2. Given x2 + ab =
2

5x, to find the values of x.

By transposition, ab=4x2;

:: ± √ ab=2x,

to find the values of x and y.

From the second equation, xby,

Substituting this value in the first equation,

therefore, extracting the square root, x= ±3,

2x 2y 4: 2

.. (Alg. 184.) x y 2: 1,
and x = 2y.

Substituting this value of x in the second equation,

.. 8 y3 — y3=56,

or 7y3 = 56;

To the second equation, adding twice the first,
x2 + 2xy + y2=s2 + 2 a2;

.. extracting the square root, x+y= ± √ s2+2a2; and from the second, subtracting twice the first,

x2 - Qxy + y2 = s2 — 2 a2;

2

.. extracting the square root, r-y=± √√/s2 − 2a2;

2

-

but x+y= ±√√s2 + 2 a2;

.. by addition, 2x= ± √s2+2a3±√√√/s2 - 2a2,

S

s2 - 2a2).

by subtraction, 2y= ± √ s2+2a2 + √

and y = ±(√√s2+2a2·

Substituting this value of x in the second equation,

х

Substituting this value for x in the second equation,

.. extracting the square root, x+y=±6.

11. Given x+y=s, to find the values of x and y.

2

and x2 — y2 = d3)'

Since x-y=(x+y). (x − y) = s. (x − y);

d2

s2+d2

S

12. Given x+y=s
and xy = a2 }'
Squaring the first equation, x2+2xy + y2 = s2,
and from the second, 4xy =4a;

to find the values of x and y.

2

.. by subtraction, x2-2xy+y=s2 - 4a3, and extracting the square root, x−y= ± √√/s2 - 4a3,

--

by subtraction, 2y=s/s2-4a2;

• . y = { (s + √√√/ 5° — 4a3).

13. Given x+y=s|

2

and a2 + y2 = a2)'

to find the values of x and y.

Squaring the first equation, x2+2xy + y2 = s2,

and doubling the second, 2x2

+2y2 = 2a2;

£

S

2

.. by subtraction, x2-2xy + y2=2a2·

and extracting the square root, x-y= ± √2a* — s*;

but x+y=s;

2

.. by addition, 2x=s± √√2a2 —s3,

and x=(s± √2a* − s*);

also by subtraction, 2y= s√√2 a* — s° :

S

y= (s = √2a - s).