And then xva +x3 =a2—x2

And x3×a2+x2=a3—x3 \3=a*—2a2x2+x•
Or a3x+x=a*—2a3x3+x*

• Whence a3x3+2a3x2=a*

Or 3a x=a*

3. Given 6-2x+10=20—3x-2; to find x.

a

4. Given 3ux+ --3-bx-a; to find x.

2

Ans. x=2.

8. Given Va2+x2=b*+**; to find x.

Ans. x=4.

9. Given x+ Va3 +xV63+x2; to find x.

REDUCTION OF TWO, THREE, OR MORE, SIMPLE EQUATIONS, CONTAINING TWO, THREE, OR MORE, UNKNOWN QUANTITIES.

PROBLEM I.

To exterminate two unknown quantities, or to reduce the two simple equations, containing them, to one.

RULE 1.

1. Observe which of the unknown quantities is the least involved, and find its value in each of the equations, by the methods already explained.

2. Let the two values thus found be made equal to each other, and there will arise a new equation with only one unknown quantity in it, whose value may be found as before,

x+y=a

}

;

2. Given ab to find x and y.

From the first equation x=a—y,
And from the second, x=b+y,
Therefore a-y-b+y, or 2y=a—b,

And consequently, y= a b

2

4. Given 4x+y=34, and 4y+x=16; to find x and y.

Ans. x=8, and y=2.

4+=1; to find x and y.

5

Ans. x=1, and y=}•

6. Given x+y=s, and x3-y=d; to find x and

1. Consider which of the unknown quantities you would first exterminate, and let its value be found in that equation, where it is least involved.

2. Substitute the value thus found for its equal in the other equation, and there will arise a new equation with only one unknown quantity, whose value may be found as before.

From the first equation x=17-2y,

And this value, substituted for x in the second, gives

17—2yX3-y=2,

Or 51-6y-y=2, or 51—7y=2

That is, 7y=51-2=49;

Whence y==7, and x=17-2y=17—14=3.

2. Given {x=3
x + y = 13 };

x-y=

; to find x and

From the first equation x=13-y,

y.

And this value, being substituted for x in the second,

Gives 13-y-y=3, or 13-2y=3;

That is, 2y=13-3=10,

Or y=10=5, and x=13-y=13-5-8.

And this value of x, substituted in the second,

x

+7y=99, and 2+7x=51 ; to find x and y.

4. Given+7y=99, and

3

6. Given a b::x:y, and x3-y3=d; to find x and y.

Let the given equations be multiplied or divided by such numbers, or quantities, as will make the term, which contains one of the unknown quantities, to be the same in both equations; and then by adding or subtracting the equations, according as is required, there will arise a new equation with only one unknown quantity, as before.

EXAMPLES.

3x+5y=40

3x+21=14

1. Given x+2y=14; to find x and y.

First, multiply the second equation by 3,
And we shall have 3x+6y=42 ;

Then, from this last equation subtract the first,
And it will give 6y-5y=42-40, or y=2,
And therefore, x=14-2y-14-4-10,