elementary algebra.. please help. is an CPT question
As you learned back in grammar school, you add fractions together by first making sure they have the same denominator. Your denominators here are y, x, and xy. So change the first two to xy.
(x/y) + (24/x) - (3 / xy) =
(x^2 / xy) + (24y / xy) - (3 / xy)
Now combine the numerators:
(x^2 + 24y - 3) / (xy)
is that the entire equation? there should be a number after the =.
If there is no value given after the = then I guess you have to just simplify as much as possible.
add x/y+24/x
= (x^2+24y)/xy -3/xy
then add (x^2 + 24y)/xy to -3/xy
= (x^2+24y-3)/xy
XY â 0, then x/y + 24/x - 3/xy = ?
x²/yx + 24y/xy - 3/xy = (x²+21)/xy
=(x^2+24y-3)/(xy)
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Verified answer
As you learned back in grammar school, you add fractions together by first making sure they have the same denominator. Your denominators here are y, x, and xy. So change the first two to xy.
(x/y) + (24/x) - (3 / xy) =
(x^2 / xy) + (24y / xy) - (3 / xy)
Now combine the numerators:
(x^2 + 24y - 3) / (xy)
is that the entire equation? there should be a number after the =.
If there is no value given after the = then I guess you have to just simplify as much as possible.
add x/y+24/x
= (x^2+24y)/xy -3/xy
then add (x^2 + 24y)/xy to -3/xy
= (x^2+24y-3)/xy
XY â 0, then x/y + 24/x - 3/xy = ?
elementary algebra.. please help. is an CPT question
x²/yx + 24y/xy - 3/xy = (x²+21)/xy
=(x^2+24y-3)/(xy)