Question 155058

Let n = # of nickels and d = # of dimes


Since "Joe has a collection of nickels and dimes that is worth $5.65", this means that the first equation is {{{0.05n+0.10d=5.65}}}


{{{0.05n+0.10d=5.65}}} Start with the first equation



{{{5n+10d=565}}} Multiply <b>every</b> term by 100 to move the decimal point two places to the right (effectively eliminating the decimal)


Also, since "the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45", this tells us that {{{0.05(n+8)+0.10(2d)=10.45}}}


{{{0.05(n)+0.05(8)+0.10(2d)=10.45}}} Distribute



{{{0.05n+0.4+0.20d=10.45}}} Multiply



{{{5n+40+20d=1045}}} Multiply <b>every</b> term by 100 to move the decimal point two places to the right to make the decimal numbers whole.



{{{5n+cross(40-40)+20d=1045-40}}} Subtract 40 from both sides



{{{5n+20d=1005}}} Subtract. So this is our second equation


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So we have the system of equations:


{{{system(5n+10d=565,5n+20d=1005)}}}



{{{5n+10d=565}}} Start with the first equation



{{{5n=565-10d}}} Subtract {{{10d}}} from both sides


Notice how there's a "5n" in the second equation. So simply replace the "5n" in the second equation with {{{565-10d}}}


{{{(565-10d)+20d=1005}}} Plug in {{{5n=565-10d}}}



{{{10d+565=1005}}} Combine like terms.



{{{10d+cross(565-565)=1005-565}}} Subtract 565 from both sides.



{{{10d=440}}} Combine like terms



{{{cross(10/10)d=440/10}}} Divide both sides by 10



{{{d=44}}} Reduce



So Joe has 44 dimes.



{{{5n+10d=565}}} Go back to the first equation



{{{5n+10(44)=565}}} Plug in {{{d=44}}}



{{{5n+440=565}}} Multiply



{{{5n=125}}} Subtract 440 from both sides



{{{n=25}}} Divide both sides by 5 to isolate "n"


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Answer:


So Joe has 25 nickels and 44 dimes