QUESTION 4. What is that number which being multiplied by 6, the product increased by 36, and that sum divided by 18, the quotient shall be 20? Solu. Let x= the number sought. Then 6x= the number multiplied by 6. And 6.x +36 18 the product increased by 36, ≈ that sum divided by 18 QUESTION 5. A post is one-fifth in the earth, three-sevenths in water, and 13 feet out of the water; what is the length of the post? Solu. Let x Then 3r Also 7 And 13 the part of it in the water. the part of it out of the water; But the part in earth + part in water + part out of water the whole post. Therefore+ 3r 7 + 13 = x. QUESTION 6. Having paid away one-fourth and one-seventh of my money, I had eighty-five pounds left in my purse; how much money was in it previously to these disbursements? Solu. Let = the money in purse at first; then Paid away Now, by the question, + ༡ is the money +380 Therefore, r= =140€ The money at first in purse. 17 Examples for practice. 7. The difference of two numbers is 10, and if to their sum 15 be added, the whole will be 43; it is required to find those two numbers? Ans. 9 and 19. 8. The difference of two numbers is 14, and if 9 times the less be subtracted from 6 times the greater, the remainder will be 33; what are those numbers? Ans. 17 and 31. 9. If to a certain number I add 20, and from two-thirds of this sum I take away 12, the remainder will be 10; what is that number? Ans. 13. 10. What number is that, of which if I add one-third, one-fourth, and two-sevenths together, the sum shall be 73 ? Ans. 84. 11. What number is that the third part of which exceeds its fifth by 72 ? Ans. 540. 12. Two merchants, A and B, lay out equal sums of money in trade? A gains 120l., and B loses Sol.; and now A.'s money is treble of B.'s-Pray what sum had each at first? Ans. 180. (66.) The equations which we have hitherte considered contained but one unknown quantity; but, in the solutions of Algebraic Problems, equations frequently arise containing two or more unknown quantities; as, ax + by = c, ax + by + cz d, &c. From the first of these it is evident, by the rules already laid down, that ax=- -by or, x = c—by' = In this equation the value of x, which is by > be found in terms of y; but y is an unknown quantity, consequently z still remains undetermined; whence it is manifest, that from the equation ax + by = c, neither x nor y can be found. But if in addition to ax + by = c, we have another equation, as, dx + ey =ƒ, from which x may be found as before c — by _ ££ ey, a d f ey = d each being equal to x. From this last equation, y may be found by the foregoing rules; and when y is found x = C- -by becomes known. From what has been said, it appears that in order to determine the values of two unknown quantities, two equations must be given. In = like manner, it may be shewn that three equations must be given to determine three unknown quantities; and so on. It might be further remarked, that these equations must be independent one upon the other. Thus, ax + by = c, and dr + ey f are independent, but ax + by =c, and ax = c — - by, are not independent, for the one may be derived from the other; and the two, though of a different form, amount, in effect, to no more than one equation. What we have now said will lead us to PROBLEM I. To exterminate two unknown quantities, or to reduce the two simple equations containing them, to a single one. (67.) Rule 1.-1. Find the value of one of the unknown quantities in terms of the other, from each of the two given equations. 2. Make the two values thus found equal to each other, and there will arise a new equation containing only one unknown quantity, whose alue may be found as before. Or 115-15y=20+4y, Or 19y=115—20=95; 2. Given {+3} it is required to find x and y. x+y= a From the first equation x=a-y, and from the second x=b+y. a-b Therefore a-y=b+y, or 2y=a-b, consequently y= ·, and 2 and y. 4. Given 4x+y = 34, and 4y+x = 16, it is required to find r Ans. x= 8, and y = 2. 6. Given x+y=s, and x2-y=d, it is required to find x and y. Ꮖ Ans, x= 2+d and y= s2-d 2s 7. Given x-y = d, and xy::n: m, it is required to find x and y. Rule 2.-1. Consider which of the unknown quantities must be first terminated, and find its value in that equation where it is least involved. 2. Substitute the value, thus found, for its equal in the other equation, and a new equation will arise with only one unknown quantity, whose value may be found as before Ex. 1. Given { x+2y=172 3x-y= 25 it is required to find x and y. From the first equation x=17-2y. And this value substituted for in the second, gives 3(17-2y)-y=2, Or 51-6y-y=2, or 51-7y=2; that is 7y = 51—2—49 ; And this value, substituted for x in the 2d, gives 13-y—y=3, o 13-2y=3, That is, 2y=13—3=10, 10 Whence y==5, and x=13-y=13—5—8, 3. Given+y=c( Sa:by it is required to find x and y. a'y The first analogy turned into an equation is bx=ay, or x= b c, or And this value of a substituted in the 2d, gives()+y=c, 4. Given 2x+3y=16, and 3x-2y=11, it is required to find x and y. 5. Given+7y=99, and+7x=51, it is required to find x and y 7. Given a : b: :x : y, and z3-y3=d, it is required to find x and y. Ans. ( da3 Rule 3.-1. Multiply or divide the given equations by such numbers or quantities as will make the term which contains one of the unknown. quantities the same in both equations. 2. Then, by adding or subtracting the equations, according as may oe required, a new equation will arise with only one unknown quantity as before. S3x+5y=402 Ex. 1. Given 2x+2y=145 it is required to find x and y. First, multiply the 2d equation by 3, and it will give 3x+6y= 42. Then from the last equation subtract the first, and it will give бy-5y=42-40, or y=2, x=14-2y=14-4=10. 2. Given (2318} it is required to find x and y. Let the first equation be multiplied by 2, and the 2d by 5, and we shall have 10x · 6y=18, 10x+25y=80; And if the former of these be subtracted from the latter, it will give Let the first equation be multiplied by 5, and the second by 3, and we shall have {26x+15y=48 Now, let these two equations be added together, and the sun will be 31x=93, |