How do you solve the system of equations #x+3y=97# and #43x-19y=105#?

1 Answer
Jun 16, 2017

See a solution process below:

Explanation:

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

#x + 3y = 97#

#x + 3y - color(red)(3y) = 97 - color(red)(3y)#

#x + 0 = 97 - 3y#

#x = 97 - 3y#

Step 2) Substitute #(97 - 3y)# for #x# in the second equation and solve for #y#:

#43x - 19y = 105# becomes:

#43(97 - 3y) - 19y = 105#

#(43 * 97) - (43 * 3y) - 19y = 105#

#4171 - 129y - 19y = 105#

#4171 - 148y = 105#

#-color(red)(4171) + 4171 - 148y = -color(red)(4171) + 105#

#0 - 148y = -4066#

#-148y = -4066#

#(-148y)/color(red)(-148) = (-4066)/color(red)(-148)#

#(color(red)(cancel(color(black)(-148)))y)/cancel(color(red)(-148)) = 2033/74#

#y = 2033/74#

Step 3) Substitute #2033/74# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = 97 - 3y# becomes:

#x = 97 - (3 * 2033/74)#

#x = 97 - 6099/74#

#x = (74/74 xx 97) - 6099/74#

#x = 7178/74 - 6099/74#

#x = 1079/74#

The solution is: #x = 1079/74# and #y = 2033/74# or #(1079/74, 2033/74)#