00:01
This problem tells us isabel paid for her lunch with $1 .75, and it was 87 coins that made up the $1 .75 that she used to pay for lunch.
00:09
And if all the coins were nickels and pennies, how many of each type were there? so we're going to say that x is our number of nickels and y is our number of pennies.
00:17
And first, ignoring the cost of the lunch, we know that the total number of coins we had was 87.
00:24
So however many nickels plus however many pennies we have should be equal to 87.
00:28
The next equation we're going to write is going to be equal to the total that we paid for the lunch, $1 .75.
00:34
And since now we're comparing this to price, we are going to include the price per number of nickels and per pennies.
00:40
So a nickel is worth five cents.
00:42
So that's 0 .05 times x for every number of nickels we have.
00:47
And then a penny is worth one cent.
00:49
So that's 0 .01 times y for every penny we have.
00:53
And now we have a system of equations that we can solve with elimination or substitution.
00:58
I'm going to solve with substitution by isolating my x value in my first equation and saying that x is equal to 87 minus y.
01:07
And if x equals 87 minus y in the first equation, that means x in the other equation has to equal the same thing.
01:14
So we'll plug in 0 .05 times the x substitution expression of 87 minus y, still plus 0 .01y, equal to $1 .75.
01:26
And now we have an equation that just has one variable and we can begin to solve for that variable...