00:01
All right, so we have the following problem where james collects nickels and dines worth up to $5 .15.
00:07
And we're going to be using a linear equation to solve this.
00:11
And it's linear because the unknown value that we're looking for is raised to the power of one.
00:16
And we're also using the fact that addition and subtraction under each other and multiplication and division also under each other.
00:23
So james has a total of $5 .15.
00:27
Cents and we can draw an equation five cents times the number of nickels he has plus ten cents times the number of dimes that he has is worth up to five dollars and fifteen cents and we're given another piece of information then he basically has five fewer nickels than dimes so if he has five fewer nickels and dines then that means that n is equal to d minus five and we're trying to solve for d.
01:03
We're trying to figure out how many times does james have.
01:07
So in this equation, we could simply plug in this n value into the original equation and get and solve for d.
01:19
So we'll just plug in our n or replace it with d minus 5.
01:34
And now we're left with this.
01:36
And now we're going to distribute the 5 cents to d and to minus 5...