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how to rearrange? and get x
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p29
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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
−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
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
Or multiply through the equations by the denominators. Then proceed to solve for x.
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p29
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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.
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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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?
how do you add/subtract the fractions when there are different variables in the denominator?
If you let a, b, c and d equal anything you want; then
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Zacken
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#5
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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
Just algebraic manipulation.
If you let a, b, c and d equal anything you want; then
EDIT:You just changed it.
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RDKGames
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#7
p29
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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
no the -4 is on the fraction
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p29
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#10
(Original post by Zacken)
Okay, same advice applies though.
Okay, same advice applies though.
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