24

60. If x: 12 : 2:3, required the value of a ?

Thus (xx3=12×2, or) 3 x=24, whence by division x=

(—=—=)8, the answer.

61. If z- 2: 2025, required z?

Thus (z-2×5=20×2, or) 5 z-10=40, whence 5z-50,

87. When the same quantity, with the same sign, is on both sides of the equation, it may be taken away from both; and if every term of an equation be multiplied by any (the same) quantity, or divided by any (the same) quantity, such multiplier, or divisor, may be taken away from them all.

the product of the means; this will be obvious from any example of four numbers, which are known to be proportionals;

Thus if 2: 4:3: 6, then will 2×6=4 X 3.
And if 12: 4 :: 21; 7, then will 12 X 7 = 4 × 21.

And the same will appear in every instance.

This is merely taking equals from equals, dividing equals by equals, or multiplying equals by equals ; wherefore in each case the results will be equal.

Leaving out the common multiplier a, we shall have 3z+

=2b-3, which by transposition becomes 3 z=2b−3·

common divisor 5, we have 2x-3=4, whence by transposition

2x=7, and by division x= ( = ) 3, the answer.

70. Given2x+3y=x+3y+8, to find x.

Ans. x=S.

71. Given 3y+3=3y+5x, to find x. Ans, x=

1+3 X 4

3
5

72. Given

to find r. Ans. x=1.

88. PROMISCUOUS EXAMPLES FOR PRACTICE,

1. Given 6 - 5 =3 x + 4, to find .

By Art. 75. 3x=9.

4

3

2

By Art. 83. y=- =1.1547, &c. the answer.

√3

8. Given 13-√/3x=√13+3x, to find x.

169-26/3x+3x=13+3x.

169-26/3x=13.

By Art. 81.

By Art. 87.

18. Given b+a+, to find x. Ans. x= x

89. REDUCTION OF SIMPLE EQUATIONS, INVOLVING

TWO OR MORE UNKNOWN QUANTITIES. When there are two or more unknown quantities, whose values are required, in order to obtain determinate answers, it is necessary that there should be as many equations (independent of each other) given, as there are unknown quantities to be found; thus, when there are two unknown quantities, two equations must be proposed; for three unknown quantities, there must be three equations; for four, four equations, &c.

If there are more equations than unknown quantities, the superfluous equations will either coincide with some of the others, or contradict them; wherefore, in the former case they are unnecessary, and in the latter, detrimental; rendering the proposed solution impossible.

If there are fewer equations than unknown quantities, the problem will admit of many answers; thus, let x+y=4, here is but one equation, and two unknown quantities to be found; now may equal 3, then y=1; if x= 2, then y=2; if x= 1, then y=3; wherefore, in this instance, both x and y admit of three interpretations, using whole numbers only; if fractions be admitted, the values of x and y will be innumerable.

In all these cases there is one general mode of procedure, namely, we exterminate all the unknown quantities from the operation, except one, the value of which is to be found by the foregoing methods; having obtained the value of this, the value of each of the other unknown quantities will be readily found by means of it, from some of the preceding equations.

For two unknown quantities.

90. First Method.

RULE 1. Find the value of one unknown quantity in each of the equations, by the foregoing rules; it may be either of the two quantities at pleasure, but must be the same unknown quantity in both equations.

II. Put the two values thus found equal to each other; this equation will then contain but one unknown quantity, the value of which is to be found by the preceding rules.

III. Having thus found the value of one unknown quantity, substitute it for that quantity in either of the preceding equations, and the value of the other unknown quantity will be found. EXAMPLES.

1. Given x+y=13, and x-y=3, to find the value of x and y. First, to find the value of x in each equation.

From the first, x=13-y; and from the second, x=3+y; these two values of x are evidently equal to each other; wherefore 13—y=3+y, whence 2y=10, and y=5; now substitute this value of y in either of the former equations, suppose in x=3+y, and it becomes x= (3+y=) 3+5=8.

Wherefore x=8, and y=5.

2. Given 2x+3y=17, and 3 x-2y=6, to find x and y. From the first equation, 2x=17-3y,

These two values being equal to the same unknown quantity, are evidently qual to one another: the unknown quantity, whose two values (or rather, two ifferent expressions of the same value) are thus found, is said to be extermiated, because it does not appear in the resulting equation.