how to rearrange? and get x

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p29
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#1
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#1
1.
−4−(3−𝑥)/(2−𝑎)=(6−5𝑥)/(−𝑎+1)
2.
(2𝑥+5)/3+(𝑥−2)/5= 2𝑥
3.
4/(2𝑥−3)−𝑦/(2−𝑥)= 3𝑥 − 1
plzzz help
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RDKGames
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#2
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#2
(Original post by p29)
1.
−4−(3−𝑥)/(2−𝑎)=(6−5𝑥)/(−𝑎+1)
2.
(2𝑥+5)/3+(𝑥−2)/5= 2𝑥
3.
4/(2𝑥−3)−𝑦/(2−𝑥)= 3𝑥 − 1
plzzz help
Add/subtract the fractions before multiplying both sides by the denominator.

Or multiply through the equations by the denominators. Then proceed to solve for x.
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p29
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#3
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#3
(Original post by RDKGames)
Add/subtract the fractions before multiplying both sides by the denominator.

Or multiply through the equations by the denominators. Then proceed to solve for x.
how do you add/subtract the fractions when there are different variables in the denominator?
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RDKGames
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#4
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#4
(Original post by p29)
how do you add/subtract the fractions when there are different variables in the denominator?
Just algebraic manipulation.

If you let a, b, c and d equal anything you want; then \displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}
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Zacken
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#5
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#5
(Original post by p29)
1.
−4−(3−𝑥)/(2−𝑎)=(6−5𝑥)/(−𝑎+1)
Assuming you mean -4 - \frac{3-x}{2-a} = \frac{6-5x}{-a + 1}

Multiply both sides by (2-a)(-a + 1).
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BrexitBojo
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#6
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#6
(Original post by RDKGames)
Just algebraic manipulation.

If you let a, b, c and d equal anything you want; then \displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{cd}
Isn't it?
\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}

EDIT:You just changed it.
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RDKGames
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#7
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#7
(Original post by alexjones1994)
Isn't it?
\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}
Typo, I edited it before you posted that lol.
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p29
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#8
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#8
(Original post by Zacken)
Assuming you mean -4 - \frac{3-x}{2-a} = \frac{6-5x}{-a + 1}

Multiply both sides by (2-a)(-a + 1).
no the -4 is on the fraction
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Zacken
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#9
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#9
(Original post by p29)
no the -4 is on the fraction
Okay, same advice applies though.
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p29
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#10
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#10
(Original post by Zacken)
Okay, same advice applies though.
thank you
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