How do you solve the simultaneous equations #y=3x - 5# and #x^2 + y^2 = 25#?

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Answer 1 · 2018-03-28T23:42:48

Solution pairs of #(x, y)# are #(0, -5)# and #(3, 4)#

After plugging first equation to second one,

#x^2+(3x-5)^2=25#

#x^2+9x^2-30x+25=25#

#10x^2-30x+25=25#

#10x^2-30x=0#

#x^2-3x=0#

#x*(x-3)=0#

Hence #x_1=0# and #x_2=3#. So #y_1=-5# and #y_2=4#. Thus, solution pairs of #(x, y)# are #(0, -5)# and #(3, 4)#