Subtract (2) from the square of (1) and divide by 2,

Substituting these values of y successively in (1), we get

3x2 + 7xz + z2 = 11 (3x+2)...........................................(4),

......

= 11(x+2) — 36 .........................(5).

Multiply (5) by 3 and subtract from (4),

=

222 + 108;

112 + 54

22

Substituting in (x + z)2 — 11 (x + z) = xz − 36,

65.

x + y + z = 13

x2 + y2+z2 = 91 };
; find x, y, and z.

y2 =

= x2

Subtract the 2nd from the square of the 1st, and divide by 2,

66.

16.

Subtract the 2nd from the square of the 1st, and divide by 2,

Multiply the 1st by x, subtract the 3rd and substitute for

..y +z = 9, taking the former value of x,
yz = 18;

y=3, or 6, and z=6, or 3.

Ex. 43.

Let x+y=the hypothenuse,

and x - y = one side.

Then by the question, 4xy = 2......

and 4{(x+y)2 + (x − y)2} = 5 {(x + y) + (x − y)} ....(2),

yz,

whence 4x4-53+1=0, oг 4x3 (x − 1) — (x3 — 1) = 0;

x = 1;

..x+y=2, the hypothenuse,

From (2) and (3), (x+w) (y + z) = 484.
From (1), y+2=44(x+w);

:. (x+w)2 − 44 (x + w) + 484=0;

.. x + w = 22.

From (2) and (4), (x + z) (y+w) = 475⋅
Then, as before, x+z=25.

From (3) and (4) (x+y) (z+w) = 459,
hence x+y=27;

:: (x+w) + (x + z) + (x + y) = 22 + 25 +27 = 74;

which is the case since n-2n+1 or (n − 1) is essentially positive whatever be the value of n;

9. Show that (a+b+c)3> 27abc, but <9 (a3 +b3 + c3).

• abcd < (a + b c + d)2.

2

:. a3 +b3+zab (a+b) or (a+b)3 <a3 +b3 + 3 (a+b)3 ;

4

11. x2+ y2> 2xy; .. x3+xy2 > 2x2y, and xy + y3> 2xy2;

••• x3+ y3 + x3y + xy2 > 2x2y + 2xy3,

or x3 y3x3y + xy3.