SOlving Simultaneous Equations - HELP!?

Question

x - y = 7 and 2x^2 - 3y^2 = 5

THANKS.. i'll give BEST ANSWER....

I would like to know how it is solved..... and the answer finally... thanks.

Answer 1

x - y = 7 (1)
2x^2 - 3y^2 = 5 (2)


first, rearrange equation (1) to find an equation for x;

x = 7 + y (1*)

then, substitute this x equation into (2);

2x^2 - 3y^2 = 5
2(7 + y )^2 - 3y^2 = 5

then solve for y;

2(y^2 + 14y + 49) - 3y^2 = 5
[2y^2 + 28y + 98] - 3y^2 = 5
-y^2 + 28y + 98 = 5
y^2 - 28y = 98 - 5
y^2 - 28y = 93
y^2 - 28y - 93 = 0
(y +3)(y - 31) = 0
y = -3 or 31
Now solve for x by substituting in the y values into any of the two original equations (1)/(2);

y = -3;

x - y = 7
x = 7 + (-3)
x = 4

y = 31;

x - y = 7
x = 7 + 31
x = 38

the answers => (x = 4, y = -3) or (x = 38, y = 31)

hope this helps:-)

Answer 2

Given eqns are :

x - y = 7 .......... (1)

2 x^2 - 3 y^2 = 5 .... (2)

From Eqn (1) we get --- x = y + 7

Plugging (y +7) for x in eqn (1) gives -

2 ( y + 7)^2 - 3y^2 = 5

Expand ( y + 7)^2 to give -

2 [ y^2 + 14y + 49 ] - 3 y^2 = 5

=> 98 + 2 y^2 + 28 y - 3 y^2 = 5

=> - y^2 + 28 y + 93 = 0

Multiply both the sides by ( - 1 ) :

=> y^2 - 28 y - 93 = 0

=> y^2 - 31 y + 3 y - 93 = 0

=> y ( y - 31 ) + 3 ( y - 31 ) = 0

=> ( y + 3 ) ( y - 31) = 0

ie y = [ -3, 31 ] and x = [ 4 , 38 ]

[ x = 4 , y = -3 ] Or [ x = 38, y = 31 ].......... Answer

Answer 3

x=7+y
substitute and get
2(7+y)^2-3y^2=5
solve and get
y^2-28y-93=0
(y+3)(y-31)=0
y=-3,31
x=4,38
solution set(4,-3),(38,31)

Answer 4

x = 7 + y
2(7 + y)² - 3y² = 5
2(49 + 14y + y²) - 3y² = 5
98 + 28y + 2y² - 3y² = 5
y² - 28y - 93 = 0
(y + 3)(y - 31) = 0
y = - 3 , y = 31
x = 4 , x = 38

(4 , - 3) , (38 , 31)