1 answer

How do you solve this system of equations: #(x - 3) ^ { 2} + ( y + 2) ^

Question:

How do you solve this system of equations: #(x - 3) ^ { 2} + ( y + 2) ^ { 2} = 16 and x - 2y = 10#?

Answers

Solution : # x~~ 0.23 , y~~ -4.89 and x~~ 6.97 , y~~ -1.52 #

Explanation:

#(x-3)^2+(y+2)^2=16 and x-2y=10:.x=(2y+10)#

Putting #x=(2y+10)# in #(x-3)^2+(y+2)^2=16# we get

#(2y+10-3)^2+(y+2)^2=16 or (2y+7)^2 +(y+2)^2 =16#

or #4y^2+28y+49 +y^2+4y+4 =16# or

#5y^2+32y+37=0; a=5 , b=32,c=37# Quadratic formula is

#y= (-b+- sqrt(b^2-4ac))/(2a)= (-32 +- sqrt (32^2-4*5*37))/10#

or #y=-3.2+- sqrt 284/10= -3.2+- 1.685#

#:.y ~~-4.885 , y ~~ -1.515 ; # when

#y=-4.885 :. x=2*(-4.885)+10 ~~ 0.23 # and when

#y=-1.515 :. x=2*(-1.515)+10 ~~ 6.97 #

Solution :# x~~ 0.23 , y~~ (-4.89) and x~~ 6.97 , y~~(-1.52) #[Ans]

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