How do you solve #1.3x+y=3# and #7x+2y=1# using substitution?

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Answer 1 · 2016-03-31T21:24:23

#y= 197/44#
#x=-50/44#

Given:

#1.3x+y=3#......................(1)
#7x+2y=1#.......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider equation (1)

Subtract 1.3x from both sides giving

#y=3-1.3x" ".....................(1_a)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute #(1_a)# in (2) giving

#7x+2(3-1.3x)=1#

#7x+6-2.6x=1#

#4.4x+6=1#

Subtract 6 from both sides

#4.4x=-5#

Multiply both sides by 10

#44x=-50#

#color(blue)(x=-50/44)#..................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~

Substitute (3) into (2)

#7(-50/44)+2y=1#

#y=1/2+7/2( 50/44)#

#color(blue)(y=4 21/44 = 197/44)#