12. Given a2+x2=4/b1+x, to find the value of x. f 13. Given Va+x+√a−x=√/ax, to find the value 15. Given a+x=√a2+x√(b2+x2), to find the value of x. b2 Ans. x a 4α 16. Given x2+3a2 — }√√x2 ~3a2=x/a, to find the value of x. 9a3 Ans. x=/ 4-4a 17. Given a++x=b, to find the value of x. 18. Given +x+/a-x=b, to find the value of x 19. Given ✔a+√x=✔ax, to find the value of x. 21. Given ✔a+ax—a—√/a3 —ax, to find the value of x. 22. Given a—x2 +x√/a2 —1—a2/1-x2, to find the value of x. Ans. x= a2 a2+3, Given ✓x+α=c-x+b, to find the value of x. Ans. x= (c2+6 -α) 2 .b Of the resolution of simple equations, containing two When there are two unknown quantities, and two independent simple equations involving them, they may be reduced to one, by any of the three following rules: RULE 1. Observe which of the unknown quantities is the least involved, and find its value in each of the equations, by the methods already explained; then let the two values, thus found, be put equal to each other, and there will arise a new equation with only one unknown quantity in it, the value of which may be found as before. (b) (b) This rule depends upon the well known axiom, that things which are equal to the same thing, are equal to each other; and the two following methods are founded on principles which are equally simple and obvious. Or 115-15y=20+4y, or 19y-15-20-95, 23-3y 10+2y = 2 5 And from the second, Here, from the first equation, x=α- -y, +b+y, 2y Here, from the first equation, x=14 3' 3y And from the second, x=24 Therefore, by equality, 14- 23-24 Or, by multiplication 84-4y=144-9y; Find the value of either of the unknown quantities in that equation in which it is the least involved; then substitute this value in the place of its equal in the other equation, and there will arise a new equation with only one unknown quantity in it; the value of which may be found as before. y. and From the first equation, x=17-2y; which value, being substituted for x, in the second, 2. Given {x+y=13} x=3 to find the values of x and y. From the first equation, x=13-y; which value being substituted for x, in the second, Gives 13-y-y=3, or 2y=13-3=10, 10 Whence y= =5, and x=13-y=8. Here the analogy in the first, turned into an equa And this value, substituted for x in the second, gives (12)2+y2=c, or a2 y2 62 Whence we have a2y2+b2 y2=b2c, or y2= a2+ 62 RULE III. Let one or both of the given equations be multiplied, or divided, by such numbers, or quantities, as will make the term that contains one of the unknown quantities the same in each of them; then, by adding, or subtracting, the two equations thus obtained, as the case may require, there will arise a new equation, with only one unknown quantity in it, which may be resolved as before. EXAMPLES. 1. Given 3x+5=14 to find the values of and y. First, multiply the second equation by 3, and it will give 3x+6y=42. Then, subtract the first equation from this, and it will give 6y5y=42 - 40, or y=2. Whence, also, x=14-2y=14-4=10. |
