SAT (Fall 2023)
- Solving linear equations and linear inequalities — Basic example
- Solving linear equations and linear inequalities — Harder example
- Interpreting linear functions — Basic example
- Interpreting linear functions — Harder example
- Linear equation word problems — Basic example
- Linear equation word problems — Harder example
- Linear inequality word problems — Basic example
- Linear inequality word problems — Harder example
- Graphing linear equations — Basic example
- Graphing linear equations — Harder example
- Linear function word problems — Basic example
- Linear function word problems — Harder example
- Systems of linear inequalities word problems — Basic example
- Systems of linear inequalities word problems — Harder example
- Solving systems of linear equations — Basic example
- Solving systems of linear equations — Harder example
- Systems of linear equations word problems — Basic example
- Systems of linear equations word problems — Harder example
Watch Sal work through a harder Solving linear equations problem.
Want to join the conversation?
- When Sal was at 10x-2=ax+x-2, why didn't he just drop the -2 from both sides and say 10x=ax+x?
Then it would be 10x=ax+1x because that is what x is.
Subtract 1x from both sides and you get 9x=ax so a=9.
Tell me if I'm wrong?(26 votes)
- I having a hard understanding the question and what's being asked for example i would have never guess that infinitely many solution meant x=x(31 votes)
Infinite solutions means that there are an infinite number of values for 'x' which will fulfill the equation
For any value of 'x' it will always be equal to itself.
So x=x will have infinite solutions.(4 votes)
- I want to practice but there tutorial videos(12 votes)
- If you are meaning that there is only videos and no practice, try looking up "solving linear equations and linear inequalities practice problems" on Google and see what you find.(7 votes)
- the phrase "infinitely many solutions" is not explained by solving for a = 9. Otherwise, the problem is straightforward, just solve for a.(7 votes)
- The key thing we are supposed to demonstrate here (and in similar "infinite solutions" problems) is that we realize that in a problem with infinitely many solutions, both sides of the equation must be the same when simplified, so that any value for x will give the same result. In order to demonstrate that, here we have to realize that we need a value for a that gives the same value as the other side of the equation when simplified.(15 votes)
- If a was equal to anything but 9, wouldn't the equation be invalid and have no solutions?(9 votes)
- Actually, the equation wouldn't be invalid and have no solutions, it just would not have infinitely many solutions. Instead, the equation would only be true when x = 0.(6 votes)
- Just a simple step: substract 3 from -5 at left and then the eqn looks like this:
since -2 is common on both sides u can equate 10x with (a+1)x
so , 10x=(a+1)x
or, 10x-x =ax
It's just 2 step eqn.(10 votes)
- Hello! Fellow test takers, just like you all I am also preparing for my test, I wish you all ,and wish me too
Good luck out there test takers you've got this
Just completed this section It's easy if you have any doubts go ahead and drop'em ill try to answer to it as simply as possible(10 votes)
- why did he distribute the x backwards?(4 votes)
- [Instructor] In the equation shown above, a is a constant. For what value of a does the equation have infinitely many solutions? So you end up with infinitely many solutions if your equation simplifies to something like x is equal to x, or one is equal to one, something that's true that's going to be true for any x that you pick. So let's see what we could do with this thing right over here. These are obviously not, if you got 100 equals 100, that would be the same, that would have an infinitely many solutions. Zero equals zero. These were all be situations where you have an infinite number of solutions. So when I look at this thing up here, my first instinct is, well let's just see if I can simplify this a little bit. I'll leave the a in there and then see if I can get to a point where it's gonna have an infinite number of solutions. So let me just rewrite it. So we're gonna have three plus 10x minus five is equal to a plus one times x minus two. So let's see, on the left-hand here I can add the three and the negative five. Or I could take three minus five. That would be negative two. So I get 10x minus two is equal to, let me distribute the x. So it's gonna be ax plus x. All I did here is I distributed the x, minus two. Now let's see, what happens if, let's see, I could get rid of both of these negative twos if I add two to both sides. So if I just, remember anything I do to one side I've gotta do to the other one if I wanna hold the equality to continue to be true. So I just added two to both sides. And I'm left with 10x is equal to ax plus x. Let's see. Let's subtract x from both sides. So if I subtract x. Actually I could write it like this. I could subtract x from both sides. On the left-hand side I'm gonna get nine x. On the right-hand side I'm gonna get ax. So how could I have an infinite number of solutions, an equation that's gonna be true for any x? Well if a was equal to nine, because if a is equal to nine I'm gonna have a situation. So if a is equal to nine then you're gonna have a situation where nine x is going to be equal to, instead of a, I'd write nine. Is going to be equal to nine x. Well that's going to be true for any x. Any x times nine is going to be equal to that same x times nine again. You're gonna have an infinite number of solutions. And so a needs to be equal, a it needs to be equal to, a is equal to nine. Now what's really interesting here is think about what would happen if a is any of these, if a is any of these other things right over here. Then you're going to force a different solution. But anyway, we'll leave that for another video.