How do you solve the system of equations #d+n =70# and #.1d+.05n = 5.5#?

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Answer 1 · 2018-01-24T13:06:23

See a solution process below:

Step 1) Solve the second equation for #d#:

#0.1d + 0.05n = 5.5#

#0.1d + 0.05n - color(red)(0.05n) = 5.5 - color(red)(0.05n)#

#0.1d + 0 = 5.5 - 0.05n#

#0.1d = 5.5 - 0.05n#

#color(red)(10) xx 0.1d = color(red)(10)(5.5 - 0.05n)#

#1d = (color(red)(10) xx 5.5) - (color(red)(10) xx 0.05n)#

#d = 55 - 0.5n#

Step 2) Substitute #(55 - 0.5n)# for #d# in the first equation and solve for #n#:

#d + n = 70# becomes:

#(55 - 0.5n) + n = 70#

#55 - 0.5n + n = 70#

#55 - 0.5n + 1n = 70#

#55 + (-0.5 + 1)n = 70#

#55 + 0.5n = 70#

#55 - color(red)(55) + 0.5n = 70 - color(red)(55)#

#0 + 0.5n = 15#

#0.5n = 15#

#color(red)(2) xx 0.5n = color(red)(2) xx 15#

#1n = 30#

#n = 30#

Step 2)

Substitute #30# for #n# in the solution to the second equation at the end of Step 1 and calculate #d#:

#d = 55 - 0.5n# becomes:

#d = 55 - (0.5 xx 30)#

#d = 55 - 15#

#d = 40#

The Solution Is:

#d = 40# and #n = 30#