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what is the standard form for 2d/d-5 + 1/d-3=3

Sagot :

Cheahjimmyidn
2d/d-5 + 1/d-3 = 3
then
(d-5)(d-3)[(2d/d-5+1/d-3)] =  3(d-5)(d-3)        multiply the equation by( d-5)(d-3)
 2d(d-3)+ 1(d-5) =  3d²-24d+45
 2d²-5d-5 =     3d² -24d+45
=    d²-19d+50 = 0

Riza1idn
[tex]\frac{ 2d}{d-5 } +\frac{ 1}{d-3} =3 \\ \\d-5\neq 0 \ \ and \ \ d-3 \neq 0 \\\\ d \neq 5 \ \ and \ \ d \neq 3 \\\\ D=R\setminus \left \{ 3,5 \right \}[/tex]

[tex]\frac{ 2d(d-3)+(d-5)}{(d-5) (d-3)}-3 =0 \\ \\\frac{ 2d^2-6d+ d-5-3(d-5)(d-3) }{(d-5) (d-3) } =0\\\\\frac{ 2d^2-6d+ d-5-3(d^2-3d-5d+15) }{(d-5) (d-3) } =0\\\\\frac{ 2d^2-6d+ d-5-3 d^2+9d+15d-45 }{(d-5) (d-3) } =0[/tex]

[tex]\frac{ -d^2+19d -50 }{(d-5) (d-3) } =0 \\\\ -d^2+19d -50 =0 \ \ / \cdot (-1)\\\\ d^2-19d +50 =0[/tex]