Given:
#color(red)(6x=29-8y , 3y = 1 -x#
Write the equations separately:
#6x=29-8y# Equation (1)
#3y = 1 -x# Equation (2)
Consider
#6x=29-8y# Equation (1)
Add #color(blue)(8y# to both sides of the equation.
#6x + color(blue)(8y)=29-8y+color(blue)(8y)#
#6x + color(blue)(8y)=29cancel(-8y)color(blue)cancel(+8y)#
#color(blue)(6x + 8y = 29)# Equation (3)
Consider
#3y = 1 -x# Equation (2)
Add #color(blue)(x# to both sides of the equation.
#3y + color(blue)(x) = 1 -x+color(blue)(x#
#3y + color(blue)(x) = 1 -cancel(x)+color(blue)(cancel(x)#
#color(blue)(x + 3y = 1)# Equation (4)
Consider:
#color(blue)(6x + 8y = 29)# Equation (3)
#color(blue)(x + 3y = 1)# Equation (4)
Multiply both sides of the Equation (4) by 6.
This resultant equation is Equation (5)
#color(blue)(6x + 8y = 29)# Equation (3)
#color(blue)(6x + 18y = 6)# Equation (5)
Subtract Equation (5) from Equation (3)
#cancel(6x) + 8y = 29# Equation (3)
#-cancel(6x) - 18y = -6# Equation (5)
You get
#-10y=23#
Divide both sides by 10 to get
#(-cancel(10)y)/cancel(10)=23/10#
#=-y = 23/10#
Multiply both sides by #(-1)# to get
#=(-y)(-1) = (23/10)(-1)#
#color(green)(y=-23/10#
Substitute this value of #y# in #color(blue)(x + 3y = 1)# Equation (4) to get
#x+3(-23/10) = 1#
#rArr x-69/10 = 1#
Add #69/10# to both the sides to get
#rArr x-69/10 + 69/10 = 1+69/10#
#rArr x-cancel(69/10) + cancel(69/10) = 1+69/10#
#rArr x = 1+69/10#
Take the common denominator of the numeric expression at the Right-Hand Side(RHS) to simplify:
#rArr x = (10+69)/10#
#color(green)( x = (79)/10#
Hence, values of #x and y# are:
#color(green)( x = (79)/10#
#color(green)(y=-23/10#
The system of linear equations is solved.