Makayla:
This is a classic example of a "Systems of Equations" Problem.
First, let's dissect what we are presented with;
We have a total of 20 coins; n nickels and d dimes. So, the nickels and dimes that I own are 20 altogether or n + d =20
This is my first equation.
My second piece of information tells me that the value of my coins equals $1.40. We know that a nickel is worth $0.05 and a dime is worth $0.10 so we can express our unknown nickels and dimes by the following equation to model the value of my coins: 0.05n + 0.10d = 1.40
So our systems of equations are as follows:
n + d = 20
0.05n + 0.10d = 1.40
The way we are going to solve this systems of equations is by way of the substitution method. (There are a number of other methods, but for the sake of simplicity I will only be using the substitution method. Feel free to message if you would like to see the other methods.)
The substitution method effectively solves one equation for a variable and SUBSTITUTES that expression into the other function, giving a numerical value for a variable. We can then use this value to solve for the other variable respectively.
Take the following equation from our problem; n + d = 20 and solve for a variable; n or d; it does not matter which.
EXAMPLE:
n + d = 20---Solve for d
-n -n---Subtract n from both sides
d = 20-n or (-n + 20)
So, we are now saying that d is equivalent to 20-n. We will now substitute this expression in the other equation as follows
EXAMPLE
0.05n + 0.10d = 1.40
0.05n + 0.10(20-n) = 1.40---substitute (20-n) in place of d
0.05n + 2-.1n = 1.40---Apply the distributive property
0.05n + 2 -.1n = 1.40---Combine like terms
-0.05n + 2 = 1.40---Solve for n by subtracting 2 from both sides
-0.05n + 2 = 1.40
- 2 - 2
-0.05n = -0.6---Divide by -0.05
-0.05 -0.05
n = 12
Therefore, I have 12 nickels.
Now, one could conclude that if I have 12 nickels and I have 20 nickels and dimes altogether, I should have 8 dimes, but lets check this algebraically.
Take the value equation and solve for d
0.05n + 0.10d = 1.40
we know that we have 12 nickels so we will substitute that in for n
0.05(12) + 0.10d = 1.40
Multiply
0.60 + 0.10d = 1.40
solve for d
0.60 + 0.10d = 1.40
-0.60 -0.60
0.10d = 0.80
0.10 0.10
d = 8
So, I have 12 nickels and 8 dimes altogether I have 20 coins (12) + (8) = 20
Totaling 1.40
0.05(12) + 0.10(8) = 1.40--->0.60 + 0.80 = 1.40
