(44.) 4(2 — x) — 3 (3 — 2x) = 9.

(45.) 6(x − 3) — 9 (x — 4) — (x — 6).

2

(57.)(x-1)-(x) = 15

(58.) (4+ ) — (2x — }) = 1};

28

(60.) (x — 3) — ¦ (x−4) = ¦(x−1}). x = 2.

- 15

x = 8.

43

x=

30

31

2a b

(43.) 2x-3(2-x) = 4.

Ans. x 2.

(44.) 4(2-x)-3(3—2x) = 9.

(45.) 6(x-3)= 9 (x —4) — (x — 6).

x = 6.

3

(60.) ( x − 3) — }(x — 4) = (x−1}). z = 2.

-

(78.) x+a: xa:: a+2b: a-3b..

(80.) x+a:x+b:: (2x+a+c)2: (2x+b+c)2. x =

a2 b2
За

- ab

2c

SIMPLE EQUATIONS CONTAINING TWO
UNKNOWN QUANTITIES.

41. When one unknown quantity occurs in the expression for another unknown quantity, it is plain that the value of neither of them can be determined, unless some other condition be given.

In order to limit such questions, there must always be as many conditions or independent equations given as there are unknown quantities to be determined; and as these equations coexist, or are all true at the same time, they are called simultaneous equations.

To solve such equations it is necessary to eliminate one of the unknown quantities, that is, to derive from the given equations another equation in which one of the unknown quantities does not occur, and which shall therefore contain only one unknown quantity.

The best methods of elimination are the following:

First, If the coefficients of the quantity to be eliminated be the same in both equations, take the difference or sum of the equations, according as the coefficients have the same or contrary signs, and the quantity will be eliminated. Thus, Given 2x + 3y = 22

and 5x

3y = 13.

By Addition we get 7x which y is eliminated.

35; an equation from

If the coefficients of one of the quantities be not the same in both equations, multiply one or both equations by such numbers as will make them the same, and proceed as above,

Second Method.-Find an expression for one of the unknown quantities in both equations, and put the one expression equal to the other. Thus,