Question

Lucy has 34 coins consisting of nickels and quarters amounting to $2.90. How many coins of each kind does she have?

Answer 1

Let Nickels be x and let quarters be y.

$\sf{\underline{Since:}}$

Lucy has 34 coins total,

$\sf{\underline{We\:can\:make\:this\:equation:}}$

$\boxed{\sf{x + y = 34}}$

$\sf{\underline{Note:}}$ The second part is about the value of the coins.

$\sf{\underline{Since:}}$

Nickels are worth 5 cents:

$\sf{\underline{That\:is:}}$ $\boxed{\sf{\frac{5}{100} =0.05} }$

Quarters are worth 25 cents:

$\sf{\underline{That\:is:}}$ $\boxed{\sf{\frac{25}{100} =0.25}}$

$\sf{\underline{We\:can\:make\:this\:equation:}}$

$\boxed{\sf{0.05x + 0.25y = 2.90}}$

$\sf{\underline{Now:}}$

We have to multiply this whole equation by 100.

$\sf{\underline{Now\:we\:get:}}$

$\boxed{\sf{5x + 25y= 290}}$

$\sf{\underline{Now:}}$

Let us rearrange the first equation, in order to solve for one variable:

$\implies$ $\sf{x + y = 34}$

$\implies$ $\sf{x= 34 - y}$

Substitute that into the second equation and solve for y.

$\implies$ $\sf{5x + 25y = 290}$

$\implies$ $\sf{5(34 - y) + 25y = 290}$

$\implies$ $\sf{170 - 5y + 25y= 290}$

$\implies$ $\sf{170 + 20y = 290}$

$\implies$ $\sf{20y= 120}$

$\implies$ $\sf{y= \frac{120}{20} }$

$\implies$ $\boxed{\sf{y = 6}}$

$\sf{\underline{So:}}$ Lucy has 6 quarters.

$\sf{\underline{Now:}}$

Substitute y into the first equation to solve for x.

$\implies$ $\sf{x = 34 - y}$

$\implies$ $\sf{x = 34 - 6}$

$\implies$ $\boxed{\sf{x = 28}}$

$\sf{\underline{So:}}$ Lucy has 28 nickels.

$\sf{\underline{Final\:answer:}}$

Lucy has 6 quarters and 28 nickels.