Question

Can someone help me understand problems like this P = 2l + 2w, for w.? I don't know how to do them for the like of me i cannot figure them out if anyone can give me the steps to the problem and go over how you do it that would be nice.

Answer 1

That sounds like P is the perimeter of a rectangle with length l and width w. To solve it for w:

\[P = 2l + 2w\]Subtract 2l from both sides\[P-2l = 2l - 2l + 2w\]\[_-2l = 2w\]Divide both sides by 2\[P/2 - 2l/2 = 2w/2\]\[P/2 - l = w\]\[w = P/2 -l\]
Remember, you can do the same thing to both sides of an equation without damaging it.

Answer 2

We can test our solution by substituting it in the original formula:

\[P = 2l + 2w\]\[P=2l + 2(P/2 - l) = 2l + 2P/2 -2l = 2l + P - 2l = P\checkmark\]

Answer 3

Why do you divide by 2 where does 2 come from?

Answer 4

your third line
*_\(2l = 2w\)

should be \(P-2l=2w\)

Answer 5

Sorry, I seem to have formatted that badly. Let me try again!

\[P = 2l + 2w\]Subtract 2l from both sides\[P-2l = 2l - 2l + 2w\]\[P -2l = 2w\]We'll switch sides because we are solving for w\[2w = P-2l\]Now we want just w =, not 2w =, so we divide both sides by 2\[2w/2 = P/2 - 2l/2\]\[w = P/2 -l\]

Answer 6

If it came out to be 9.376w = we would divide by 9.376. We're just trying to make the coefficient of w = 1, so that we can read out the value directly. We make the coefficient of w = 1 by dividing by the coefficient if it isn't 1.

Answer 7

okay, so how would i solve these problems V = 1/3 lwh, for h.?

Answer 8

\[V = {\frac{1}{3}lwh}\]Just divide both sides by everything that isn't \(h\). First I would multiply both sides by 3 to remove the fraction
\[\frac{1}{3}lwh = V\]\[lwh = 3V\]Now divide both sides by \(lw\) to get\[\frac{lwh}{lw} = \frac{3V}{lw}\]Cancel matching items in numerator and denominator\[h = \frac{3V}{lw}\]

Answer 9

Just divide both sides by everything that isn't h, so basically divide each side by 1/3lw? dont understand that part right there

Answer 10

You had \[V = \frac{1}{3}lwh\] and wanted to solve for \(h\). The way to isolate the variable you want (\(h\) in this case) is to move everything to the other side of the equals sign. If you have multiple terms (terms being things separated by +, -) you add or subtract them from both sides to remove them from the side with the variable you want to isolate. If you just have one term containing the variable, then you need to divide by everything else in the term. That's why I said "divide by everything that isn't h" — we needed to divide each side by the rest of the term containing h. If you have a fraction, many people find it easier to multiply both sides by the denominator of the fraction (3 in this case) than to actually divide by the fraction.

Answer 11

Getting any clearer?

Answer 12

Yes thanks
so the steps should be in this order

V = 1/3lwh

3 x v = 1/3lwh x 3

lw divided by 3v = lwh divided by lw

lwh over lw = 3v over lw, Correct?

Answer 13

I didn't quite follow your next-to-last step.

\[V = \frac{1}{3}lwh\]Multiply both sides by 3\[3*v = 3*\frac{1}{3}lwh\]\[3V=lwh\]Divide both sides by l\[\frac{3V}{l}=\frac{lwh}{l}\]\[\frac{3V}{l}=wh\]Divide both sides by w\[\frac{3V}{lw} = \frac{wh}{w}\]\[\frac{3V}{lw}=h\]Switch sides\[h=\frac{3V}{lw}\]

Answer 14

If we started out with the variable we wanted to isolate on both sides of the equation, we would add or subtract the term to both sides to move it to the other side. For example
\(3x + 2y = 4x - y + 16\), solve for \(y\):
\[3x+2y=4x-y+15\]Add y to both sides
\[3x+2y+y=4x-y+15+y\]\[3x+3y=4x+15\]Now subtract 3x from both sides\[3x+3y-3x=4x+15-3x\]\[3y=x+15\]Divide both sides by 3\[\frac{3y}{3} = \frac{x}{3}+\frac{15}{3}\]\[y=\frac{x+15}{3}\]