Question #f47ec

4 Answers

Answer:

The key is to multiply one or both equations by values that will cause one variable to become zero.

#x=-5 and y=-6#

Explanation:

In this example multiplying one of the equations by #-1# will result in one positive #x# term and one negative #x# term.
When added they will equal #0x #

.Step #1.# multiply the first equation by #-1#

# -1( -x -y =11) = + x + y = -11 #

Step #2# add the two equations to get a #0# value for the #x# term.

# +x + y = -11 #
#ul(-x +3y = -13)" "# which gives
# 0x + 4y = -24 #

Step #3.# solve for the other variable. #(y)#

# (4y) /4 = (-24)/4# so

#y = -6 #

step #4.# substitute the value for one variable into one of the equations and solve for the other variable.

# -x - (-6) = 11 " "# add #-6# to both sides

# -x + 6 - 6 = 11 - 6 # which gives

# -x = 5# divide both sides by #-1#

# (-x)/-1 = 5/-1# so

# x = -5 #

Aug 20, 2017

Answer:

# x = - 5, y = - 6#

Explanation:

#-x - y= 11. #

This means:

#-x = 11+y#

Now we substitute #- x# in the second equation:

#- x + 3y = -13#

We have:

#11+ y+ 3y= -13#

#11+ 4y = -13#

#4y = -13 -11#

#y = -24/4 = -6#

Now we substitute #y# in either one of the two equations:

#- x -(-6) = 11#

#- x +6 = 11#

#- x = 11 - 6#

# x = - 5#

We can now substitute #x# and #y# in either equation to be sure that the answer is right.

Aug 20, 2017

Answer:

#(-5,-6)#

Explanation:

The goal here is to solve for one variable by first eliminating one variable, hence, the elimination method.

Given:

#-x-y=11#

#-x+3y=-13#

We can eliminate the #x# variable since they have the same coefficient by subtracting both equations. Thus,

#" "cancel(-x)-y=11#
#-#
#" "cancel(-x)+3y=-13#

This yields:

#-4y=24#

Now we solve for the variable #y#

#cancel(-4)/cancelcolor(red)(-4)y=24/color(red)(-4)#

#y=-6#

The next step is to find the value for the variable #x# by substituting the value for #y# into one of the original equations. I will use the first equation.

We substitute #-6# for #y# in the first equation and solve for #x#

#-x-(color(red)(-6))=11#

#-x+6=11#

#-x+6color(red)(-6)=11color(red)(-6)#

#-x=5#

#x=-5#

We now have the solutions #x=-5# and #y=-6# or #(-5,-6)# but we must check if the solution checks out but plugging both values into both equations.

Equation 1:

#-(-5)-(-6)=11#

#5+6=11#

#11=11# TRUE

Equation 2:

#-(-5)+3(-6)=-13#

#5-18=13#

#13=13# TRUE

The values check out. The solution #(-5,-6)#

You can learn more about this method here: http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L2_T2_text_final.html

Aug 20, 2017

Answer:

#x =-5 and y=-6#

Explanation:

The most important concept with the ELIMINATION method concerns ADDITIVE INVERSES.

Values which have the same number but opposite signs are called additive inverses. Their sum is #0#

#(+5)+(-5) = 0," " (-12)+(+12) =0#

To eliminate one of the variables, create additive inverses.
Other contributors have eliminated #x#, but let's look
at the #y#-terms instead

Notice that the #y#-terms already have opposite signs, but have different numbers. Create additive inverses.

#color(white)(xxxx)-xcolor(blue)(-y) =+11 " "A#
#color(white)(xxxx) -xcolor(blue)(+3y)=-13" "B#

#A xx3: -3xcolor(blue)(-3y)=+33" "C" "# now there are
#color(white)(xxx.xx) -xcolor(blue)(+3y)=-13" "B" "# additive inverses

Add the equations together:

#C+B:" "-4x " "= 20" "larr# only #x# values,
#color(white)(xxxxx.xxxx) x=-5#

That that you know the value for #x#, substitute to find #y#

#" "-x -y = 11#

#" "-(-5)-y = 11#

#5-11 = y#

#-6=y#