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(2.) If equal quantities be taken from equal quantities, the remainders will be equal.

(3.) If equal quantities be multiplied by the same or equal quantities, the products will be equal.

(4.) If equal quantities be divided by the same or equal quantities, the quotients will be equal.

(5.) If the same quantity be added to and subtracted from another, the value of the latter will not be altered. (6.) If a quantity be both multiplied and divided by the same quantity, its value will not be altered.

(7.) The same powers and roots of equal quantities are equal.

From the above axioms the following rules for the resolution of equations can be derived :—

63. RULE 1. Any quantity can be taken from one side of an equation to the other, by changing its sign.

This rule is derived from axioms 1st and 2d, as will appear evident from the following example:

Let 3x-4=2x+6; if 4 be added to both sides the equality will still exist by axiom first, but the equation will become 3x=2x+6+4; where the 4 has been taken to the other side and its sign changed; so that taking a minus quantity from one side to the other, and changing its sign, is equivalent to adding that quantity to both sides; if now 2x be taken from both sides, the equation will become 3x-2x-10, where by taking 2x from both sides of the equation, it has disappeared from the second side, but has reappeared on the first; hence, taking a plus quantity from one side, and placing it on the other, with its sign changed, is equivalent to subtracting equals from equals, and it has just been shown, that taking a minus quantity from one side to the other, and making it plus, is equivalent to adding equals to equals. The solution of the above equation will now stand as under:

3x-4=2x+6, the given equation.

3x=2x+10, by transposing-4 and adding 4 and 6. 3x-2x-10, by transposing 2x.

x=10, by performing the subtraction on the first side. 64. RULE 2. If, after all the unknown quantities are transposed to the first side, and the known ones to the second, the unknown quantity have a coefficient, it may be taken away by dividing both sides of the equation by it. This rule is evidently the same as axiom 4th.

EXAMPLE. Given 4x+27=48-3x, to find the value of x.

4x+27=48-3x, given equation.

4x+3x=48-27, by transposing -3x and 27.
7x=21, by collecting the terms.

..x=3, by applying the rule.

65. RULE 3. If there are fractions in any of the terms, they may be taken away by multiplying all the terms by each of the denominators in succession; or by multiplying all the terms at once, by the least common multiple of all the denominators.

It is evident that this rule is merely an application of axiom 3d, and points out when that axiom may be applied.

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EXAMPLE. +++4=2x, to find the value of x.

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x

x++ï+4=2x, the given equation.

4x+2x+x+16-8x, by multiplying both sides by 4. 7x+16=8x, by collecting the like terms.

.. 16, by subtracting 7x from both sides. 66. RULE 4. If the value of any root of the unknown quantity can be found from the equation, raise both sides to the power denoted by the root, and the value of the unknown quantity will be found. This is evident from axiom 7th.

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EXAMPLE. Given x2+x=+4; to find the value of x,

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x+x+4, the given equation.

2x+x=x+8, by multiplying by 2. Rule 3d.

2x=8, by taking x from both sides.

x=4, by dividing by 2. Rule 2d. ..x=16, by the Rule.

67. RULE 5. If, after the equation has been reduced to its simplest form by the preceding rules, the value of some power of the unknown quantity is obtained, its value may be found by extracting the corresponding root of both sides. This is also evident from axiom 7th.

x2+3x
3

EXAMPLE. Given =x+12; to find the value of x.

x2+3x =x+12, the given equation.

3

x2+3x=3x+36, by multiplying by 3. Rule 3d.
x2-36, by subtracting 3x from both sides.

..x=6, by extracting the square root.

The previous rules will be found sufficient for the solution of equations containing only one unknown quantity; the following solutions are added as examples of their application.

MISCELLANEOUS EXAMPLES.

1. Given 5x+34=4x+36; to find the value of x. By transposition, 5x-4x-36-34. Rule 1st. .. by collecting the terms, x=2.

2. Given 4ax-5b3dx+2c; to find the value of x. By transposition, 4ax-3dx=2c+56. Rule 1st. By collecting, (4a—3d)x=2c+5b.

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Rule 2d.

3. Given +12—19; to find the value of x.

3

3

By multiplying all the terms by 6, the least common multiple of the denominators, it becomes 2x-10+3x=72 -2x+20. Rule 3d.

And by transposition, 2x+3x+2x=72+20+10. Rule 1st.

Hence, by collecting the terms, it becomes 7x=102. .. by division, by Rule 2d,

x=144. 68. SCHOLIUM 1. If the relation between x and the known quantities be not given in the form of an equation, but of a proportion, it may be changed into the form of an equation, by making the first term divided by the second. the third divided by the fourth,-see definitions ;—or by making the product of the extremes that of the means.

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For when a : b::c: d, by definition,=, and multiplying both sides by bd, it becomes ad bc; or the product of the extremes is equal to that of the means.

5x+4 18-x

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2

4

4. Given
By multiplying extremes and means, 10x+8=

:: 7:4; to find the value of x.

Multiplying by 4 it becomes 40x+32=126—7x. .. by transposition, 47x=94. Rule 1st.

And hence,

x=2. Rule 2d.

126-7x

4

69. SCHOLIUM 2. When an equation appears under the form of the equality of two fractions, it may frequently be solved with much elegance, by making the sum of the numerator and denominator on the first side, divided by their difference, equal to the sum of the numerator and denominator on the second, divided by their difference. If only one side be a fraction, it may be reduced to the above form by writing 1 for a denominator on the integral side. The above principle may be demonstrated thus:

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... (3.) And by dividing (2) by

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11. Given *++*+2=14+5—; to findx. Ans.x=13.

3

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Ans. x=

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14. Given + =12

15. Given +=c; to find x.

16. Given ax+b2=bx+a2; to find x.

17. Given bx+2x−a=3x+2c; to find x.

ab ac-1

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PROBLEMS IN SIMPLE EQUATIONS.

EXAMPLE 1. What number is that to which, if its half and its fifth part be added, the sum will be 34?

Let x represent the number sought; then its half will be

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and multiplying by 10; 10x+5x+2x=340, Rule 3d;

hence by collecting the terms,

17x=340,

... x= 20, by Rule 2d. 2. What number is that whose third part exceeds its seventh part by 4?

Let x represent the number sought; then its third part

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