# how to rearrange? and get x

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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
0
5 years ago
#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.
0
#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?
0
5 years ago
#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 0
5 years ago
#5
(Original post by p29)
1.
−4−(3−𝑥)/(2−𝑎)=(6−5𝑥)/(−𝑎+1)
Assuming you mean Multiply both sides by .
0
5 years ago
#6
(Original post by RDKGames)
Just algebraic manipulation.

If you let a, b, c and d equal anything you want; then Isn't it? EDIT:You just changed it.
0
5 years ago
#7
Typo, I edited it before you posted that lol.
0
#8
no the -4 is on the fraction
0
5 years ago
#9
(Original post by p29)
no the -4 is on the fraction
0
#10
(Original post by Zacken)
thank you
0
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