Question #17602

3 Answers
Jan 4, 2018

Use the elimination or substitution methods

Explanation:

Here I would use the elimination method and get rid of the y variable. The first equation can be multiplied by 3 to make the equation #-9x-6y=-15#

Then add the two equations together to get rid of the y variable and solve for the #x #
# -9x-6y=-15#
#ul(+ 7x+6y=+1 larr" Add")#
#-2x+color(white)(".")0=-14#

#=>color(red)(x=7)#

Plug in for #y#

#7( color(red)(7) )+6y=1#

#49+6y=1#

#6y=-48#

#y=-8#

Double check by plugging in both solutions
The answer is (7, -8)

Jan 4, 2018

#x=7#
#y=-8#

Explanation:

#{(-3x-2y=-5),(7x+6y=1):}#

Let's procede finding the x of the second equation:
#{ (-3x-2y=-5),(7x=-6y+1):}#

#{(-3x-2y=-5),(x=-\frac{6}{7}y+\frac{1}{7}):}#

Now we can substitute the x that we found in the first equation:
#{(-3(-\frac{6}{7}y+\frac{1}{7})-2y=-5),(x=-\frac{6}{7}y+\frac{1}{7}):}#

#{(\frac{18}{7}y-\frac{3}{7}-2y=-5),(x=-\frac{6}{7}y+\frac{1}{7}):}#

#{(\frac{18}{7}y-2y=-5+\frac{3}{7}),(x=-\frac{6}{7}y+\frac{1}{7}):}#

#{(\frac{18-14}{\cancel{7}}y=\frac{3-35}{\cancel{7}}), (x=-\frac{6}{7}y+\frac{1}{7}):}#

#{(4y=-32),(x=-\frac{6}{7}y+\frac{1}{7}):}#

#{(y=-8),(x=-\frac{6}{7}y+\frac{1}{7}):}#

Now let's substitute #y=-8# in the second equation:
#{(y=-8),(x=-\frac{6}{7}(-8)+\frac{1}{7}):}#

#{(y=-8),(x=\frac{48}{7}+\frac{1}{7}):}#

#{(y=-8),(x=\frac{49}{7}=7):}#

Final result:
#x=7#
#y=-8#

Jan 4, 2018

Have a look at https://socratic.org/help/symbols Note the hash symbols at the beginning and end of the entered text.

#x=7 # and #y=-8#

Explanation:

Given:
#color(green)(-3x-2y=-5" "....................Equation(1))#
#color(green)(+7x+6y=+1" "....................Equation(2))#

#[7xxEqn(1)]+[3xxEqn(2)]#
#color(white)("dddddddddd") darr#

#color(green)(-21x-14y=-35" "....Equation(1_a))#
#color(green)(ul(+21x+18y=+color(white)(5)3)........Equation(2_a))#
#color(green)(color(white)("dd")0color(white)("d.")+color(white)("d")4y=-32)#

#color(green)( y=-8 )#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I choose #Eqn(2)#
Substitute for #color(red)(y)# in #Eqn(2)# giving:

#color(green)(+7x+6color(red)(y)=+1color(white)("ddd")->color(white)("ddd")7x+[6color(red)(xx(-8))]=1)#

#color(white)("ddddddddddddddd.d")color(green)(->color(white)("ddd")7x-color(white)("ddd")48color(white)("dddd")=1)#

#color(white)("ddddddddddddddd.d")color(green)(->color(white)("ddddddddddddd.d")7x=49)#

#color(green)(color(white)("ddddddddddddddddd")->color(white)("ddddddddddddddd") x=49/7=7 ) #