How do you solve the system #6x^2-5x+8y^2+y=23# and #y=x-1#?

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How do you solve the system #6x^2-5x+8y^2+y=23# and #y=x-1#?

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Answer 1 · 2016-07-06T06:55:34

The Soln. Set #={(2,1), (-4/7,-11/7)}.#

Explanation:

We substitute #y=x-1.......(I)# in the qudr. eqn., to get,

#6x^2-5x+8(x-1)^2+x-1=23.#
# rArr 6x^2-5x+8x^2-16x+8+x-1-23=0.#
#rArr 14x^2-20x-16=0, # or, #7x^2-10x-8=0.#
#rArr 7x^2-14x+4x-8=0.... ..[7xx8=14xx4, 14-4=10]#
#rArr 7x(x-2)+4(x-2)=0.#
#rArr (x-2)(7x+4)=0.#
#rArr x=2, x=-4/7#
#(1)#, then#rArr y=1,y=-11/7#

These satisfy the given eqns. Hence,

The Soln. Set #={(2,1), (-4/7,-11/7)}.#

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How do you solve the system #6x^2-5x+8y^2+y=23# and #y=x-1#?

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