92 THE APPLICATION OF ALGEBRA 
Let x = the number of clays in which B overtake.-, 
A: then the miles travelled by B, in that time, will be 
20fc; and those travelled by A, 28 f 26 f 24 4- 22, 
&c. continued to .r terms'; where the last term (by 
Section 10, Theorem 3) will be equal to 28—2 x x— 1, 
or 30 — 2x; and therefore the sum of the whole pro 
gression eq.ual to 28 t- 30 — 2x x \x y or 2g.r— x 2 (by 
Theorem 4). Hence we have 20.r — 2gx —x 2 ; whence 
20 = 2g— x, and x = g: therefore 20 x g = 180 
is the distance which was to be found. 
PROBLEM XXXir» 
To find three numbers, so that 4 the first, of the se 
cond, and y of the third, shall be equal to 62; of the 
first, y of the second, and \ of the third, equal to 47; 
and y of thefrst, Y of the second, and ~ oj the third, equal 
to 38. 
Put a — 62, b =47, and r = 38, and let the num 
bers sought be denoted by x, y, and z; then the condi 
tions of the problem, expressed in algebraic terms, will 
stand thus, 
Which, cleared of fractions, become 
6x 4- 4y 4- 3~ = 12tf, 
SOX 4- 15y 4- 12~ = 605, 
15X f 12 y 4- 103 = 60C, 
And, by subtracting the second of these equations 
from the quadruple of the first, (in order to exterminate: 
z) we have 4X 4- y = 48a — 60b; moreover by taking 
3 times the third from 10 times the first, we have 15x l 
4?/ = 120a— 180c; this subtracted from 4 times the 
last,' leaves x = 72a— 240b f isoc = 24; whence 
V (48a 
60b 
4x). — 60, and z ( 
12a — 6.r 
4 // 
= 120. 
3