How do you solve the following system: #x - 3y = -1 , 7x+15y=32 #?

1 Answer
May 8, 2017

Answer:

See a solution process below:

Explanation:

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

#x - 3y = -1#

#x - 3y + color(red)(3y) = -1 + color(red)(3y)#

#x - 0 = -1 + 3y#

#x = -1 + 3y#

Step 2) Substitute #-1 + 3y# for #x# in the second equation and solve for #y#:

#7x + 15y = 32# becomes:

#7(-1 + 3y) + 15y = 32#

#(7 * -1) + (7 * 3y) + 15y = 32#

#-7 + 21y + 15y = 32#

#-7 + (21 + 15)y = 32#

#-7 + 36y = 32#

#color(red)(7) - 7 + 36y = color(red)(7) + 32#

#0 + 36y = 39#

#36y = 39#

#(36y)/color(red)(36) = 39/color(red)(36)#

#(color(red)(cancel(color(black)(36)))y)/cancel(color(red)(36)) = (13 xx 3)/color(red)(12 xx 3)#

#y = (13 xx color(red)(cancel(color(black)(3))))/color(red)(12 xx color(black)(cancel(color(red)(3))))#

#y = 13/12#

Step 3) Substitute #13/12# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = -1 + 3y# becomes:

#x = -1 + (3 * 13/12)#

#x = -1 + 39/12#

#x = (12/12 * -1) + 39/12#

#x = -12/12 + 39/12#

#x = 27/12#

#x = (3 xx 9)/(3 xx 4)#

#x = (color(red)(cancel(color(black)(3))) xx 9)/(color(red)(cancel(color(black)(3))) xx 4)#

#x = 9/4#

The solution is: #x = 9/4# and #y = 13/12# or #(9/4, 13/12)#