Question

Question #83656

Answer 1

A cassette costs $ 7 and a CD costs $ 15.

Explanation:

Let's assume that the cost of a cassette is `c` and the cost of a CD is `d`

We can thus form 2 equations:

Equation 1: `5c + 2d = 65`

Equation 2: `3c + 4d = 81`

If we change equation 2 to make `d` the subject of the formula, we get:

`d = (81 - 3c)/4`

Substituting this equation into the first one, we get:

`5c + 2((81-3c)/4) = 65)`

Which is `5c + (81 - 3c)2 = 65`

Multiplying the whole equation by 2 gives

`10c + 81 - 3c = 130`

Solving for `c` will give us
`7c = 130 - 81`
So `c = 49/7 = 7`

Substituting `c` in equation 1:

`d = (81 - 3(7))/4 = (81 - 21)/4 = 15`

So the cost of a cassette is $ 7 and that of a CD is $ 15.

Answer 2

Cassettes are `$7` and CDs are `$15`.

Explanation:

Set up a system of equations, with x as cassettes and y as cds.
`5x+2y = 65` `rarr` equation 1
`3x+4y = 81` `rarr` equation 2

Now, use either substitution or elimination to solve. I show substitution below, but you get the same answer either way.
equation 1
`5x+2y=65` `rarr` subtract `5x` from both sides
`2y=65-5x` `rarr` plug into equation 2
equation 2
`3x+2(65-5x) = 81` `rarr` distribute
`3x+130-10x = 81` `rarr` get x terms on one side, anything else on the other
`-7x=-49` `rarr` divide by -7
`x=7` `rarr` plug back into equation 1
equation 1
`5(7)+2y=65` `rarr`distribute
`35+2y=65` `rarr`subtract 35 from both sides
`2y=30` `rarr`divide by 2
`y=15` `rarr` use to check answers

CHECK
`5(7)+2(15) = 65`
`35+30=65`
`65=65`

`3(7)+4(15) =81`
`21+60 = 81`
`81=81`