Question

Solve the equation: xy=6 x^2+y^2=13

Answer 1

solve first equation for y

y=6/x

plug that expression in for y in the second equation

x^2+(6/x)^2=13

now can you try and solve for x?

Answer 2

\(\bf x^2+\left(\cfrac{6}{x}\right)^2=13x^2\implies x^2+\cfrac{36}{x^2}=13\\ \quad \\
\textit{multiplying both sides by }x^2\\ \quad \\
x^2\left(x^2+\cfrac{36}{x^2}\right)=x^2(13)\\ \quad \\\implies x^4+36=13x^2\implies [x^2]^2-13[x]^2+36=0\)

factor the quadratic to get the 4 values for "x"

Answer 3

hmm added a premature \(x^2 \).

Answer 4

\(\bf x^2+\left(\cfrac{6}{x}\right)^2=13\implies x^2+\cfrac{36}{x^2}=13\\ \quad \\
\textit{multiplying both sides by }x^2\\ \quad \\
x^2\left(x^2+\cfrac{36}{x^2}\right)=x^2(13)\\ \quad \\\implies x^4+36=13x^2\implies [x^2]^2-13[x]^2+36=0\)

Answer 5

factor it to get the values