Question

How do you solve `s+b=28` and `16s+19b=478`?

Answer 1

There are several ways to solve this type of problem. I think the substitution method to be a good choice for this particular problem.

Explanation:

Substitution method:

Rewrite the first equation as:

`s = 28-b`

Substitute `28-b` for s into the second equation:

`16(28-b) + 19b = 478`

Distribute the 16:

`448-16b + 19b = 478`

Subtract 448 from both sides:

`-16b + 19b = 30`

Combine like terms:

`3b = 30`

Divide both sides by 3

`b = 10`

Substitute 10 for b into the first equation:

`s + 10 = 28`

`s = 18`

Check `s = 18 and b = 10` in both equations:

`s+b=28`
`16s+19b=478`

`18+10=28`
`16(18)+19(10)=478`

`28=28`
`288+190=478`

`28 = 28`
`478 = 478`

This checks.

Answer 2

See the entire solution process below:

Explanation:

Step 1) Solve the first equation for `s`:

`s + b = 28`

`s + b - color(red)(b) = 28 - color(red)(b)`

`s + 0 = 28 - b`

`s = 28 - b`

Step 2) Substitute `28 - b` for `s` in the second equation and solve for `b`:

`16s + 19b = 478` becomes:

`16(28 - b) + 19b = 478`

`(16 xx 28) - (16 xx b) + 19b = 478`

`448 - 16b + 19b = 478`

`448 + (-16 + 19)b = 478`

`448 + 3b = 478`

`-color(red)(448) + 448 + 3b = -color(red)(448) + 478`

`0 + 3b = 30`

`3b = 30`

`(3b)/color(red)(3) = 30/color(red)(3)`

`(color(red)(cancel(color(black)(3)))b)/cancel(color(red)(3)) = 10`

`b = 10`

Step 3) Substitute `10` for `b` in the solution to the first equation at the end of Step 1 and calculate `s`:

`s = 28 - b` becomes:

`s = 28 - 10`

`s = 18`

The solution is: `b = 10` and `s = 18`