Question 189726
Let n=# of nickels, d=# of dimes, and q=# of quarters



Since "He found six more dimes than nickels but three times as many quarters as dimes", this means that {{{d=n+6}}} and {{{q=3d}}}. We'll call these equations 1 and 2.



Furthermore, because "The total value of the coins was $11.40", this means that {{{0.05n+0.10d+0.25q=11.40}}}



Note: the total value of the nickels alone is {{{0.05n}}} (ie the value of ONE nickel multiplied by the number of nickels). The same is applied to the dimes and quarters. These expressions are then added up to get the total value. This is probably where you're stuck.



{{{0.05n+0.10d+0.25q=11.40}}} Start with the last equation.



{{{100(0.05n)+100(0.10d)+100(0.25q)=100(11.40)}}} Multiply EVERY term by 100 to make every number a whole number



{{{5n+10d+25q=1140}}} Multiply. Let's call this equation 3.




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So we have the equations



Equation 1: {{{d=n+6}}}


Equation 2: {{{q=3d}}}


Equation 3: {{{5n+10d+25q=1140}}}



You may see this as a system of equations, which looks like


{{{system(d=n+6,q=3d,5n+10d+25q=1140)}}}




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{{{5n+10d+25q=1140}}} Start with the third equation



{{{5n+10d+25*highlight((3d))=1140}}} Plug in {{{q=3d}}}. Notice how the variable "q" is no longer in the equation.



{{{5n+10d+25(3d)=1140}}} 



{{{5n+10d+75d=1140}}} Multiply



{{{5n+10*highlight((n+6))+75*highlight((n+6))=1140}}} Plug in {{{d=n+6}}}. Now the variable "d" is gone. 



Now we're left with a simple equation with one unknown variable.



{{{5n+10(n+6)+75(n+6)=1140}}}



{{{5n+10n+60+75n+450=1140}}} Distribute.



{{{90n+510=1140}}} Combine like terms on the left side.



{{{90n=1140-510}}} Subtract {{{510}}} from both sides.



{{{90n=630}}} Combine like terms on the right side.



{{{n=(630)/(90)}}} Divide both sides by {{{90}}} to isolate {{{n}}}.



{{{n=7}}} Reduce.



So this means that there are 7 nickels.



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{{{d=n+6}}} Go back to the first equation



{{{d=7+6}}} Plug in {{{n=7}}}



{{{d=13}}} Add



So there are 13 dimes



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{{{q=3d}}} Move onto the second equation



{{{q=3(13)}}} Plug in {{{d=13}}}



{{{q=39}}} Multiply



So there are 39 quarters.




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Answer:


So there are 7 nickels, 13 dimes and 39 quarters.