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

2 Answers
Apr 2, 2017

Answer:

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.

Apr 2, 2017

Answer:

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#