solve: (11/4 y^1/4) - (3/2y^-3/4) = 0
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Originally Posted by samtheman17 solve: (11/4 y^1/4) - (3/2y^-3/4) = 0 if you mean ... $\displaystyle \frac{11}{4} y^{\frac{1}{4}} - \frac{3}{2} y^{-\frac{3}{4}} = 0$ then start by multiplying every term by $\displaystyle y^{\frac{3}{4}}$
okay.......... so far i have (22/4 y ^3/16) - (6/2y ^ -9/16) now i make the others have the same denominator so..... (22/4 y ^ 3/16) - (12/4y ^ -9/16) ...... now im not sure how to subtract exponents
Originally Posted by samtheman17 okay.......... so far i have (22/4 y ^3/16) - (6/2y ^ -9/16) now i make the others have the same denominator so..... (22/4 y ^ 3/16) - (12/4y ^ -9/16) ...... now im not sure how to subtract exponents you do not multiply the exponents. note the following ... $\displaystyle y^{\frac{3}{4}} \cdot y^{\frac{1}{4}} = y^{\frac{3}{4} + \frac{1}{4}} = y $ $\displaystyle y^{\frac{3}{4}} \cdot y^{-\frac{3}{4}} = y^{\frac{3}{4} - \frac{3}{4}} = y^0 = 1 $
ohh okay, so (11/4 y) - (3/2) = 0 then solve for y and get 6/11 thanks you !
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