Question 1204507
 

let dime be {{{d}}} and nickel {{{n}}}

{{{1d}}}=${{{0.10}}}
{{{1n}}}=${{{0.05}}}

if  Rachelle has ${{{4.45}}} in dimes and nickels, we have

{{{0.10d+0.05n=4.45}}}.....eq.1

if the number of nickels is twenty more than the number of dime, we have

{{{n=d+20}}}...eq.2

go to

{{{0.10d+0.05n=4.45}}}.....eq.1, substitute {{{n}}}

{{{0.10d+0.05(d+20)=4.45}}}.....solve for {{{d}}}

{{{0.10d+0.05d+0.05*20=4.45}}}

{{{0.15d+1=4.45}}}

{{{0.15d=4.45-1}}}

{{{0.15d=3.45}}}

{{{d=3.45/0.15}}}

{{{d=345/15}}}

{{{d=23}}}


go to


{{{n=d+20}}}...eq.2, substitute {{{d}}}

{{{n=23+20}}}

{{{n=43}}}

Rachelle has {{{23}}} dimes and {{{43 }}}nickels.