Question 151577
Let {{{x=nickels}}}
{{{y=dimes}}}
so, {{{x+y=42}}}, right? ----------------> eqn 1
Also, {{{0.05x+0.10y=3.40}}} ------------> eqn 2
in eqn we get, {{{y=42-x}}}, substitute in eqn 2 we have,
{{{0.05x+0.10(42-x)=3.40}}}
{{{0.05x+4.2-0.10x=3.40}}}
{{{4.20-3.40=0.10x-0.05x}}}
{{{0.80/0.05=cross(0.05)x/cross(0.05)}}} -----------> {{{x=16}}}, # of dimes
For # of nickels go back eqn 1:
{{{16+y=42}}}, {{{y=42-16=26}}}, # of nickels
In doubt? Go back eqn 2;
{{{0.05(16)+0.10(26)=3.40}}}
{{{0.80+2.60=3.40}}}
{{{3.40=3.40}}}
Thank you,
Jojo