Simultaneous equation difficulty minus { <mtable columnalign="left left" rowspacing=".2

rs450nigglix2

rs450nigglix2

Answered question

2022-05-23

Simultaneous equation difficulty minus
{ 9 x y = 1   . . . . . . ( 1 ) 6 x + 3 y = 10   . . . . . . ( 2 )
( 1 ) × 3
27 x 3 y = 3   . . . . . . ( 3 )
( 3 ) ( 2 )
21 x = 7
3 x = 1

Answer & Explanation

Alessandra Weeks

Alessandra Weeks

Beginner2022-05-24Added 5 answers

From the second two equations: Yes,
21 x = 7 3 x = 1 x = 1 3
Now solve for y.
In the first two equations, you should have
27 x = 7 3 x = 1 x = 1 3
Now solve for y.
Note that the two systems you posted are identical, save for the equation 27 3 y = 3 in the second "pair". I think you must have multiplying the first equation by 3, but dropped a "sign". You should have gotten
27 3 y = 3 ,
in which case you'll find that x = 1 3 is the solution for x...

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