College AlgebraD.C. Heath & Company, 1890 - 577 páginas |
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Página 86
... equation contains but one unknown quantity , any value of the unknown quantity which satisfies the equation is called a Root of the equation . The solution of an equation containing but one unknown quantity is the process of finding its ...
... equation contains but one unknown quantity , any value of the unknown quantity which satisfies the equation is called a Root of the equation . The solution of an equation containing but one unknown quantity is the process of finding its ...
Página 89
... equation by the product of all the denominators . " SOLUTION OF SIMPLE EQUATIONS 184. 1. Solve the equation 7 x 6 - 5 3 = 3x 1 5 4 Clearing of fractions by multiplying each term of the equation by 60 , the L.C.M. of 6 , 3 , 5 , and 4 ...
... equation by the product of all the denominators . " SOLUTION OF SIMPLE EQUATIONS 184. 1. Solve the equation 7 x 6 - 5 3 = 3x 1 5 4 Clearing of fractions by multiplying each term of the equation by 60 , the L.C.M. of 6 , 3 , 5 , and 4 ...
Página 90
... equation ; then 812 35 5 3 12 - 3 4 ' in the given or , which is identical . We may now give a rule for the solution ... Solve the equation 3x - 1 4x - 5 7x + 5 = 4 + 4 5 10 Multiplying each term by 20 , the L.C.M. of 4 , 5 , and 10 ...
... equation ; then 812 35 5 3 12 - 3 4 ' in the given or , which is identical . We may now give a rule for the solution ... Solve the equation 3x - 1 4x - 5 7x + 5 = 4 + 4 5 10 Multiplying each term by 20 , the L.C.M. of 4 , 5 , and 10 ...
Página 91
Webster Wells. 3. Solve the equation 6x + 1 15 2x + 4 7x + 16 = 2x - 1 5 = 6x - 3 . Multiplying each term by 15 , 6x ... equation by the L.C.M. of the monomial denominators . 4. Solve the equation 20 x + 2b a2 + b2 = x - a x + a x2 - a2 ...
Webster Wells. 3. Solve the equation 6x + 1 15 2x + 4 7x + 16 = 2x - 1 5 = 6x - 3 . Multiplying each term by 15 , 6x ... equation by the L.C.M. of the monomial denominators . 4. Solve the equation 20 x + 2b a2 + b2 = x - a x + a x2 - a2 ...
Página 92
Webster Wells. 5. Solve the equation .2 x.01.03x = .113x + .161 . .2 x .03x.113x = = .01 + .161 . Transposing , Uniting terms , Dividing by .057 , .057x = .171 . x = 3 . 185. A simple equation containing but one unknown quan- tity cannot ...
Webster Wells. 5. Solve the equation .2 x.01.03x = .113x + .161 . .2 x .03x.113x = = .01 + .161 . Transposing , Uniting terms , Dividing by .057 , .057x = .171 . x = 3 . 185. A simple equation containing but one unknown quan- tity cannot ...
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Términos y frases comunes
a₁ ab² absolute value Algebra arithmetical arithmetical means b₁ Binomial Binomial Theorem coefficient common factor Commutative Law Compare Art complex number continued fraction convergent cube root decimal denominator denote determinant Dividing divisible divisor equal EXAMPLES exponent Extracting the square figures Find the numbers follows from Art geometrical progression given equation Hence imaginary number infinite series last term letters logarithm mantissa Multiplying Note nth root number Art number of terms obtained P₁ partial fractions perfect power polynomial positive integer prove pure imaginary quadratic equation quotient radical sign rational and integral rational numbers real number remainder represented result rule of Art second term Solve the equation square root Sturm's Theorem Substituting Subtracting surd Theorem third unknown quantities Whence zero
Pasajes populares
Página 41 - The square of the sum of two numbers is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second.
Página 270 - In any proportion the terms are in proportion by Composition; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term.
Página 271 - In a series of equal ratios, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Let a: 6 = c: d = e :/. Then, by Art.
Página 269 - If the product of two quantities is equal to the product of two others, one pair may be made the extremes, and the other pair the means, of a proportion. Let ad = ос.
Página 268 - The terms of a ratio are the two numbers to be compared; thus, in the above ratio, 20 and 4 are the terms. When both terms are considered together, they are called a couplet ; when considered separately, the first term is called the antecedent, and the second term the consequent. Thus, in the ratio 20 : 4, 20 and 4 form a couplet, and 20 is the antecedent, and 4 the consequent.
Página 140 - ... from the given expression. Divide the first term of the remainder by twice the first term of the root, and add the quotient to the root and also to the divisor.
Página 137 - Arts. 200 and 201 we derive the following rule : Extract the required root of the numerical coefficient, and divide the exponent of each letter by the index of the root.
Página 38 - Divide the first term of the dividend by the first term of the divisor, giving the first term of the quotient. Multiply the whole divisor by this term, and subtract the product from the dividend, arranging the remainder in the same order of powers as the dividend and divisor.
Página 79 - Multiply the numerators together for the numerator of the product, and the denominators together for the denominator of the product.
Página 270 - In any proportion the terms are in proportion by Alternation ; that is, the first term is to the third as the second term is to the fourth.