# Thread: Systems Of Linear Equations in Three Variables

1. ## Systems Of Linear Equations in Three Variables

I really hate word problems, and all of a sudden my math teacher throws in three of them in which may seem easy for you but i can't seem to get the systems of equations to find the solution:

1.Barbara has nickles, dimes, and quaters worth 2.35 in her purse. the number of dimes is three less than the sum of the number of nickles and quarters. how many of each type of coin does she have if there are 19 coins in all?

2.. David paid 9.50 for some .15, .25, and .45 stamps. he bought 38 in all. the number of .25 stamps was 8 more than twice the number of .45 stamps. how mny of each type of stamps did he buy?

3. when 3 large diamonds are weighed in pairs, the masses of the pairs are found to be 6 crats, 10 carats, and 12 carats. find the mass of each diamond.

i thank those in advance for helping me...i feel as if im a bother, and i apologize in advance

2. Originally Posted by >_<SHY_GUY>_<
I really hate word problems, and all of a sudden my math teacher throws in three of them in which may seem easy for you but i can't seem to get the systems of equations to find the solution:

1.Barbara has nickles, dimes, and quaters worth 2.35 in her purse. the number of dimes is three less than the sum of the number of nickles and quarters. how many of each type of coin does she have if there are 19 coins in all?

...
Let n denote the number of nickels, d the number of dimes and q the number of quarters. Then you have:

$\displaystyle \left|\begin{array}{lcr}n+d+q&=&19 \\ 5n+10d+25q&=&235 \\ d&=& n+q-3\end{array}\right.$ . I don't know which method you prefer to solve a system of linear equations. As solution I've got: (n, d, q) = (6, 8, 5)

3. Originally Posted by >_<SHY_GUY>_<
...
3. when 3 large diamonds are weighed in pairs, the masses of the pairs are found to be 6 crats, 10 carats, and 12 carats. find the mass of each diamond.
You have three diamonds A, B and C which have the weights a, b and c. According to your problem you get:

$\displaystyle \left|\begin{array}{lcr}a+b&=&6\\a+c&=&10\\b+c&=&1 2 \end{array}\right.$ ........ I don't know which method you use to solve such a system of linear equation. I've got: (a, b, c) = (2, 4, 8)

4. Originally Posted by earboth
Let n denote the number of nickels, d the number of dimes and q the number of quarters. Then you have:

$\displaystyle \left|\begin{array}{lcr}n+d+q&=&19 \\ 5n+10d+25q&=&235 \\ d&=& n+q-3\end{array}\right.$ . I don't know which method you prefer to solve a system of linear equations. As solution I've got: (n, d, q) = (6, 8, 5)

How were you able to get 5n+10d+25q=235...
i still am confused , the third one i got, but im stuck on one and two