Elementary Algebra |
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Page 95
... REDUCTION OF FRACTIONS . 120. To prove that By b by ma mb a ma = b mb ' where a , b , m are positive integers . we mean a equal parts , b of which make up the unit ... ( 1 ) ; But ma mb b parts in ( 1 ) = mb parts in ( 2 ) ; ... 1 part ...
... REDUCTION OF FRACTIONS . 120. To prove that By b by ma mb a ma = b mb ' where a , b , m are positive integers . we mean a equal parts , b of which make up the unit ... ( 1 ) ; But ma mb b parts in ( 1 ) = mb parts in ( 2 ) ; ... 1 part ...
Page 96
... Reduce to lowest terms 24a3c2x2 24a3c2x2 18a3x2-12a2x3 * 24a3c2x2 18a3x - 12a x3 6a2x2 ( 3a - 2x ) 4ac2 = 3a - 2x Example 2. Reduce to lowest terms 6x2-8xy 9xy - 12y2 ' 6x2-8xy 2x ( 3x - 4y ) 2x 9xy - 12y23y ( 3x - 4y ) — By NOTE . The ...
... Reduce to lowest terms 24a3c2x2 24a3c2x2 18a3x2-12a2x3 * 24a3c2x2 18a3x - 12a x3 6a2x2 ( 3a - 2x ) 4ac2 = 3a - 2x Example 2. Reduce to lowest terms 6x2-8xy 9xy - 12y2 ' 6x2-8xy 2x ( 3x - 4y ) 2x 9xy - 12y23y ( 3x - 4y ) — By NOTE . The ...
Page 97
... reduced to its lowest terms by dividing both numerator and denominator by the highest common factor , which may be found by the rules given in Chap . XII . Example . Reduce to lowest terms 3x3 - 13x2 + 23x - 21 15x3- 38x2 - 2x + 21 ...
... reduced to its lowest terms by dividing both numerator and denominator by the highest common factor , which may be found by the rules given in Chap . XII . Example . Reduce to lowest terms 3x3 - 13x2 + 23x - 21 15x3- 38x2 - 2x + 21 ...
Page 98
... Reduce to lowest terms : a3 - a2b - ab2 - 2b3 a3 + 3a2b + 3ab2 + 2b3 * a3 + 2a2-13a + 10 x3 - 5x2 + 7x - 3 2 . x3 - 3x + 2 2x3 + 5x2y – 30xy2 + 27y3 3 . 4 . a3 + a2-10a +8 4x3 + 5xy2 - 21y3 4a3 + 12a2b- ab2 – 15b3 1 + 2x2 + x + 2x4 5 ...
... Reduce to lowest terms : a3 - a2b - ab2 - 2b3 a3 + 3a2b + 3ab2 + 2b3 * a3 + 2a2-13a + 10 x3 - 5x2 + 7x - 3 2 . x3 - 3x + 2 2x3 + 5x2y – 30xy2 + 27y3 3 . 4 . a3 + a2-10a +8 4x3 + 5xy2 - 21y3 4a3 + 12a2b- ab2 – 15b3 1 + 2x2 + x + 2x4 5 ...
Page 102
... take the lowest common denominator , which is the lowest common multiple of the denominators of the given fractions . RULE I. To reduce fractions to their lowest common deno- 102 FRACTIONS . Addition and Subtraction 77 86.
... take the lowest common denominator , which is the lowest common multiple of the denominators of the given fractions . RULE I. To reduce fractions to their lowest common deno- 102 FRACTIONS . Addition and Subtraction 77 86.
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Common terms and phrases
a+3b a+b+c a₁ a²+b² arithmetic means arithmetical arranged ascending powers b₁ beginner Binomial Theorem cents CHAPTER coefficients column compound expression continued fraction convergent cube root decimal denote digits dimes Divide dividend division divisor equal EXAMPLES XI Find the highest find the number Find the square Find the sum find the value following expressions given expressions greater harmonic mean Hence highest common factor integer less letters logarithm lowest common multiple method miles an hour Multiply number of terms numerator and denominator obtain partial fractions prefixed prove quadratic quadratic equation quotient ratio remainder Resolve into factors result rule of signs second term Simplify SIMULTANEOUS EQUATIONS solution square root subtraction Suppose surds symbols Transposing unknown quantity walk whence write yards zero
Popular passages
Page 331 - The logarithm of a product is equal to the sum of the logarithms of its factors.
Page 256 - In a quadratic equation wJiere the coefficient of the first term is unity, (i) the sum of the roots is equal to the coefficient of x with its sign changed ; (ii) the product of the roots is equal to the third term.
Page 168 - Thus the 4th root (2x2) = the square root of the square root ; the sixth root (3x2) = the cube root of the square root, or the square root of the cube root.
Page 178 - A basket of oranges is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder and six more. How many did the basket contain at first ? 17.
Page 179 - Two vessels contain mixtures of wine and water ; in one there is three times as much wine as water, in the other five times as much water as wine. Find how much must be drawn off from each to fill a third vessel which holds seven gallons, in order that its contents may be half wine and half water.
Page 280 - The pressure of wind on a plane surface varies jointly as the area of the surface, and the square of the wind's velocity. The pressure on a square foot is 1...
Page 213 - Art. 167 we saw that if the number of unknown quantities is greater than the number of independent equations, there will be an unlimited number of solutions, and the equations will be indeterminate. By introducing conditions, however, we can limit the number of solutions. When positive integral values of the unknown quantities are required, the equations are called simple indeterminate equations. The introduction of this restriction enables us to express the solutions in a very simple form. Ex. 1....