Question

Can I get some help on these please

Answer 1

Ex 5

Simply multiply the equation by 6, in order to get rid of the denominators:

$\dfrac{x}{2}+\dfrac{y}{3} = 4 \iff 3x+2y=24$

Ex 6

We have to rearrange the equation in the form

$(x-x_0)^2+(y-y_0)^2=r^2$

So that $r$ is the radius, and $2r$ will be the diamater. We have to complete the squares: rewrite the equation as

$x^2-6x+y^2+8y=144$

You can see that $x^2-6x$ is the beginning of $x^2-6x+9 = (x-3)^2$

Similarly,  $y^2+8y$ is the beginning of $y^2+8y+16 = (y+4)^2$

So, if we add 16 and 9 to both sides, the equation becomes

$x^2-6x+9+y^2+8y+16=144+9+16 \iff (x-3)^2+(y+4)^2=169$

So, now we know that $r^2=169$, and thus the radius is 13, so the diameter is 26.