Finding mistakes in one-step equations | 6th grade | Khan Academy

We're told that Lisa tried to solve an equation: see, 42 is equal to 6a, or 6 times a. Then we can see her steps here, and they say where did Lisa make her first mistake. So pause this video and see if you can figure that out. It might be possible she made no mistakes.

All right, well we know she ends up with seven equals six, which is sketchy. So let's see what happened here. So right over here, it looks like, well, she did something a little bit strange. She divided the left-hand side by 6 and the right-hand side by a. You don't want to divide two sides of an equation by two different things. Then it's no longer going to be an equation; the equality won't hold. An algebraically legitimate thing is to do the same thing to both sides, but she didn't do it here. So this is where she made her first mistake.

Let's give another example here. So here it says that Jin tried to solve an equation: all right, x plus 4.7 is equal to 11.2. Where did Jin make his first mistake? Pause this video and try to figure it out.

All right, so it looks like in order to isolate the x on the left-hand side, Jin is subtracting 4.7 from the left, and then also subtracting 4.7 from the right, so that is looking good: doing the same thing to both sides, subtracting 4.7 from both sides. Then over here on the left-hand side, these two would cancel, so you'd be left with just an x.

And let's see, 11.2 minus 4.7: 11.2 minus 4 would be 7.2, and then minus the 0.7 would be 6.5. So this is where Jin made his mistake on the calculating part.

Let's do another example; this is a lot of fun. So here we are told that Marina tried to solve an equation, and we need to figure out where Marina made her first mistake. All right, 1/6 is equal to two-thirds y.

So the first step, or the first thing that Marina did right over here is to multiply both sides of this equation by the reciprocal of two-thirds, which is three-halves. She multiplied the left-hand side by three-halves, multiplied the right-hand side by three-halves, which is a very reasonable thing to do: we’re doing the same thing to both sides, multiplying by three-halves.

Then when we go over here, let's see: three-halves times one-sixth, we could divide the numerator and the denominator by three, so it's going to be one over two. So that indeed is going to be one-half times one-half, which is one-fourth, so that checks out.

And on this side, if you multiply three-halves times two-thirds, that's going to be one, so this checks out. So it actually looks like Marina did everything correctly: no mistake, no mistake for Marina.

Let's do one last example. So here, Taylor is trying to solve an equation, and so where did Taylor first get tripped up? n minus 2.7 is equal to 6.7.

In order to isolate this n over here, I would add 2.7 to both sides, but that's not what Taylor did. Taylor subtracted 2.7 from both sides. So the first place that Taylor starts to trip up or move in the wrong direction is right over here.

Now what Taylor did is not algebraically incorrect; you would end up with n minus 5.4 is equal to 4. But it's not going to help you solve this equation. You just replace this equation with another equivalent equation that is no simpler than the one before. And then, of course, instead of getting n minus 5.4 equals four, Taylor calculated incorrectly as well.

But where they first started to get tripped up, or at least not move in the right direction, would be right over here.