Page:Elementary algebra (1896).djvu/76

58 Rh Ex. 2. Solve 5x—(4x — 7)(3x — 5)= 6 — 3(4x^2 — 9)(x— 1).

Simplifying, we have

5x —(12x^2— 41x +35)= 6 —3(4x^2— 13x+9),

and by removing brackets, 5x—12x^2 + 41x — 35= 6 — 12x^2 + 39x — 27.

Erase the term — 12x^2 on each side and transpose;

thus 5x+ 41x — 39x=6 — 27 + 35;

collecting terms, 7x= 14 x = 2.

Nore. Since the — sign before a bracket affects every term within it, in the first line of work of Ex. 2, we do not remove the brackets until we have formed the products.

Ex. 3. Solve 7x — 5[x — {7 — 6(x — 3)}]=3x+1. Removing brackets, we have

7x — 5[x —{7-6x+18]=3x+1,

7x —5[x —25+6x]=3x+1,

7x—5x+ 125 —30x=3x+1;

transposing, 7x —5x—30x—3x%=1—125; collecting terms, —31x=— 124; x=4.

81. It is extremely useful for the beginner to acquire the habit of occasionally verifying, that is, proving the truth of his results. Proofs of this kind are interesting and convincing; and the habit of applying such tests tends to make the student self-reliant and confident in his own accuracy.

In the case of simple equations we have only to show that when we substitute the value of a in the two sides of the equation we obtain the same result.

Ex. To show that x = 2 satisfies the equation

52\x—(4x—7)(3x—5)=6—3(4x—9)(x—1). Ex. 2, Art. 80.

When x=2, the left side 5x—(4x—7)(8x—5) =10—(8—7) (6—5) =10-1=9.

The right side 6 — 3(4x — 9)(x— 1)=6 — 3(8 — 9) (2 — 1) =6—3(—1)=9.

Thus, since these two results are the same, x=2 satisfies the equation.