=0.
What do you think would be the answer to the "?" in 3x + 6 = 3(x + ?) to have infinite solutions?(2 votes)
- when you multiply negative and positive numbers what answer do you get?(5 votes)
- You get a negative number- negative*positive=negative.
For an example: -2*3=-6.(5 votes)
- I am working multistep equations with variables on both sides and I do not understand how to work the problems and am very confused. I have problems for example like f(b)=2b+6
Can you please show me how to work problems such as this?(4 votes)
- Why do we need to know how many solutions there are to each equation??
P.S. I’m not meaning to be sarcastic or rude, I’m genuinely asking!(3 votes)
- When you are asked to solve an equation, you are being asked to find all values that will make the equation be true. Equations with one variable that are linear equation have 3 possible solution scenarios.
1) The variable has one solution
2) The equation is a contradiction (always false), so it has no solutions.
3) The equation is an identity (always true), so the variable has a solution set of all real numbers. In other words, any number you can imagine will make the equation be true. In this scenario, there are infinite solutions.
Understand the number of solutions helps you to identify what is the solution set to the equation.(4 votes)
- I don't get what he says at0:49because if 4x-8+x wouldn't it be 4x minus a positive x because if we remove the 8 then it would be 4x-x and that would give us 3x. I need help. I have a math test about equations please help!(2 votes)
- Well this reply will be too late. But when you remove/transfer a number, you take the operation sign with it:
4x+x=5xI added 8 to both sides,They cancelled out
- Why so many questions?(3 votes)
- why did sal add 4x + x? i know like x is 1 but the equation shows 4x - 8 +x like that doesn't alow it in math?(2 votes)
- It is not clear what you think is not allowed.
The commutative property of addition, lets use reorder the terms in 4x-8+x to get 4x+x-8. Then, combine (add) 4x+1x = 5x. So the expression becomes: 5x-8
Hope this helps.(3 votes)
- my computer Is broken so I cant watch this(2 votes)
We're asked to use the drop-down to form a linear equation with infinitely many solutions. So an equation with infinitely many solutions essentially has the same thing on both sides, no matter what x you pick. So first, my brain just wants to simplify this left-hand side a little bit and then think about how I can engineer the right-hand side so it's going to be the same as the left no matter what x I pick. So right over here, if I distribute the 4 over x minus 2, I get 4x minus 8. And then I'm adding x to that. And that's, of course, going to be equal to 5x plus blank. And I get to pick what my blank is. And so 4x plus x is 5x. And of course, we still have our minus 8. And that's going to be equal to 5x plus blank. So what could we make that blank so this is true for any x we pick? Well, over here we have 5 times an x minus 8. Well, if we make this a minus 8, or if we subtract 8 here, or if we make this a negative 8, this is going to be true for any x. So if we make this a negative 8, this is going to be true for any x you pick. You give me any x, you multiply it by 5 and subtract 8, that's, of course, going to be that same x multiplied by 5 and subtracting 8. And if you were to try to somehow solve this equation, subtract 5x from both sides, you would get negative 8 is equal to negative 8, which is absolutely true for absolutely any x that you pick. So let's go-- let me actually fill this in on the exercise. So I want to make 5-- it's going to be 5x plus negative 8.