3y-x 5 Ans. x=9, and y=4. 7x+6 4y-9 11 3 and 3x+42y—3:: 5 : 3, to find the values of x and y. 3x 13. 2 Ans. x=7, and y=9. Ans. x=7, and y=9. to find the values to find the values of x and y. Ex. 20. Given 3x-2y=15, and y+10x-15: 7:3, of x and y. Ans. x 45, and y=60. Ex. 21. Given x+150: y-50: : 3:2, and x the values of x and y. Ans. x=300, and y=350. Ex. 22. Given (x+5).(y+7)=(x+1)(y−9)+ 112, and 2x+10=3y+1, to find the values of x and y. Involving three or more unknown Quantities, 252. When there are three independent simple equations involving three unknown quantities. RULE. From two of the equations, find a third, which involves only two of the unknown quantities, by any of the rules in the preceding Section; and in like manner from the preceding equation, and one of the other, another equation which contains the same two unknown quantities may be deduced, Having therefore two equations, which involve only two unknown quantities, these may be determined; and, by substituting their values in any of the original equations, that of the third quantity will be obtained. 253. If there be four unknown quantities, their values may be found from four independent equations. For from the four given equations, by the rules in the last Section, three may be deduced which involve only three unknown quantities, the values of which may be found by the last Article ; and hence the fourth may be found by substituting in any of the four given equations, the values of the three quantities determined. If there be n unknown quantities, and n independent equations, the values of those quantities may be found in a similar manner. For from the n given equations, n-1 may be deduced, involving only n-1 unknown quantities; and from these n-1, n-2 may be obtained, involving only n-2 unknown quantities; and so on, till only one equation remains, involving one unknown quantity; which being found, the values of all the rest may be determined by substitution. y+2z=33. (A). Multiplying the third equation by 12, the least common multiple of 2, 3, and 4, 6x+4y+3x=120 multiplying the 1st equ" by 6, 6x+6y+6z=174; .. by subtraction, 2y+3z=54; but, multiplying equation (A) by 2, 2y+4z=66; .. by subtraction, z=12. From equation (A), by transposition, y=33-2z; .. by substitution, x=29-9-12, and x=29-21, .'. x=8. In like manner, had the first equation been multiplied by 2, and subtracted from the second, an equation would have resulted, involving only x and z; and had it been multiplied by 4, and subtracted from the third when cleared of fractions, another equation would have been obtained, involving also x and z; whence by the preceding rules, the values of x and 2 could be found, and consequently the value of y also, by substitution. Or if the first equation be multiplied by 3, and the second subtracted from it, an equation would arise, involving only x and y ; and if the first when multiplied by 3, be subtracted from the third when cleared of fractions, another would arise involving only x and y; whence the values of x and y might be determined. And hence the third, that of z might be found. SECOND METHOD. From the first equation, x=29-y-z; substituting this value of x in the second equation, 29-y-z+2y+3z=62; .. by transposition, y=33-2z. Also substituting, in the third equation, the value of x found from the first, |