76. (a2m + 1) (x3 − 1)2 = 2 ́(x + 1).

Ans. x =

m

2

2 m

77. (-9)-3= 11 (x2 - 2). Ans.

5, or ±2.

78. √{x + √(2 x − 1)} − √ {x — √ (2 x − 1)} =

86. (x + 3)2 — 2 (x2 + 3) = 2 x (x + 1)2.

±√37

2

.a.

This manner of proceeding may be adopted whenever the sum and product of the two unknown quantities are given

This mode of proceeding may be adopted whenever the difference and product of the two unknown quantities are

given.

3. x2 + y2

(1).

(2).

2y2 = 404

x2 + 2xy + y

Double (1) and square (2).

=

400}

Subtract.

y2 = 4

=±2)
= 20

+ 2y=788)

11, or 9. y= 9, or 11. Double (1) and square (2).

}

(1).

(2).

Subtract.

This mode of proceeding may always be adopted when the sum of the squares and the sum or difference of the two unknown quantities are given.

n being an odd number, or any equations similar to these were given, we might proceed by dividing one equation by the other, as in this example.

x2 + 4x3y + 6x2y2 + 4xy2+ y1 — 2401

+y+: 337) Subtract.

=

Taking xy

txy

=

49

Multiply by 2xy, and transpose.

Ex. 1, we have

xy

98xy — 1032. Complete the square.

98 xy +49

49

xy

2401 1032 1369,

= ± 37;

4937= 86, or 12.

12, and x + y = 7, and proceeding
= as in
= 4, or 3; y = 3, or 4.

2xy = 800. Adding this to (1),

x2 + 2xy + y2+x+y=1722. Complete the square.

and ·.· xy = 400, by proceeding as in Ex. 1, we have

≈= = 25, y = 16.