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Ex. 19. Given 5ax-26+4bx=23+5c to find the value of x.
5c+26 Ans. 23
32-11 5x–5 97–70. Ex. 20. Given 21+ 16 = 8 t- to find the value of x.
Ans. x=9. (99.) An equation may always be cleared of fractions by multiplying each member into all the denominators, according to Art. 97. But sometimes the same object may be attained by a less amount of multiplication.
Thus, in the preceding example, the equation may be cleared of fractions by multiplying each term by 16 instead of 16x8x2, and it is important to avoid all useless multiplication. In general, it is sufficient to multiply by the least common multiple of all the denominators.
A common multiple of two or more numbers is any number which they will divide without a remainder ; and the least common multiple is the least number which they will so divide. Thus 16 is the least common multiple of 16, 8, and 2.
2x-4. Ex. 21. Given x+- =12
to find the value of 2.
2.6-4 The minus sign prefixed to the fraction
QUEST.-How may an equation be cleared of fractions with the least amount of multiplication? What is a common multiple? What is the loast common maltiple of several numbers ?
shows that the value of the entire fraction is to be subtracted from 12; and since a quotient is subtracted by changing its sign, the equation, cleared of fractions, will stand
6x+9x—15=72—4x+8, which gives
5x+14 1 Ex. 22 Given 3x-2 the value of x.
Ans. x=7. 17—30 4c+2
Ex. 23. Given 5
find the value of x.
Ex. 24. Given **=15–482 to find the val
ue of x.
Ans. x=11. Ex. 26. Given -272+ = -7 to find the value of .
- Ans. x=20. Ex. 27. Given 2 + = to find the value of x.
QUEST.-What does a minus sign prefixed to a fraction show ?
(100.) PROBLEMS PRODUCING EQUATIONS OF THE FIRST DEGREE CONTAINING BUT ONE UNKNOWN QUANTITY.
It has already been observed, Art. 13, that the solution of a problem by algebra consists of two distinct parts :
1st. To express the conditions of the problem algebraically; that is, to form the equation.
2d. To solve the equation.
(101.) The second operation has been fully explain. ed; but the first is often more embarrassing to begin. ners than the second. Sometimes the statement of a problem furnishes the equation directly; and sometimes it is necessary to deduce from the statement new conditions, which are to be expressed algebraically.
It is impossible to give a general rule which will enable us to translate every problem into algebraio
QUEST.—The solution of an algebraic problem consists of what parts 1 How do we reduce a problem to an equation ?
language. The ability to do this with facility can only be acquired by reflection and practice.
(102.) In most cases, however, we may obtain the desired equation by applying the following
RULE. Denote one of the required quantities by x; and then indicate by means of the algebraic signs the same operations on the known and unknown quanti. ties as would be necessary to verify the value of the unknown quantity, if it were already known.
Problem 1. What number is that to the double of which, if 20 be added, the sum is equal to four times the required number?
Let x represent the number required.
The problem is now translated into algebraic language, and it only remains to solve the equation in the usual way. Transposing, we obtain
10= x, or,
x=10. To verify this number, we have but to double 10, and add 20 to the result; the sum is 40, which is equal to four times 10, according to the conditions of the problem.
Prob. 2. What number is that the double of which exceeds its half by 9?
QUEST.-Give the rule for forming the equation.