Question

help please @ganeshie8 @Jamierox4ev3r

Answer 1

i know its b or d

Answer 2

substitute each point and check to see its equal to the right hand side of each equation.

Answer 3

i got b

Answer 4

If you are given no choices then you can solve as given below.

Answer 5

b is the correct solution

Answer 6

x+y=17
squaring both sides
\[x ^{2}+y ^{2}+2xy=289,169+2x y=289,2x y=289-169=120\]
xy=60
y=60/x
\[x+\frac{ 60 }{x }=17,x ^{2}+60=17x, or x ^{2}-17x+60=0\]
\[x ^{2}-12x-5x+60=0,x \left( x-12 \right)-5\left( x-12 \right)=0,\left( x-12 \right)\left( x-5 \right)=0\]
it gives x=12,5
when x=12,y=17-12=5
when x=5,y=17-5=12