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    −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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    (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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    (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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    (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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    (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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    (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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    (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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    (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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    (Original post by p29)
    no the -4 is on the fraction
    Okay, same advice applies though.
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    (Original post by Zacken)
    Okay, same advice applies though.
    thank you
 
 
 
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