Question

xy=18 x^2 +y^2 = 45 then find x+y​

Answer 1

Step-by-step explanation:

Given:

x²+y²=45

xy=18

\begin{gathered}\frac{1}{x}+\frac{1}{y} \\\\=\frac{y+x}{xy}\end{gathered}

x

1

+

y

1

=

xy

y+x

Taking square

\begin{gathered}=\frac{(y+x)^{2} }{(xy)^{2} } \\\\=\frac{x^{2}+y^{2}+2xy }{(xy)^{2} } \\\end{gathered}

=

(xy)

2

(y+x)

2

=

(xy)

2

x

2

+y

2

+2xy

Substituting the known values

\begin{gathered}=\frac{45+(2\times 18)}{18\times 18} \\\\=\frac{45+36}{324} \\\\=\frac{81}{324} \\\\=\frac{1}{4}\end{gathered}

=

18×18

45+(2×18)

=

324

45+36

=

324

81

=

4

1

Learn More:

If X + Y + Z is equal to 1, xy + yz plus zx is equal to -1 and x y z is equal to -1, find the value of x cube + y cube + Z cube

Answer 2

Answer:

3(x

2

+7)

\longrightarrow (3 \times x^2 ) +(3 \times 7)⟶(3×x

2

)+(3×7)

\longrightarrow 3x^2 + 21⟶