How do solve the following linear system?: # x= y - 11 , y = 4x + 4 #?

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Answer 1 · 2016-06-29T19:29:02

Soln. is #(x,y)=(7/3,40/3).#

Submit the value of #y,# taken from #2^(nd)# eqn., in the #1^(st)# eqn. to get,

#x=4x+4-11 rArr x-4x=4-11 rArr -3x=-7 rArr x=-7/-3=7/3.#

Hence, by #2^(nd)# eqn., #y=4(7/3)+4=28/3+4=40/3.#

Soln. is #(x,y)=(7/3,40/3).#

Answer 2 · 2016-06-29T19:30:01

#x = 7/3 = 2 1/3 and y = 40/3 = 13 1/3#

There are several methods you can use, but there is a single #y# term in each equation, so equating them is probably the easiest method.

Change the first equation to: # y = x+11#

Now, the y in each equation is the same, so;

if: #" " y = y #
then:#" "4x + 4 = x + 11#

#" "4x - x = 11-4#
#" " 3x = 7#

#x = 7/3 = 2 1/3#

To find the value of #y#, substitute #x = 7/3# into either of the equations above ..... or both, to check that they give the same answer.

#y = 4xx 7/3 +4 = 40/3 " and " y = 7/3 +11 = 40/3#