be subtracted from that of the minuend, (2), immediately above it, we will therefore, borrow a unit from the figure in the place of hundreds, (3), which, as we have already seen, will make ten when brought to the place of tens, to which, if we add the figure (2), already in the place of tens, we shall have twelve, from which, if 4 be taken, the remainder will be 8, which we will place below the line, immediately under the figure subtracted, (4), and proceed to the next, or the figure in the place of hundreds, (2). This, it might at first seem, should be taken from the figure (3), above it; but when we remember that a unit, (one,) was borrowed from this, to enlarge its neighbor on the right, (2), that it might satisfy the demand of a greater figure, (4), we shall see why it should be one less than 3, or 2. Now, if 2 be taken from 2, nothing (0), will remain; hence, 0 must be placed under the figure (2), subtracted, as before. We now come to the period of thousands, the units figure of which is 5, this taken from the figure (6), above it, leaves 1, to be placed under it; next, the figure (3), in the place of tens, when, taken from that above it, (5), leaves 2 for the remainder; and, lastly, by taking the figure (4), in the place of hundreds, from the figure (7), above it, we have a remainder of 3, to place below it, and therefore, these several remainders, as they stand below the line, all occupying periods and places similar to the figures which produced them, will (when read in connection,) give the difference required, viz. 321082. Wherefore, 756325-435243321082, which was to be done. Again; In the equation 4653006-3425728d., (the minuend of which contains ciphers,) we proceed as follows: set down the numbers as before. 4653006 minuend. 1227278 difference. First, We must take 8 from 6, which is impossible, unless it be increased, which cannot be done by borrowing from the places of tens, or hundreds, since they are occupied by ciphers, (00); we must, therefore, borrow a unit from the number formed by the next three figures on the left, (300), which will reduce it to 299; this unit added to the six, will increase it to 16, from which, if 8 be taken, 8 will remain; hence, the next figure (2), must be taken from 9, leaying 7; also the next (7), must be taken from 9, leaving 2; but 5 cannot be taken from 2, we must, therefore, borrow one of its left hand neighbor, (5), which will make it 12, from which, if 5 be taken, 7 remains; having borrowed a unit from 5, it is reduced to four, from which, if 2 be taken, 2 remains; and, lastly, 4 from 6 leaves 2, and 3 from 4 leaves 1; which remainders, if read as they stand, will give 1,227,278 s., the difference sought; hence, the given equation becomes, 4653006 — 3425728=1,227,278d., which is the solution required. We have thus attempted to explain the principles upon which the laws of subtraction are founded. With these principles the pupil should acquaint himself thoroughly. The plan pursued heretofore, although strictly in accordance with reason, is not so well adapted to facilitate the operations of subtraction, as that which we will now explain. Let it be required to solve the equation 64048-57302=d. Thus, 64048 minuend. 6746 difference. Here we have, first, 2 from 8 and 6 remains: 0 (nothing,) from 4 and 4 remains; we are now required to take 3 from 0, which is impossible; instead, therefore, of formally borrowing a unit from 4, we will conceive it to be done, and simply add 10 to 0, which will make 10, from which take 3, and 7 remains; again, instead of treating the 4 as less by a unit, (3), we will add one to the figure under it, (7), which amounts to the same thing, making it 8; we now have 8, to subtract from 4, which is impossible, we will, therefore, increase it by ten, as before, making 14, from which take 8, and 6 remains; and, lastly, carry one to 5 makes 6, and 6 from 6, leaves 0, (nothing.) These remainders, read in connection, give 6746, for the difference required; hence, 64048-573026746. Let us now perform one more example, in which we will go through the exercise in the shortest manner. Solve the equation 30,605,060-26058073= d. First, Place the subtrahend under the minuend, as in the previous examples. Thus, 30605060 minuend. We will now proceed by saying, 3 from 10 leaves 7, one to 7 is 8, and 8 from 16 leaves 8; one to 0 (nought,) is 1, and 1 from 10 leaves 9; one to 8 is 9, and 9 from 15 leaves 6; one to 5 is 6, and 6 from 10 leaves 4; one to 0 (nought,) is 1, and 1 from 6 leaves 5; one to 6 is 7, and 7 from ten leaves 3; one to 2 is 3, and 3 from 3 leaves 0 (nothing.) Each remainder being placed below the line, and read as before, gives 3546987 d., or the difference sought; hence, 30605060-260580733546987 d., which was to be done. = Before proceeding further, the learner should be able to furnish (mentally, without hesitation,) the quantity denoted by d., in the following equations, by mentioning, first, the number on the left; second, the sign (-), (by calling it minus, or less ;) third, the quantity on the right; fourth, the sign of equality (=), (by saying equals ;) and, lastly, the difference (d). Thus, 5-3-2, or five less three, equals two; 7-3=4, read 7 less 3, equals 4, &c.; as, 4-3=d; 5— 3=d; 6—2=d; 7—4=d; 8—3=d; 7—2=d; 9—5=d; 8-4-d; 6-4-d; 7-3=d; 8-2=d; 9—3=d; 10— 7=d; 12-8=d; 13-7=d; 14—6=d; 15—6=d; 16— 7=d; 17—9=d; 18-9-d; 19—10=d; 17—8—d; 15— 7=d; 14-8=d; 16-9=d, &c. Equations in subtraction, may be solved by subtracting the subtrahend from the minuend, according to the following RULE for subtracting integers. 1. Place the subtrahend under the minuend, as in addition, and draw a line under the former. 2. Begin with the units figure of the period of units, and subtract each successive figure of the subtrahend from that immediately above it in the minuend, and place the remainder below the line, directly under the figure subtracted; but if the figure in the minuend be smaller than that below it, in the subtrahend, increase the former by adding ten, and subtract from their sum, setting down the remainder as before, observing to add a unit to the next figure in the subtrahend, before subtracting it, (to balance the unit borrowed from the figure, or number above it, as has been explained.) This repeated for each successive figure of the subtrahend, and the remainders read in order as they stand, will be the difference (d) required. PROOF. Add the difference to the subtrahend, and if their sum agree with the minuend, the work is supposed to be right, because the difference added to the lesser number, is always equal to the greater. EXAMPLES PRODUCING EQUATIONS. Solve the following equations: 1st. 679843-548721=d. 2d. 706-689 d. 3d. 504891-450720 d. 4th. 1001 d. 5th. 746875-1001 d. 6th. 1000-9=d. Find the value of d, in the following: 1st. 605040302— 504030201 d. 2d. 100000000-991=d. 909090 d. 4th: 75621-100= d. 3d. 999999 Express the difference between seven millions, seven thousand and seven; less, five millions, fifty thousand and five hundred, in the form of an equation. APPLICATION. 1. A father left to his oldest son 3728 dollars, and to his daughter 2967 dollars; what was the difference in their portions? The equation is 3728-2967-d., in which the value of d. is required. 2. How many more people are there in New York than in Philadelphia; the former having 312000 inhabitants, and the latter 228000? 3. Florida has been settled 281 years, and Ohio 58 years; how much longer has Florida been settled than Ohio? 4. Ohio has 1519467 inhabitants, and Florida 54477; what is the difference in their population? It will now be most convenient, for general purposes, to give particular names to such quantities as have the sign + (plus,) or - (minus,) prefixed; the former we will call positive, and the latter negative quantities; hence, I. Positive quantities are such as have a positive, or additive value; they are known in equations by having no sign, or the sign (plus) prefixed; as, 8+5=13, in which 8, 5, and 13 are each of them positive quantities. II. Negative quantities are such as have a negative, or subtractive value; as, 8—5—3, in which —5 only is a negative quantity, or 5-8-3, in which 5 is a positive quantity; but —8 and -3 are negative quantities. We will now solve some equations containing positive and negative quantities. Solve 48-6+7—16+5—31+6—54+18—9+25=x. Here the unknown quantity is represented by x, the value of which may be found, as follows: Add together all the posi tive quantities, and also the negative quantities, and having found the sum of each, subtract the smaller from the greater, prefixing to the remainder the sign —, if the excess be nega tive, and the difference will be the value (x) required; thus, POSITIVE AND NEGATIVE QUANTITIES. 31 48+7+5+6+18+25=109. Again —6—16—31—54— 9=—116. Hence, 109–116= −7=x, the value required. It may seem strange to the learner, that the quantities to be subtracted should amount to more in the aggregate than those from which they are to be taken; but this apparent contradiction may be reconciled by supposing that A. keeps an account of his income and expenditures; that the positive quantities are his income, for a certain period, in dollars, and the negative quantities, his outgo, for the same time; in this case, it is plain, that his expenditures must have exceeded his income by 7 dollars; hence, as it is common to express it, he comes out minus seven (7) dollars: but had his income exceeded his outgo, by the same quantity, he would have been plus seven (7) dollars. This, it may be said, establishes the expediency of such a mode, but does not reconcile the inconsistency of taking a larger number from a smaller one, as in the equation 109–116— —7; but this is so only in appearance, since the positive quantity is in fact, taken from the negative; but, since positive quantities are, more appropriately, the representatives of responsibility, they should be the LEADING terms of an equation; for, if we should place the negative quantity (—116), on the left, making it the leading term, and represent the positive quantity (109), as being subtracted from it, the equation would stand —116—109— -7, which is absurd; since two or more quantities, on the same side of the equation, having like signs, should be added, but having different signs, they should be subtracted, i. e., the quantity, or quantities containing a LESs number of units, should be taken from those containing MORE; thus, 8—5—3, but 5—8— -3. Here 3 is the remainder in both cases; but in the former it becomes a positive, and in the latter a negative quantity. From what has been said we learn that, I. A less number may be taken from a greater, by giving the difference a positive sign, or which is the same thing, by giving it no sign; and, vice versa, a greater number may be taken from a less, by giving the difference a negative sign; as, 9-7-2, or +2; but 7—9——2. II. A negative remainder always indicates a deficiency in the minuend, to the extent of its own value; but a positive remainder shows an excess in favor of the minuend, to the |