# What are possible value(s) of x and y if y^2=x^2-64 and 3y=x+8??

Mar 25, 2018

#(x, y) = (-8, 0), (10, 6)

#### Explanation:

$3 y = x + 8 \implies x = 3 y - 8$

${y}^{2} = {x}^{2} - 64$
${y}^{2} = {\left(3 y - 8\right)}^{2} - 64$
${y}^{2} = 9 {y}^{2} - 48 y + 64 - 64$
$8 {y}^{2} - 48 y = 0$
$8 y \left(y - 6\right) = 0$
$y = 0 , 6$

$x = 3 y - 8 \mathmr{and} y = 0$:
$x = 0 - 8$
$= - 8$

$x = 3 y - 8 \mathmr{and} y = 6$:
$x = 3 \times 6 - 8$
$x = 10$

$\left(x , y\right) = \left(- 8 , 0\right) , \left(10 , 6\right)$

Mar 25, 2018

$\left(- 8 , 0\right) , \left(10 , 6\right)$

#### Explanation:

${y}^{2} = {x}^{2} - 48 \to \left(1\right)$

$3 y = x + 8 \to \left(2\right)$

$\text{from equation "(2)" we can express x in terms of y}$

$\Rightarrow x = 3 y - 8 \to \left(3\right)$

$\text{substitute "x=3y-8" in equation } \left(1\right)$

$\Rightarrow {y}^{2} = {\left(3 y - 8\right)}^{2} - 64 \leftarrow \textcolor{b l u e}{\text{expand }} {\left(3 y - 8\right)}^{2}$

$\Rightarrow {y}^{2} = 9 {y}^{2} - 48 y \cancel{+ 64} \cancel{- 64}$

$\Rightarrow 8 {y}^{2} - 48 y = 0 \leftarrow \textcolor{b l u e}{\text{factorise}}$

$8 y \left(y - 6\right) = 0$

$\text{equate each factor to zero and solve for y}$

$8 y = 0 \Rightarrow y = 0$

$y - 6 = 0 \Rightarrow y = 6$

$\text{substitute these values into equation } \left(3\right)$

$y = 0 \Rightarrow x = - 8 \Rightarrow \left(- 8 , 0\right)$

$y = 6 \Rightarrow x = 18 - 8 = 10 \Rightarrow \left(10 , 6\right)$
graph{(y^2-x^2+64)(y-1/3x-8/3)((x+8)^2+(y-0)^2-0.04)((x-10)^2+(y-6)^2-0.04)=0 [-20, 20, -10, 10]}