Solving for a variable | Linear equations | Algebra I | Khan Academy
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0:01 - 0:06Solve P equals 2l plus 2w for l.
-
0:06 - 0:08So this right here,
this is just the formula -
0:08 - 0:10for the perimeter
of a rectangle. -
0:10 - 0:12Perimeter is equal
to 2 times the length -
0:12 - 0:14plus 2 times the width.
-
0:14 - 0:17But they just want us to solve
this equation right here, -
0:17 - 0:18solve for l.
-
0:18 - 0:19So let's do that.
-
0:19 - 0:25So we have P is equal to
2 times l plus 2 times w. -
0:25 - 0:27So we need to solve for l.
-
0:27 - 0:29So let's isolate all of
the l terms on one side. -
0:29 - 0:31And the best way, we
could just do that -
0:31 - 0:33by leaving it here
on the right and then -
0:33 - 0:34getting rid of this 2w.
-
0:34 - 0:35And the best way to
get rid of this 2w -
0:35 - 0:37is to subtract it
from the right. -
0:37 - 0:39But if you're going to
subtract it from the right, -
0:39 - 0:41you also have to
subtract it from the left -
0:41 - 0:43if you want this
equality to hold. -
0:43 - 0:46So you have to also
subtract it from the left. -
0:46 - 0:48And so the left-hand
side becomes P minus 2w. -
0:51 - 0:54And the right-hand
side, you get-- -
0:54 - 0:56this 2w minus 2w cancels out.
-
0:56 - 0:59You just have a 2l.
-
0:59 - 1:00And then if you
want to solve for l, -
1:00 - 1:04you just have to divide both
sides of this equation by 2. -
1:04 - 1:07You just divide both sides
of this equation by 2. -
1:07 - 1:09And we have isolated our l.
-
1:09 - 1:12We get l is equal to
P minus 2w over 2. -
1:12 - 1:14Or if we wanted to
write it the other way, -
1:14 - 1:20you could write l is equal
to P minus 2w over 2. -
1:20 - 1:23And we are done.
- Title:
- Solving for a variable | Linear equations | Algebra I | Khan Academy
- Description:
-
Solving for a Variable
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Fran Ontanaya edited English subtitles for Solving for a variable | Linear equations | Algebra I | Khan Academy | |
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Fran Ontanaya edited English subtitles for Solving for a variable | Linear equations | Algebra I | Khan Academy |