Define a variable for each type of coin. Using the given information, write equations using these variables.
Penny: 2.5g Nickel: 5g Dime: 2.3g
Practice makes perfect
Let's start by defining a variable for each type of coin. Let p, n, and d be the mass of a penny, the mass of a nickel, and the mass of a dime, respectively. Now, let's take a look at each of the given sentences.
The combined mass of a penny, a nickel,
and a dime is9.8g.
The sentence above implies that p+ n+ d=9.8. Next, let's check the second sentence.
Ten nickels and three pennies have the
same mass as25 dimes.
From the information above, we can set the second equation to be 10 n + 3 p = 25 d. Finally, let's analyze the last sentence.
Fifty dimes have the same mass as
18nickels and10pennies.
This sentence can be written algebraically as 50 d=18 n+10 p. That way, we have formed a system of three equations.
p+ n+ d=9.8 & (I) 10 n + 3 p = 25 d & (II) 50 d=18 n+10 p & (III)
Using the Commutative Property of Addition, let's rearrange the terms of the equations of the above system.
p+ n+ d=9.8 & (I) 3 p+10 n - 25 d=0 & (II) 10 p+18 n-50 d=0 & (III)
Let's solve the system using the Elimination Method. We multiply Equation (II) by -2 and add the resulting equation to Equation (III).
rc
-6 p - 20 n + 50 d =0&
+ 10 p+18 n-50 d =0&
4 p-2 n = 0& (IV)
Now, we multiply Equation (I) by 25 and add the resulting equation to Equation (II).
rc
25 p+25 n+25 d = 245&
+ 3 p+10 n - 25 d = 0 &
28 p+35 n = 245& (V)
To eliminate the variable n we multiply Equation (IV) by -7 and add the resulting equation to Equation (V).
r
- 28 p+14 n=0
+ 28 p+35 n =245
49 n = 245
From the final equation, we get that the mass of a nickel is 5g. To find the mass of a penny, we substitute n= 5 into Equation (IV).
