Ex 5
Simply multiply the equation by 6, in order to get rid of the denominators:
[tex] \dfrac{x}{2}+\dfrac{y}{3} = 4 \iff 3x+2y=24 [/tex]
Ex 6
We have to rearrange the equation in the form
[tex] (x-x_0)^2+(y-y_0)^2=r^2 [/tex]
So that [tex] r [/tex] is the radius, and [tex] 2r [/tex] will be the diamater. We have to complete the squares: rewrite the equation as
[tex] x^2-6x+y^2+8y=144 [/tex]
You can see that [tex] x^2-6x [/tex] is the beginning of [tex] x^2-6x+9 = (x-3)^2 [/tex]
Similarly, [tex] y^2+8y [/tex] is the beginning of [tex] y^2+8y+16 = (y+4)^2 [/tex]
So, if we add 16 and 9 to both sides, the equation becomes
[tex] x^2-6x+9+y^2+8y+16=144+9+16 \iff (x-3)^2+(y+4)^2=169 [/tex]
So, now we know that [tex] r^2=169 [/tex], and thus the radius is 13, so the diameter is 26.
