Question 1129508
.
<pre>
Let D be the number of dimes, and let Q be the number of quarters.


Then you have this system of 2 equations in 2 unknowns


   D +    Q = 28             (1)    (there are 28 coins, in all)
10*D + 25*Q = 490   cents    (2)    (the total)


You can solve the system by different methods.
Let me call some of them: Substitution; Elimination; Cramer's rule (using determinants).


What you choose, depends on your level in Math and on your preferences.
I don't know what your level and preferences are.
So I will choose the most simple - Substitution.


From eq(1) express  D = 28-Q and substitute it into eq(2), replacing D there. You will get


10*(28-Q) + 25*Q = 490

280 - 10Q + 25Q = 490

15Q = 490 - 280 = 210

Q = {{{210/15}}} = 14.


<U>Answer</U>.  14 quarters and (the rest) 28-14 = 14 dimes.
</pre>

Solved.


---------------------


There are different methods for solving such problems: 
&nbsp;&nbsp;&nbsp;&nbsp;- using system of two equations (as in this post);
&nbsp;&nbsp;&nbsp;&nbsp;- using single equation in one single unknown;
&nbsp;&nbsp;&nbsp;&nbsp;- and simple logical methods without using equations.


You may learn this subject and different methods from the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Three-methods-for-solving-standard-typical-coin-problem.lesson>Three methods for solving standard (typical) coin word problems</A>

in this site.



For coin problems and their detailed solutions see the lessons in this site:

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/Coin-problems.lesson>Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/More-Coin-problems.lesson>More Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/Solving-coin-problem-without-equations.lesson>Solving coin problems without using equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Kevin-and-Randy-Muise-have-a-jar.lesson>Kevin and Randy Muise have a jar containing coins</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Typical-coin-problems-from-the-archive.lesson>Typical coin problems from the archive</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Three-methods-for-solving-standard-typical-coin-problem.lesson>Three methods for solving standard (typical) coin word problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/More-complicated-coin-problems.lesson>More complicated coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Solving-coin-problems-mentally-by-grouping-without-using-equations.lesson>Solving coin problems mentally by grouping without using equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Non-typical-coin-problems.lesson>Non-typical coin problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Santa-Claus-helps-solving-coin-problem.lesson>Santa Claus helps solving coin problem</A>


You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations. 


Read them attentively and become an expert in this field.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Coin problems</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.