The above solutions have given x= bc'-b'c ́a'b-ab' ab'-a'b' ac'-a'c a'c-ac' also y=ab'a'b and a'b—ab'; these are not two values, but the same value in two different forms. SOLUTION OF PROBLEMS PRODUCING EQUATIONS CON TAINING TWO OR MORE UNKNOWN QUANTITIES. 1. Let x the first, and y= the second; x+2y=21, 2a+1=14,} by the question. 2x+4y=42, 3y=28, y= 8, value of the second. the first, and y= the second; by the question. then (1)×2 (2)x3 3 (9) X6 10 X =6, value of the first. 3. Let x=the first and y= the second; then 1 x+5=2y, 2 Зах =4+3,} by the question. From (1) we have x-2y-5, and from (2) we have x=y+1; .2y—5—4y+1, by Rule 2, and 6y-15-4y+3, by mult. by 3, 2y=18, by transp. .. y=9, value of y. x=4y+1=13, value of x. From the first, (y—2)=2x+2, and from the second, 5. Let the left hand digit be represented by x, and the right hand digit by y, then the number will be represented by 10x+y; hence 4 6. Let the notation be the same as in last example; then Transposing (1) 9x-9y=-27. (3) X9 5 90x-45y=0. Hence the number sought is 36. NOTE. The above equation may also be solved with less labour, by finding a value of y in terms of x, from the first equation, and substituting this value instead of it in the second, (Rule 3.) Thus, from the first y=2x, substituting this value of y, in the second, gives, after transposition, 9x=27, therefore x=3, consequently y=6, and the number sought is .. 36. 7. Let the digits be represented by x, y, and z, respectively; then 1 x+y+2=7, =100%+10y+x, 4 99x-99z=297, 6 Transp. and dividing (6) 7 (11)÷198 2y=14-5y. 99x+992-495, substi- 9 198x=792. 10 x=4. 11 1982-198. Hence the number sought is 421. 8. Let the first be x, the second y, and the third z; then 9. Since the digits are in arithmetical progression, if the middle digit be represented by y, and the common difference by x, the others can be represented by y+x and y—x; and since by subtracting from the number, the digits were inverted, the greatest digit must be in the place of hundreds, and the least in the unit's place; therefore we have the following equations:— Therefore the first figure being (y+x)=4, and the third being (y-x)=2, and the number sought is 432. 10. Let x = the time in which A could finish the whole, y= the time in which B could do it, and z the time in which C could do it. In order to obtain a general solution of all questions similar to this, put a for 8 days, b for 9 days, and c for 10 days; then since A could do it in x days, he could do in a day, and since A and B together could do it in a days, they could do in a day, and so of the others; hence |