2. Given 5x-3y=9 to find the values of x {2x+5y=165 and y Multiply the first equation by 2, and the second by 5; then 10x-6y=18, and 10x+25y=80. And if the former of these be subtracted from the lat ter there will arise 31y=62, or y = 62 31 2. Whence, by the first equation, x=' 9+3y_15 5 =3. EXAMPLES FOR PRACTICE. 1. Given 4x+y=34, and 4y+x=16, to find the values of x and y. Ans. x=8, y=2 2. Given 2x+3y=16, and 3x-2y=11, to find the values of x and y y. 2x 3y 9 3. Given + 5 4 20 values of x and y. 4. Given+7y=99, and +7x=51, to find the val 8. Given ax+by=c, and dx+ey=f, to find the values of x and y. Ans. x= ce-bf ae- -bd' y= af-de e-bd 9. Given x+y=a, and x2 —y2=b, to find the values of x and y. Ans. x= a2+b a2 - b 2a • 10. Given x2+xy=a, and y2+xy=b, to find the values of x and y. α b Ans. x= y= " Va+b va+b Of the resolution of simple equations, containing three or more unknown quantities. When there are three unknown quantities, and three independent simple equations containing them, they may be reduced to one, by the following method. RULE. Find the values of one of the unknown quantities, in each of the three given equations, as if all the rest were known; then put the first of these values equal to the second, and either the first or second equal to the third, and there will arise two new equations with only two unknown quantities in them, the values of which may be found as in the former case; and thence the value of the third. Or, multiply each of the equations by such numbers, or quantities, as will make one of their terms the same in them all; then, having subtracted any two of these resulting equations from the third, or added them together, as the case may require, there will remain only two equations, which may be resolved by the former rules. And in nearly the same way may four, five, &c. unknown quantities bé exterminated from the same num ber of independent simple equations; but, in cases of this kind, there are frequently shorter and more commodious methods of operation, which can only be learnt from practice. 1. Given EXAMPLES. x+y+z29) x+2y+32=62 to find x, y, and z. }x+y+12=10 Here, from the first equation, x=29-y-z. From the second, x=62-2y-3z, 3 Therefore 33-2x=27—2, or z=12, Whence, also, y=33-2z=9 2x+4y-3z=22) 2. Given 4x-2y+5x=18 to find x, y, and z. 6x+7y-z 63 Here, multiplying the first equation by 6, the second by 3, and the third by 2, we shall have 12x+24y-18z=132, 12x-6y+152=54, And, subtracting the second of these equations successively from the first and third, there will arise 30y-33z=78, Or, by dividing the first of these two equations by 3, and then multiplying the result by 2, 20y-22z=52, Whence, by subtracting the former of these from the latter, we have 5z=20, or z=4. And, consequently, by substitution and reduction, y=7 and x=3. 3. Given x+y+z=53, x+2y+32=105, and x+3y +4z=134, to find the values of x, y, and z. + Ans. x=24, y=6, and z=23 4. Given x+y +2=32, 1 + z=12, to find the values of x, y, and z. 6 Ans. x=12, y=20, z=30. 5. Given 7x+5y+2z79, 8x+7y+9z = 122, and x+4y+5x=55, to find the values of x, y, and z. Ans. x=4, y=9, z=3 6. Given x+y=a, x+z=b, and y+zc, to find the values of x, y, and z. MISCELLANEOUS QUESTIONS, PRODUCING SIMPLE EQUATIONS. The usual method of resolving algebraical questions, is first to denote the quantities, that are to be found, by x, y, or some of the other final letters of the alphabet; then, having properly examined the state of the question, perform with these letters, and the known quantities, by means of the common signs, the same operations and reasonings, that it would be necessary to make if the quantities were known, and it was required to verify them, and the conclusion will give the result sought Or, it is generally best, when it can be done, to denote only one of the unknown quantities by x or y, and then to determine the expression for the others, from the nature of the question; after which the same method of reasoning may be followed, as above. And, in some cases, the substituting for the sums and differences of quantities; or availing ourselves of any other mode, that a proper consideration of the question may suggest, will greatly facilitate the solution. 1. What number is that whose third part exceeds its fourth part by 16 ? Let x the number required, Hence x=192, the number required. 2. It is required to find two numbers such, that their sum shall be 40, and their difference 16. Let x denote the least of the two numbers required, Then will x+16= to the greater number, And x+x+16=40, by the question, That is 2x=40-16, or a= 24 2 =12= least number. And x+16=12+16=28= the greater number required. 3. Divide 1000l. between A, B, and c, so that a shall have 721. more than в, and c 1007. more than A. Let x=B's share of the given sum, And x+172 c's share. = Hence their sum is x+x+72+x+172, |