How do you solve #a+3b=7# and #2a=b-7#?

1 Answer
May 18, 2018

Answer:

See a solution process below:

Explanation:

Step 1) Solve the first equation for #a#:

#a + 3b = 7#

#a + 3b - color(red)(3b) = 7 - color(red)(3b)#

#a + 0 = 7 - 3b#

#a = 7 - 3b#

Step 2) Substitute #(7 - 3b)# for #a# in the second equation and solve for #b#:

#2a = b - 7# becomes:

#2(7 - 3b) = b - 7#

#(2 * 7) - (2 * 3b) = b - 7#

#14 - 6b = b - 7#

#14 + color(blue)(7) - 6b + color(red)(6b) = b + color(red)(6b) - 7 + color(blue)(7)#

#21 - 0 = 1b + color(red)(6b) - 0#

#21 = (1 + color(red)(6))b#

#21 = 7b#

#21/color(red)(7) = (7b)/color(red)(7)#

#3 = (color(red)(cancel(color(black)(7)))b)/cancel(color(red)(7))#

#3 = b#

#b = 3#

Step 3) Substitute #3# for #b# in the solution to the first equation at the end of Step 1 and calculate #a#:

#a = 7 - 3b# becomes:

#a = 7 - (3 * 3)#

#a = 7 - 9#

#a = -2#

The Solution Is:

#a = -2# and #b = 3#