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how to rearrange? and get x

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
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.
Reply 2
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p29
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12
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?
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 ab+cd=ad+bcbd\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}
(edited 7 years ago)
Reply 4
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Zacken
22
Original post by p29
1.
−4−(3−𝑥)/(2−𝑎)=(6−5𝑥)/(−𝑎+1)


Assuming you mean −4−3−x2−a=6−5x−a+1-4 - \frac{3-x}{2-a} = \frac{6-5x}{-a + 1}

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

If you let a, b, c and d equal anything you want; then ab+cd=ad+bccd\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{cd}


Isn't it?
ab+cd=ad+bcbd\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}

EDIT:You just changed it.
Original post by alexjones1994
Isn't it?
ab+cd=ad+bcbd\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}


Typo, I edited it before you posted that lol.
Reply 7
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p29
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12
Original post by Zacken
Assuming you mean −4−3−x2−a=6−5x−a+1-4 - \frac{3-x}{2-a} = \frac{6-5x}{-a + 1}

Multiply both sides by (2−a)(−a+1)(2-a)(-a + 1).


no the -4 is on the fraction
Reply 8
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Zacken
22
Original post by p29
no the -4 is on the fraction


Okay, same advice applies though.
Reply 9
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p29
OP
12
Original post by Zacken
Okay, same advice applies though.


thank you

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