Page:Elementary algebra (1896).djvu/75

57 Rh TRANSPOSITION OF TERMS.

78. To solve 3x—8=x+12.

Here the unknown quantity occurs on both sides of the equation. We can, however, transpose any term from one side to the other by simply changing its siyn. This we proceed to show.

Subtract 2 from both sides of the equation, and we get

3x—x—8=12. . . (Axiom 2).

Adding 8 to both sides, we have 8x—x=12+8. . (Axiom 1).

Thus we see that + x has been removed from one side, and appears as —x on the other; and —8 has been removed from one side and appears as + 8 on the other.

It is evident that similar steps may be employed in all cases. Hence we may enunciate the following rule:

Rule. Any term may be transposed from one side of the equation to the other by changing its sign.

79. We may change the sign of every term in an equation; for this is equivalent to multiplying both sides by —1, which does not destroy the equality (Axiom 3).

Ex. Take the equation —3x—12=x- 24.

Multiplying both sides by — 1, 3x+ 12=— x + 24

which is the original equation with the sign of every term changed.

80. We can now give a general rule for solving a simple equation with one unknown quantity.

Rule. Transpose all the terms containing the unknown quantity to one side of the equation, and the known quantities to the other. Collect the terms on each side; divide both sides by the coefficient of the unknown quantity, and the value required is obtained.

Ex. 1. Solve 5(x —3)—7(6— x) +3 = 24 — 3(8—x).

Removing brackets, 5x— 15 — 42+ 7x+4+3 = 24-24+3x;

transposing, 5x+7x—3x=24- 24+15+ 42-3;

collecting terms, 9x = 54. x = 6.