Make x the subject: Many thanks in advance. y = (2/3)zx And this one which I have already done but I need checking: y = (6x - 7)/3 My answer: x = (3y + 7)/6
Last edited by BomberAF; March 11th 2010 at 12:23 PM.
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Originally Posted by BomberAF Make x the subject: Many thanks in advance. y = (2/3)zx And this one which I have already done but I need checking: y = (6x - 7)/3 My answer: x = (3y + 7)/6 gives Similarly gives
Originally Posted by harish21 gives Similarly gives Thanks for the answer to the first question but you have got the second one incorrect: It is (2/3)yz, the y and z are not over the 3. Could you also check this one for me: y = (1/2)xz. The answer I get is: x = (2y)/z
Originally Posted by BomberAF Thanks for the answer to the first question but you have got the second one incorrect: It is (2/3)yz, the y and z are not over the 3. Could you also check this one for me: y = (1/2)xz. The answer I get is: x = (2y)/z Okay. For the first question, i thought was in the numerator. If is in the denominator, then your question becomes which gives: For you get hope that is what you are looking for!
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