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139. Solution of Equations.-An equation of the first degree containing but one unknown quantity may be solved by transforming it in such a manner that the unknown quantity shall stand alone, constituting one member of an equation; the other member will then denote the value of the unknown quantity. Let it be required to find the value of x in the equation

4X-2 5x 3x


5 8 4 Clearing of fractions, we have

32.0-16+25x=30x+200. By transposition we obtain

32x+25x-30x=200+16. Uniting similar terms, 27x=216. Dividing each member by 27, according to Art. 134, we have

x=8. To verify this value of x, substitute it for æ in the original equation, and we shall have

32-2 40 24


6+5=6+5; that is,

11=11, an identical equation, which proves that we have found the correct value of x.

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140. Hence we deduce the following

RULE. 1. Clear the equation of fractions, and perform all the operations indicated.

2. Transpose all the terms containing the unknown quantity to one side, and all the remaining terms to the other side of the equation, and reduce each member to its most simple form.

3. Divide each member by the coefficient of the unknown quan. tity.

There are various artifices which may sometimes be em

ployed, by which the labor of solving an equation may be considerably abridged. These artifices can not always be reduced to general rules. If, however, any reductions can be made before clearing of fractions, it is generally best to make them; and if the equation contains several denominators, it is often best to multiply by the simpler denominators first, and then to effect any reductions which may be possible before getting rid of the remaining denominators. Sometimes considerable labor may be saved by simply indicating a multiplication during the first steps of the reduction, as we can thus more readily detect the presence of common factors (if there are any), which may

be canceled. The discovery of these artifices will prove one of the most useful exercises to the pupil.


EXAMPLES. 1. Solve the equation

8x+1 2x +9 3x

7 7 3 Clearing of fractions,

63x-24x-3=14x+63 +84. Transposing and reducing,

25x=150. Dividing by 25,

X=6. To verify this result, put 6 in the place of w in the original equation.

Solve the following equations: 2. 3ax-4ab=2ax-6ac.

Ans. x=4b-6c. 3. 32 – 10x=8x+x.

Ans. x=9. a(da +222) 4.

284-3 5. +6x=

Ans. = 9. 4


ab-1 6. “=bc+d+

bc+d 23+6 11x-37 7. 3x+ =5+

Ans. Q=7.

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Ans. x=

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Ans. x=5.

8. 5ax-26+4bx=2x+5c.

3x-5 2.r-4 9. x+


3 30-11 50-5 97-700 10. 21+

+ 16 8

2 5х

5x+14 1 11. 3x4

3 12

atx b-X 12. 3x-a+cx=

3 3x

2x 13. (L




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Ans. x=


4a2-36 8a+3ac-3

abcd 3bd + ad-4abd - 2ab

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Ans. x=

c+7=4x+ 14. (a+2)(6+X)—a(b+c)=7+?.

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16. X



17–33 4x+2


3r-3 20x 6x-8 4x-4


2 7 5

21 4x-113
6x+7 7x-13 2x+4

9 6+3 3
4 2 3

70ab3ac =

Ans. x= 4


8α Ans. x=


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2 24.

35 6x – 101


Ans. x=11.


9x+4 4x-19 5x+32 11x+13 25.

Ans. x=100. 5X — 48 51 17

51 4 7

37 26. +

Ans. x=1. x+2' x+3x2 + 5x+6*

1 1 1 1 27.

Ans. x=5. 2 -8

a(b-a) 28.

Ans. x=

(6+a) 3

Ans. x=12.

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ato-abra 29. (x+3)(x-3)+=(x+5)(x-—3).



25- X 16x+4}

x+1 3x+2

23 X

Ans. x=33.

2+1 +5.

Solution of Problems. 141. A problem in Algebra is a question proposed requiring us to determine the value of one or more unknown quantities from given conditions.

142. The solution of a problem is the process of finding the value of the unknown quantity or quantities that will satisfy the given conditions.

143. The solution of a problem consists of two parts:

1st. The statement, which consists in expressing the conditions of the problem algebraically; that is, in translating the conditions of the problem from common into algebraic language, or forming the equation.

2d. The solution of the equation..

The second operation has already been explained, but the first is often more embarrassing to beginners than the second. Sometimes the conditions of a problem are expressed in a distinct and formal manner, and sometimes they are only implied, or are left to be inferred from other conditions. The former are called explicit conditions, and the latter implicit conditions.

144. It is impossible to give a general rule which will enable

us to translate every problem into algebraic language, since the conditions of a problem may be varied indefinitely. The following directions may be found of some service:

Represent one of the unknown quantities by some letter or symbol, and then from the given conditions find an expression for each of the other unknown quantities, if any, involved in the problem.

Express in algebraic language the relations which subsist between the unknown quantities and the given quantities; or, by means of the algebraic signs, indicate the operations necessary to verify the value of the unknown quantity, if it was already known.

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Prob. 1. What number is that, to the double of which if 16 be added, the sum is equal to four times the required number?

Let x represent the number required.
The double of this will be 2.x.
This increased by 16 should equal 4.x.
Hence, by the conditions, 2x+16=4x.

The problem is now translated into algebraic language, and it only remains to solve the equation in the usual way. Transposing, we obtain

16=40-2x=20, and


x=8. To verify this number, we have but to double 8, and add 16 to the result; the sum is 32, which is equal to four times 8, according to the conditions of the problem.

Prob. 2. What number is that, the double of which exceeds its half by 6?

Let x=the number required.
Then, by the conditions,


2xClearing of fractions, 4x--x=12,

3x=12. Fence

x=4. To verify this result, double 4, which makes 8, and diminish

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