From simple queries to complex problems, IDNLearn.com provides reliable answers. Join our community to receive prompt, thorough responses from knowledgeable experts.
Sagot :
P = 2( L + W )
-divide both sides by 2
P/2 = L + W
-subtract L from both sides
P/2 - L = W
or
(P - L) / 2 = W
-divide both sides by 2
P/2 = L + W
-subtract L from both sides
P/2 - L = W
or
(P - L) / 2 = W
Hello,
We know that if we make the same operation on both sides, we don't affect the equation:
For example:
[tex]2=2 \\ 2\bold{+2}=2\bold{+2} \\ 4=4[/tex]
Do you see? we add 2 on both sides and we don't affect the equality
We can do the same with the multiplication, for example
4=4 (i'll multiply both sides by 1/2)
[tex]4*\bold{ \frac{1}{2}}=4*\bold{ \frac{1}{2}} \\ 2=2[/tex]
Now, we do exactly the same in your excercise:
[tex]P=2(L+W) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[multiply\,\,by\,\,1/2] \\ \\ P*\bold{ \frac{1}{2}}=2(L+W)*\bold{ \frac{1}{2}} \\ \\ \frac{P}{2}=L+W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[Subtract\,\,L]\\ \\ \frac{P}{2}-\bold{L}=L+W-\bold{L}\\ \\ \frac{P}{2}-\bold{L}=L-\bold{L}+W \\ \\ \frac{P}{2}-L=W\\ \\ \boxed{ W= \frac{P}{2}-L }[/tex]
We know that if we make the same operation on both sides, we don't affect the equation:
For example:
[tex]2=2 \\ 2\bold{+2}=2\bold{+2} \\ 4=4[/tex]
Do you see? we add 2 on both sides and we don't affect the equality
We can do the same with the multiplication, for example
4=4 (i'll multiply both sides by 1/2)
[tex]4*\bold{ \frac{1}{2}}=4*\bold{ \frac{1}{2}} \\ 2=2[/tex]
Now, we do exactly the same in your excercise:
[tex]P=2(L+W) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[multiply\,\,by\,\,1/2] \\ \\ P*\bold{ \frac{1}{2}}=2(L+W)*\bold{ \frac{1}{2}} \\ \\ \frac{P}{2}=L+W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[Subtract\,\,L]\\ \\ \frac{P}{2}-\bold{L}=L+W-\bold{L}\\ \\ \frac{P}{2}-\bold{L}=L-\bold{L}+W \\ \\ \frac{P}{2}-L=W\\ \\ \boxed{ W= \frac{P}{2}-L }[/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.