Ex. 5. Given x2+y=13, to find the values of and y5, x and y.

By addition, 2x2 = 18; .'. x=±√/3=±3. By subtraction, 2y2=8;.. y=±√√/4= ± 2. Ex. 6. Given 81x1=256, to find the values of x. By extracting the square root, 9x2=±16; By extracting again, 3x=±±16=±4, or ±4√/−1; ..x=4, or x=4-1. Ex. 7. Given x-3x2+3x2-1=27, to find the values of x.

By evolution, x-1=3; .. x2=4, and x=±2. Ex. S. Given 36x2 =a2, to find the values of x. Ans. x=a.

Ex. 9. Given x3-27, to find the value of x.

Ex. 10. Given x2+6x+9=25, to find the values Ans. x=2, or—8.

of x. Ex. 11. Given 3x2-9=21+3, to find the values

2

Ans.

of x. 11. Ex. 12. Given x3-x2+x-a. to find the value of x. Ans. x=a+3. Ex. 13. Given x2+3x+}=a2b2, to find the values of x. Ans. xab—ã• Ex. 14. Given x2+bx+1b2a2, to find the vaAns. xa-1b.

lues of x. Ex. 15. Given a1-2x2+1=9, to find the values of x. Ans. x=2, or ±√−2. Ex. 16. Given x1-4x2+4x=4, to find the values Ans. x2, or ±√0.

of x. Ex. 17. Given 5x2-27-3x2+215, to find the va

Ans. x

11.

lues of x. Ex. 18. Given 5x-1=244, to find the values

of x.

Ans. x7.

Ex. 19. Given 9x2+9=3x2+63, to find the values of x.

Ans. x=3.

Ex. 20. Given 2ax+b-4-cx2-5+d-ax3, to

Ex. 21. Given x+y=a and xyb, to find the values of x and y.

Ans. x=±√(√(2a+2b)) and y=±√(±√ (2a-2b)).

§ III. EXAMPLES IN WHICH THE PRECEDING RULES

ARE APPLIED IN THE SOLUTION OF PURE EQUATIONS.

398. When the terms of an equation involve powers of the unknown quantity placed under radical signs.

Let the equation be cleared of radical signs, as in Sect. I; then, the value of the unknown quantity will be determined by extracting the root, as in Sect. II.

And by a similar process, any equation containing the powers of a function of the unknown quantity, or containing the powers of two unknown quantities, may frequently be reduced to lower dimensions. Ex. 1. Given 3⁄4/x2=3/(a+b), to find the values equation

of x.

Here, the given quantity may be exhibited under the form (x2—8)3 = (x-3); then, by squaring both

*

2

sides, (x2 — 9)*×3 — (x — 3)2 9)4。

=

,

I

or (x2 — 9)2=x-3; by squaring again, x2-9=x2-6x+9; ... by transposition, 6x=18; and x=3. Ex. 3. Given x2 y29, and x-y=1; to find the values of x and y.

Dividing the corresponding terms of the first equation by those of the second, we have

x+y=9; adding this equation to the second, 2x=10; .. x=5, and y=9-x;..y=4. Ex. 4. Given x+y=5, to find the values of

and

y=1, x and y.

Adding the two equations, 2x=6, ..√x=3, and by involution, x=9.

Subtracting the two equations, 2y=4, and y=2;

.. by involution, y=4 to find the values of

Ex. 5. Given x2+xy=12,
and y2+xy=24, 5 x and y.

By addition, x2+2xy+y2=36; ... extracting the square root, x+y=6. Now x2+xy=x.(x+y)=±6x; ..6x=12, and x=2;

•'•y=±6=2=±4. 2α2

Ex. 6. Given x+(a2+x2)=.

to find

'√(a2 + x2)'

the values of x.

Multiplying by √(a2+x2), we have x√(a2+x2) +a2+x2=2a2;

by transposition, x/(a2+x)=a2x2; and squaring both sides, a2x2+x=a1-2a2x2+x+;

.. by addition, x2+2xy+y2=25,

by extracting the square root, x+y=±5;

but x − y = 1;

or

.. by addition, 2x=6, or 4: and x=3,
by subtraction, 2y=4, or -6; and y=2, or

Ex. 8. Given x+y

--

-2; 3.

= = 13, to find the values of

and x+y=4, x and y.
x32+y=4

Squaring the second equation, x+2x+y=25

but 23

x

+y3=18

... by subtraction, 2x*y*=12.

Subtracting this from the 1st equation,

៖

x*3 —2x31y3+y3=1 ;

I

... extracting the square root, x—y3 = ±1;

8

and y=2, or 3;

4 4

Ex. 9. Given æo+x*y*+y=273, } and' x+xy+y+=21,

values of x and y.

=

=8, or 27.

Dividing the first equation by the second, x1-¤2 ya+y1=13;

subtracting this from the second equation, 2x2y2=8 ;

... x2y2=4; by adding this equation to the second, x+2x2y2 + y1=25;

x2+y2=±5. Subtracting the equation xy24, from x-x2 y2 +y=13,

x1—2x2y2+y1—9; .'. x2—y2=±3,

... by addition, 2x2=18, and x=12, or ±2/-1; and by subtraction, 2y2±2, and y=±1, or ±√

-1.

Ex. 10. Given

value of x.

Multiply the numerator and denominator

by √(a+x)+√(a−x), [√.(a+x)+√(a−x)]2 =b,

2x

or 2a+2/(a2 —x2)=2bx; .•. √(a2 — x2)=bx—a, and squaring both sides, a2-x=b2x2-2ɑbx+a2

2ab

b2+1°

Ex. 11. Given (x2—y2)× (x−y)=3xy, ( to find and (x-y1) x (x2-y2)=45x2y2, {

the values of x and y.

3

3

Dividing the second equation by the first, (x2+y2) .(x+y)=15xy; ;·• • x2 + x3y + xy2+y3=15xy; but from the first, x3 — x3y — xy3 +y3= 3xy;

... by addition, 2x+2y3-18xy, and x+y=9xy. But by subtraction, 2x2y+2xy=12xy, and x+y=6; .. by cubing, 3 +3x2y+3xy2+y3=216;

.. by subtraction, 3x2y+3xy2-216-9xy,

or 3.(x+y).xy=3×6.xy=216-9xy; ... 27xy=216,

and xy=8.

Now x2+2xy+y2=36, and 4xy =32;

... by subtraction, x2-2xy+y2=4, and by extracting the square root, x-y=±2, but x+y= 6,

... by addition, 2x=8, or 4; and x=4, or 2 ; and by subtraction, 2y=4, or 8; ..y=2, or 4. Ex. 12. Given 2+(a3 —x2 ) _ x

lues of x.

Ex. 13. Given

values of x.

Ex. 14. Given

=, to find the va

Ans. x=√(2αb-b2).

18

x2+3x-7=x+2+1, to find the

x

Ans. x=3, or -3.

√(x+a) +2 2√(2)

to find the values of x.

=b2.