How do you solve? #x^2 + y^2 = 25# #4x - 2y = 7#

1 Answer
Apr 22, 2018

Solution: #x~~ 3.524, y ~~ 3.548# and #x~~ -0.724, y ~~ -4.948#

Explanation:

#x^2+y^2=25 (1); 4 x-2 y =7 :. 2 y= 4 x-7# or

#y=2 x - 7/2 (2)# ,putting #y=2 x - 7/2 # in equation(1)

we get, #x^2+ (2 x-7/2)^2=25# or

#x^2+ 4 x^2- cancel(2)*2 x*7/cancel(2) +49/4 =25# or

#5 x^2 -14 x +49/4 =25 # or

#20 x^2 -56 x +49=100 # or

#20 x^2 -56 x -51=0 # Comparing with standard quadratic

equation #ax^2+bx+c=0 ; a=20 ,b=-56 ,c=-51#

Discriminant # D= b^2-4 a c :.D=3136 +4080 = 7216#

Discriminant is positive, we get two real solutions,

Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (56+-sqrt 7216)/40 =7/5+- (4 sqrt 451)/40 # or

#x= 7/5+- sqrt 451/10 or x ~~ 3.524 , x~~ -0.724#

When #x ~~ 3.524 , y= 2 x-3.5=2*3.524-3.5~~3.548 #

When #x ~~ -0.724 , y= 2 x-3.5=2*(-0.724)-3.5#

#y~~ -4.948 :.# Solution: #x~~ 3.524, y ~~ 3.548# and

#x~~ -0.724, y ~~ -4.948# [Ans]