How do solve the following linear system?: # 6x=29-8y , 3y = 1 -x #?

2 Answers
Mar 9, 2018

Answer:

The solution is a point at #x=7.9# and #y=-2.3#

Explanation:

Substitution implies that we solve for either #x# or #y# in one equation and substitute that into the other. I see a lonely #x# in the second equation that should be easy to solve for, so I'll start there:

#3y=1-x#

#x = 1-3y#

Then we substitute #1-3y# for #x# in the first equation:

#6(1-3y)=29-8y#

#8y-18y = 29 - 6#

#-10y = 23#

# y = -2.3#

Then we can use our original solution for #x# to get

#x = 1- 3y = 1-3*(-2.3) = 7.9#

check:

eq1:

#6x = 29 - 8y#

#6*(7.9) = 29-8*(-2.3)#

#47.4 = 47.4 \implies true#

eq2:

#3y = 1-x#

#3*(-2.3) = 1-7.9#

#-6.9 = -6.9 \implies true#

Mar 10, 2018

Answer:

#color(green)( x = (79)/10#

#color(green)(y=-23/10#

Explanation:

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.