- #1

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## Homework Statement

So the problem is trying to isolate mA

## Homework Equations

## The Attempt at a Solution

am pretty lost. I believe you have to divide the primes? Im unsure

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- #1

- 18

- 2

So the problem is trying to isolate mA

am pretty lost. I believe you have to divide the primes? Im unsure

- #2

DrClaude

Mentor

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- #3

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You may be misinterpreting that. Ignore it and see where you get to.My teacher said because the vA and vB on the right were prime they could not be combined

- #4

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when doing this, don't you divide the ##m_\mathrm{A}## plus ##v_\mathrm{A}## prime on the right by its counterpart on the left first? even though it contains a prime?

- #5

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ok thanksYou may be misinterpreting that. Ignore it and see where you get to.

- #6

DrClaude

Mentor

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No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation likewhen doing this, don't you divide the ##m_\mathrm{A}## plus ##v_\mathrm{A}## prime on the right by its counterpart on the left first?

$$

5x -3 = 2x + 1

$$

the first step is to try and gather all terms in ##x## together, and leave the rest on the other side,

$$

\begin{align*}

5x - 3 -2x +3 &= 2x + 1 - 2x + 3 \\

3x &= 4

\end{align*}

$$

- #7

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ok so right now im a little confused because unlike the example you gave with 5x, the formula is essentially equal on both sides minus the primes. in my mind I would try to cross out the mAvA on the right, is that right?No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like

$$

5x -3 = 2x + 1

$$

the first step is to try and gather all terms in ##x## together, and leave the rest on the other side,

$$

\begin{align*}

5x - 3 -2x +3 &= 2x + 1 - 2x + 3 \\

3x &= 4

\end{align*}

$$

- #8

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I tried putting all of mA on the left and I got to mAvA - mAvA (prime) = mBvB (prime) - mBvB did I mess up?No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation like

$$

5x -3 = 2x + 1

$$

the first step is to try and gather all terms in ##x## together, and leave the rest on the other side,

$$

\begin{align*}

5x - 3 -2x +3 &= 2x + 1 - 2x + 3 \\

3x &= 4

\end{align*}

$$

- #9

verty

Homework Helper

- 2,164

- 198

I tried putting all of mA on the left and I got to mAvA - mAvA (prime) = mBvB (prime) - mBvB did I mess up?

No, that's correct. How you want to think of it is doing the same thing to both sides. So to move "m_a v'_a" to the left, subtract it from both sides. Or if it was multiplied by that, you would divide it on both sides.

- #10

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Factorise each side.I tried putting all of mA on the left and I got to mAvA - mAvA (prime) = mBvB (prime) - mBvB did I mess up?

- #11

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when you say factorise could you explain a little more pleaseFactorise each side.

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- #13

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thanks!You want to go from ##m_\mathrm{A} v_\mathrm{A} - m_\mathrm{A} v_\mathrm{A}'## to an expression where ##m_\mathrm{A}## appears only once, using the distributive property of multiplication.

- #14

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so right now im at distributing mA by itself into (vA - mAv’A) on the left of the equal sign and vice versa for mB on the right of the equal sign. am I on the right track?You want to go from ##m_\mathrm{A} v_\mathrm{A} - m_\mathrm{A} v_\mathrm{A}'## to an expression where ##m_\mathrm{A}## appears only once, using the distributive property of multiplication.

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- #16

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for that part, was I supposed to separate them so I would have mA(vA) -(mA)(v’A) ?Check that part I highlighted in red.

- #17

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