2. Given (5x-3y= 92 Let the first equation be multiplied by 2, and the second by 5, And we shall have 10x- 6y=18 10x+25y=80; And if the former of these be subtracted from the latter, Multiply the first equation by 5, and the second by 3, 6x+15y=48; Now, let these equations be added together, And it will give 31x=93, or x=}}=3, 4. Given ax+by=c, and dx+ey=f; to find x and y. To exterminate three unknown quantities, or to reduce the three simple equations, containing them, to one. RULE. 1. Let x, y, and z, be the three unknown quantities to be exterminated. 2. Find the value of x from each of the three given equations. 3. Compare the first value of x with the second, and an equation will arise involving only y and z. 4. In like manner, compare the first value of x with the third, and another equation will arise involving only y and z. 5. Find the values of y and z from these two equations, according to the former rules, and x, y, and z will be exterminated as required. NOTE. Any number of unknown quantities may be exterminated in nearly the same manner, but there are often much shorter methods for performing the operation, which will be best learnt from practice. But, from the first of these equations, y=62-29-22 33-22, First, the given equations, cleared of fractions, become 12x+ 8y+ 6z=1488 20x+15y+12z=2820 30x+24y+20z=4560 Then, if the second of these equations be subtracted from double the first, and three times the third from five times the second, we shall have 4x+y=156 10x+3y=420 And again, if the second of these be subtracted from three times the first, it will give 12x-10x=468-420, or x=48 = 24 ; 3. Given x+y+z=31, x+y=2-25, and y—2—9 6 to find x, y, and z. Ans. x 20, y=8, and z=3. 4. Given x+y=a,x+z=b, and y+z=c; to find x, y, and z. 5. Given VOL. I. ax+by+cz=m PP and Z. A COLLECTION OF QUESTIONS, PRODUCING SIMPLE EQUATIONS. 1. To find two such numbers, that their sum shall be 40, and their difference 16. Let x denote the less of the two numbers required, And x+x+16=40 by the question; That is, 2x40-16=24, Or x=3=12= less number, And x+16=12+16=28= greater number required. 2. What number is that, whose part exceeds its part by 16? Let x equal number required, I Then will its part be, and its part —; Whence x=192 the number required. 3. Divide 1000l. between A, B, and C, so that A shall have 721. more than B, and C 1001. more than A. Let x=B's share of the given sum, Then will x+72=A's share, And x+172 C's share, And the sum of all their shares x+x+72+x+172, That is, 3x=1000-244-756, Or x=26-2521. B's share, And 472-252+72=3241. = A's share, And +172-252+172=4241. C's share. 2521. 3241. 4241. 1000l. the proof. 4. A prize of 1000l. is to be divided between two persons, whose shares therein are in the proportion of 7 to 9; required the share of each. Let x equal first person's share, Then will 1000-x equal second person's share, Or 16x=7000, 7000 Whence x= =4371 10s. = first share, 16 And 1000-x=1000-4371, 10s, 5621. 10s. second share. 5. The paving of a square at 2s. a yard cost as much as the inclosing of it at 5s. a yard; required the side of the square. ed. Let x equal side of the square sought, Then 4x yards of inclosure, And yards of pavement; Whence 4xx5=20x equal price of inclosing, And 22x2 equal price of paving. 20x by the question, Therefore, x=10x, and x=10= length of the side requir 6. A labourer engaged to serve for 40 days upon these conditions, that for every day he worked he should receive 20d. but for every day he played, or was absent, he was to forfeit 8d.; now at the end of the time he had to receive 11, 11s. 8d. The question is to find how many days he worke ed, and how many he was idle, Let be the number of days he worked, Then will 40-x be the number of days he was idle; Also x 20=20x= sum earned, And 40-xx8-320— 8x= sum forfeited, Whence 20x-320-8x=380d. (=11. 11s. 8d.) by the question, that is, 20x-320+8x=380, Or 28x=380+320==700, And x=700=25= number of days he worked, |