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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? I am unsure

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- Thread starter jxj
- Start date

No, you want a single ##m_\mathrm{A}##. You need to do the inverse of distribution:$$a (b+c) = ab +...+ ac

- #1

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So the problem is trying to isolate mA

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

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- #2

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- #3

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

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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?DrClaude said:

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ok thanksharuspex said:You may be misinterpreting that. Ignore it and see where you get to.

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No, I wouldn't divide yet, I would simply start by sorting the terms. When you have an equation likejxj said: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?

$$

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*}

$$

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ok so right now I am 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?DrClaude said: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?DrClaude said: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

Homework Helper

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jxj said: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.

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

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when you say factorise could you explain a little more pleaseharuspex said:Factorise each side.

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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.jxj said:when you say factorise could you explain a little more please

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thanks!DrClaude said: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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so right now I am 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?DrClaude said: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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Check that part I highlighted in red.jxj said:so right now I am 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.

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

- #17

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No, you want a single ##m_\mathrm{A}##. You need to do the inverse of distribution:jxj said:for that part, was I supposed to separate them so I would have mA(vA) -(mA)(v’A) ?

$$

a (b+c) = ab + ac

$$

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