Mathematics 68 Online
OpenStudy (anonymous):

Joe has a collection of nickels and dimes that is worth \$7.65. If the number of dimes was doubled and the number of nickels was increased by 35, the value of the coins would be \$14.30. How many nickels and dimes does he have? saifoo.khan will u help me?

OpenStudy (anonymous):

I need help big time... there was someone helping me a few mins ago but idk where he went

OpenStudy (anonymous):

I can help.

OpenStudy (anonymous):

thank u

OpenStudy (saifoo.khan):

Let n = number of nickels Let d = number of dimes ----------------------- The question gives you info to make two equations : 0.05*n + 0.10*d = 7.65 0.05*(n +35) + 0.10*(2d) = 14.30 ------------------------ Start by simplifying and rewriting the second equation : 0.05*(n +35) + 0.10*(2d) = 14.30 0.05*n + (0.05*35) + 0.20*d = 14.30 0.05*n + 1.75 + 0.20*d = 14.30 0.05*n + 0.20*d = 12.55 ---------------------- Now the two equations are : 0.05*n + 0.10*d = 7.65 0.05*n + 0.20*d = 12.55 Subtract the second equation from the first and solve for d : - 0.10 * d = - 4.90 d = ( - 4.90 / - 0.10 ) = 49 Substitute that value of d into the first equation and solve for n : 0.05*n + 0.10*d = 7.65 0.05*n + (0.10 * 49) = 7.65 0.05*n + 4.90 = 7.65 0.05*n = ( 7.65 - 4.90 ) 0.05*n = 2.75 n = ( 2.75 / 0.05 ) = 55 ------------------ You can check to verify that the values n = 55 and d = 49 satisfy both equations.

OpenStudy (anonymous):

How can you type so fast...?!

OpenStudy (anonymous):

thank you so so soooooo much....

OpenStudy (saifoo.khan):

Abe, i have solutions of all of these with me, coz i help manyyy college people. You welcome 87

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