The sum of three number is 4. If the first is doubled and the third is tripled, then the sum is two less than the second. Four more than the first added to the third is two more than the second. Find the numbers?

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Answer 1 · 2017-05-19T22:15:14

1st #= 2#, 2nd #= 3#, 3rd #= -1#

Create the three equations:

Let 1st #= x#, 2nd #= y# and the 3rd = #z#.

EQ. 1: #x + y + z = 4#

EQ. 2: #2x + 3z + 2 = y" "=> 2x - y + 3z = -2#

EQ. 3: #x + 4 + z -2 = y " "=> x - y + z = -2#

Eliminate the variable #y#:

EQ1. + EQ. 2: #3x + 4z = 2#

EQ. 1 + EQ. 3: #2x + 2z = 2#

Solve for #x# by eliminating the variable #z# by multiplying EQ. 1 + EQ. 3 by #-2# and adding to EQ. 1 + EQ. 2:

(-2)(EQ. 1 + EQ. 3): #-4x - 4z = -4#

#" "3x + 4z = 2#
#ul(-4x - 4z = -4)#
# -x " "= -2 " " => x = 2 #

Solve for #z# by putting #x# into EQ. 2 & EQ. 3:

EQ. 2 with #x: " "4 - y + 3z = -2 " " => -y + 3z = -6#

EQ. 3 with #x: " "2 - y + z = -2 " " => -y + z = -4#

Multiply EQ. 3 with #x# by #-1# and add to EQ. 2 with #x#:

#(-1)( -y + z = -4) => y -z = 4#

#" " -y + 3z = -6#
#" "ul(+y -z = " "4)#
#2z = -2 " " => z = -1#

Solve for #y# , by putting both #x " and " z# into one of the equations:

EQ. 1: #" "2 + y - 1 = 4#

#y = 3#

Solution: 1st #= 2#, 2nd #= 3#, 3rd #= -1#

CHECK by putting all three variables back into the equations:

EQ. 1: #" "2 + 3 -1 = 4" "# TRUE

EQ. 2: #" "2(2) + 3 (-1) + 2 = 3" "# TRUE

EQ. 3: #" "2 + 4 -1 -2 = 3" "# TRUE