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

Apr 2, 2017

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:

$16 \left(28 - b\right) + 19 b = 478$

Distribute the 16:

$448 - 16 b + 19 b = 478$

Subtract 448 from both sides:

$- 16 b + 19 b = 30$

Combine like terms:

$3 b = 30$

Divide both sides by 3

$b = 10$

Substitute 10 for b into the first equation:

$s + 10 = 28$

$s = 18$

Check $s = 18 \mathmr{and} b = 10$ in both equations:

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

$18 + 10 = 28$
$16 \left(18\right) + 19 \left(10\right) = 478$

$28 = 28$
$288 + 190 = 478$

$28 = 28$
$478 = 478$

This checks.

Apr 2, 2017

See the entire solution process below:

#### Explanation:

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

$s + b = 28$

$s + b - \textcolor{red}{b} = 28 - \textcolor{red}{b}$

$s + 0 = 28 - b$

$s = 28 - b$

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

$16 s + 19 b = 478$ becomes:

$16 \left(28 - b\right) + 19 b = 478$

$\left(16 \times 28\right) - \left(16 \times b\right) + 19 b = 478$

$448 - 16 b + 19 b = 478$

$448 + \left(- 16 + 19\right) b = 478$

$448 + 3 b = 478$

$- \textcolor{red}{448} + 448 + 3 b = - \textcolor{red}{448} + 478$

$0 + 3 b = 30$

$3 b = 30$

$\frac{3 b}{\textcolor{red}{3}} = \frac{30}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} b}{\cancel{\textcolor{red}{3}}} = 10$

$b = 10$

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 $s = 18$