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39ab-14a2

bx

(3be+ad)x

(55) Given

Ans. x= 27ab-9b+12'

5ab (3bc-ad)x 5a(2b-a)

2b-a2ab(a+b)3c-d2ab(a—b)

a2-b2
5a(2b-a)
3c-d

(58) Given (3a-x)(a−b)+2ax=4b(a+x).

Ans. x=

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4a2-3ab

5a-b 7ab-3a2

Ans. x=

a-36

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(70) Given (m+p)(‡x—3r)=({m+2p)(x-7r).

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When an equation can never be verified, whatever value we put in the place of the unknown quantity, it is said to be impossible; and when an equation is always verified, whatever value be put for the unknown quantity, it is said to be indeterminate.

CASES OF IMPOSSIBILITY AND INDETERMINATION IN EQUATIONS OF THE FIRST DEGREE.

I. PROBLEM. To find a number such that the third of it, augmented by 75, and five twelfths of it, diminished by 35, shall make three quarters of it, added to 49.

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An absurdity. There is, therefore, no value of r which can satisfy the equation [1].

The impossibility may be rendered evident in the equation [1] itself by reducing the similar terms in the first member; thus,

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It is evident that the two members will always differ by 9, whatever be the value of x.

II. PROBLEM. To find a number such that, adding together the half of it increased by 10, two thirds of it increased by 20, and five sixths of it diminished by 34, the sum shall be equal to twice the excess of this number over 5.

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The unknown x is, therefore, altogether indeterminate; that is to say, it may be taken equal to 2 or 3, or any number whatever.

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