## Related questions with answers

Question

Find the exact solution(s) of the system of equations.

$\begin{aligned} & 3 x^2-20 y^2-12 x+80 y-96=0 \\ & 3 x^2+20 y^2=80 y+48 \end{aligned}$

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 5Given the system of equations,

$\begin{aligned} 3x^2-20y^2-12x+80y-96&=0&&\text{Eq.1}\\ 3x^2+20y^2&=80y+48&&\text{Eq.2} \end{aligned}$

Isolate the constant on the other side of the equation. Equation 1.

$\begin{aligned} 3x^2-20y^2-12x+80y-96&=0\\ 3x^2-20y^2-12x+80y-96+(96)&=0+(96)&&\text{APE}\\ 3x^2-20y^2-12x+80y&=96\\ \end{aligned}$

Equation 2.

$\begin{aligned} 3x^2+20y^2&=80y+48 3x^2+20y^2+(-80y)&=80y+(-80y)+48&&\text{APE}\\ 3x^2+20y^2-80y&=48 \end{aligned}$

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