sp=84, and you will have s2—2 sp+p2=25; whence by evolution =-p=±5: this equation added to, and subtracted from s+p= 9, gives 2s= (19+5=) 24 or 14, and s=12 or 7; and likewise p= (19+5=) 14 or 24, and p=7 or 12; therefore x+y= s=) 12 or 7, and xy= (p=) 7 or 12. Subtract four times the ast from the square of the last but one, and there remains x2. xy+y2= (17)2—28, or 7)°—48=) 116 or 1; whence, by evolu=on, x−y=±√116, or+1; add this to, and subtract it from +y=12 or 7, and we shall have 2x=±√116+12, or ±1+ = (by taking the later value only) 8 or 6, whence x=4 or 3; kewise 2y= (7F1=) 6 or 8, whence y=3 or 4; if we make ǹ =4, then y=3; if x=3, then y=4 °. 13. Given x2 y2 + 9, and x+y=6, to find x and y. y х Let x=z+v, and y=z—v, then by adding these two equations gether, (r+y=) 2 z=6, and z=3; whence x=3+v, and y= -v. Multiply the first given equation by xy, and x3 +y3=9xy; hich by substituting 3+v for x, and 3-v for y, becomes 3+ v3 3—v3=9×3+v×3-v; this by involution, multiplication, d addition, becomes 54+18 v2=81−9 v2, whence by transposion 27 v2=27; therefore v2=1, and v=+1, whence_x=(z+w =3+1) 4 or 2; and y= (z-v=31=) 2 or 4; if x=4, en y=2; but if x=2, then y=4, 14. Given x2+8x=65, to find x. Ans. x=5, or —13. 15. Given y2-12 y=540, required the value of y? Ans. y= 16. Given z2-20 z=-91, what is z equal to? or 7. 17. Given 3x2 —21 x−450=6000, to find x. -43. 18. Given z2+z=2, required the value of z? -2. Ans. z Ans. x=50, Ans. z=1, If the former values of x-y, namely, ± √116 or +10.77 be taken, n for the affirmative values = 11.385, &c. and y =,615, &c. and for the gative values x =.615, &c. and y=11.385, &c. both of which values wer the conditions of the question equally with those given above. It bears from the solution, that this example (which was inserted by mise) is not an adfected, but a pure quadratic, and therefore is misplaced: same may be said of the thirteenth example. 19. Given 5 y2—25 y+40=10, to find y. Ans. y=3, or 2. 24. Given 2x2+x-y=3y2+2y, and x+y=7, to find s Ans. x=42 or 4, y=−35 or 3. and y. 98. By this rule may be solved all equations whatever, wherein there are only two different dimensions of the unknown quantity, provided the index of the one be exactly double that of the other. RULE. Having completed the square and extracted the root as before, transpose the known quantities which are on the same side with the unknown one, and then extract the root implied by the index of the unknown quantity, from both sides of the equation P. 25. Given +6 x2=72, to find the values of x. By completing the square x2+6 x2+9=81. By evolution x2+3=±9. By transposition x2= (±9—3=) 6 or—12. By evolution x=(±√/6=) +2.4494897428, or± √−12, the latter of which are impossible. ▸ Every equation will have as many roots as the unknown quantity has dimensions; thus, in the 25th example, ≈ being in the fourth power,' the equation will have four roots, as appears by the solution. In example 26, y comes out equal to 49 or 25, the latter of which being substituted in the equation, will not answer, except -5 be substituted for the square root instead of +5, the reason of which is obvious, since the 25 arose from -5X -5. One root only is required in the following examples, as finding the rest would, in many cases, require rules which have not yet been given. By transposition y= (+7+2=) 9, or -5. 28. Given x+2x2=24, to find one value of x. 29. Given y6-4 y3-32, to find y. Ans. x=2. Ans. y=2. Ans. z=4. Ans. x=4. 1+ √4a+1 32. Given x2-x"a, to find x. Ans. x-" 2 - 99. PROBLEMS. Every problem proposed to be solved algebraically, contains some conditions laid down, which are called the data; from whence one or more quantities are required to be found, called the quæsita. The first thing necessary to be done preparatory to the solution, is to understand clearly the import and signification of the problem: it must be freed from every thing ambiguous and un a The word problem is derived from the Greek golλnua. An algebraic problem is a proposition wherein some unknown truth is required to be investigated or discovered, and the truth of the discovery demonstrated. necessary; the conditions, and the manner of their dependance on each other, must be clearly ascertained and stated, and they must be carefully distinguished from the quantities proposed to be found: when this is accomplished, the conditions of the proposed problem will be exhibited under the form of one or more equations; namely, as many equations as there are unknown quantities, the solution of which is the subject of the preceding rules. Much depends on a proper substitution for the quantities required: no general rule for this can be given; sometimes a letter must be put for each; frequently, having substituted a letter for one of the unknown quantities, expressions for the others may be derived by means of this and the conditions proposed, without the aid of new letters; sometimes a substitution for the sum, difference, product, quotient, roots, powers, &c. of the unknown quantities, may be conveniently made; but the proper application of these must be learned by experience and practice. The following modes of substitution will apply in many cases. For one unknown quantity put x, for two put x and y, x being the greater, y the less; for their sum x+y, for their difference x-y, for the square of the greater x2, for the cube root of the less 3 3y, for the sum of their squares x2+y2, for the difference of their squares x2-y2, for the square of the sum x+y2, for the cube root of their difference 3/x-y, for their product xy, their quotient where the greater is proposed to be divided by the y У less, or, where the less is proposed to be divided by the greater. In general, the sum of any two quantities is represented by interposing the sign + between them; the difference by the sign, the product by the sign x, or by placing them as coefficients to each other, and the quotient by placing the dividend above the divisor; following in every case the method applicable to it, as proposed in algebraic notation. 1. What number is that, to which 9 being added, the sum will be 23? Let the required number be represented by x. To which adding 9, the sum will be x+9. Wherefore by transposition x= (23—9=) 14, the answer required '. 2. What number is that, from which 27 being subtracted, the remainder is 41? Let the number required be called x. From which subtracting 27, the remainder will be x—27. Therefore, by transposition x= (41+27=) 68, the answer required 3. 3. What number is that, which being multiplied by 4, and 5 being subtracted from the product, the remainder will be 6? Let the required number be x. This multiplied by 4, is 4x. From which subtracting 5, the remainder is 4 x—5. This remainder by the problem equals 6, wherefore 4x-5=6 11 4 And by division x= =) 24, the answer '. 4. What number is that, which being divided by 7, with 8 added to the quotient, the sum will be 9 ? Let the required number be called x. This divided by 7, the quotient is Which sum, by the problem, equals 9, whence For if it be multiplied by 4, the product is 11. From this subtracting 5, the remainder is 11-5=6, according to the problem. |