x265-y2.

Putting these two values of x2 equal to each other,

784

-65-y2.

y2

Clearing this Equation of its fraction,

784=65y2—y1.

From this Equation y will be found value of x will then be found to be ±7, or ±4.

4. Find the value of x and y in the Equations
x+4y-14, and 4x-2y+y2-11.

±4, or 7, (263). The

By transposition in the first Equation,

x=14—4y.

Substituting this value of x in the second Equation, 56-18y+y211; hence

y=3, or 15.

We shall now easily find x=2, or -46.

5. Find the values of x and y in the Equations
x+2y=7, and x2+3xy-y2=23.

Transposing 2y in the first Equation,

x=7-2y.

Hence x2-49-28y+4y2, and 3xy=21y-6y2.

Substituting these values of 22 and 3xy in the second Equation, and adding similar terms in the first member of the resulting Equation,

49-7y-3y2-23; hence
y=2, or —41.

Substituting these values of y, successively, in the first Equation, we shall find x=3, or 15%.

6. Find the values of x and y in the Equations

x2 + 1/2

11, and xy+2y2=120.

Clearing the first Equation of its fraction,

2x+y=22; hence
y=22-2x.

Then xy=22x-2x2, and 2y2—968—176x+8x2.

Substituting these values of ry and 2y2 in the second Equation, transposing, and uniting similar terms,

6x2-154x-848;

from which will be found equal to 8, or 17.
The value of y will then be found to be 6, or -13.

X

Hence 10x=10y-20, and 3xy=3y2—6y.

Substituting these values of 10x and 3xy in the third Equation, and reducing, we find

17y-3y2=20; hence

y=4, or 1.

x-y 2

3.xy

10

Substituting these values of y, successively in the first Equation, we find x=2, or -§.

8. Find the values of x and y in the Equations

x+3y

=4, and y

x+2

=1.

Clearing these Equations of fractions, and reducing each resulting Equation, we have

x+y=8, and xy-y-2x-2.

Transposing y in the third Equation,

x=8-y; hence
xy=8y-y2.

Substituting these values of x and xy in the fourth Equation, and reducing,

9y-y2=18;

which will give y=6, or 3; and we shall now find x=2, or 5.

9. Find the values of x and y in the Equations
5(x+y)=13(x-y), and x+y2=52.

Developing both sides of the first Equation, and reducing,
8x=18y; hence

9y

4

From the second Equation, we have

x=25-y2.

Equating these two values of x, we have

9y

=25-y2.

4

From this Equation we shall find y=4, or −64; and the value of x will then be found to be 9, or -14.

x=

Solutions by Means of an Auxiliary Unknown Quantity.

10. Find the values of x and y in the Equations x2+xy=54, and 2xy+y2=45.

Assuming xvy, and substituting this value for x in each of the given Equations, we have

v2y2+vy2=54, and 2vy2+y2=45.

From the first of these Equations,

y2=

54
v2+v'

and from the second y2=

45

2v+1

v=2, or

Substituting these values of v, successively, for v in the fourth Equation, we have

From the second,

4y2+y2-45 and y2

The first of these Equations will give y=±3.

The second of the two will give y±√−225=±15√−1. Then, since xvy, we have x=±3×2=±6; or x=±15√-1 3

=9-1.

5

7

11. Find the values of x and y in the Equations

xy=28, and x2+y2=65.

Substituting vy for x in each of these Equations, we have vy2-28 and v2y2+y2=65.

From the first of these Equations,

y2=

y2=

Equating these two values of y2,

28

x=

3

5

=

65

v v2+1

Clearing this Equation of its fractions,

6y2

5

28

V

7

4 or

4' 7

=45.

65

v2+1

28v2+28=65v; hence

Substituting the first of these values for v in the third Equation, we shall find y±4.

Substituting the second value for v in the same Equation, we shall find y±7.

Then since x=vy, we shall have x=±4×

12. Find the values of x and y in the Equations x2+xy 12, and xy-2y2-1.

Clearing this Equation of fractions,

Substituting vy for x in each of these Equations,
v2y2+vy2=12, and vy2-2y2=1.
From the first of these Equations, we shall find
y2-
and from the second, y2=

12

=

"

v2 + v

Equating these values of y2,

12

1

=
22+v v- -2

12v-24=v2+v;

7

4

which will give v=3, or 8.

Substituting these values, successively, in the fourth Equation we find

Then x1x3±3, or x=±

=±7; or x=±7

1

√6

1

√6

×8=±

1

v-2

(209).

8

√/6

13. Find the values of x and y in the Equations
4x2-2xy=12, and 2y2+3xy=8.

Assuming yvx, and substituting this value for x in the given. Equations,

4x2-2vx2-12, and 2v2x2+3vx2=8.

Finding the values of 22 from these Equations, and putting them equal to each other,