Bradbury's Elementary Algebra : Designed for the Use of High Schools And Academies
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In our eReader you can find the full English version of the book. Read Bradbury's Elementary Algebra : Designed for the Use of High Schools And Academies Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Bradbury's Elementary Algebra : Designed for the Use of High Schools And Academies
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23 — 4 a; y = — ^ — EQUATIONS OF THE FIRST DEGREE. 105 There are three methods of elimination : — I. By substitution.
II. By comparison.
111. By combination.
CASE I.' 112. Elimination by substitution.
1. Given -[^'' + ^^ = ^^1, to find a and 2/.
OPERATION.
4a: + 5y=23 (1) 5a: + 4y= 22 (2) (3) 6x + 4('-?^)= 22 (4) 25X + 92 — 16a: = 110 (5) y = ^ = 3 0) ^= 2 (6) Transposing 4 a; in (1) and dividing by 5, we have (3), which gives an expression for the value of y. Substituting this value of y in (2...), we have (4), which contains but one unknown quantity ; i. e. y has been eliminated. Reducing (4) we obtain (6), or ar,= 2. Substituting this value of x in (3), we obtain (7), or y = 3. Hence, RULE.
Find an expression for the value of one of the unknown quantities in one of the equations, and substitute this value for the same unknown quantity in the other equation.
Note. — After eliminating, the resulting equation is reduced by the rule in Art. 102. The value of the unknown quantity thus found must be substituted in one of the equations containing the two unknown quantities, and this reduced by the rule in Art.
What to read after Bradbury's Elementary Algebra : Designed for the Use of High Schools And Academies?
You can find similar books in the "Read Also" column, or choose other free books by Bradbury, William Frothingham, 1829-1914 to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
23 — 4 a; y = — ^ — EQUATIONS OF THE FIRST DEGREE. 105 There are three methods of elimination : — I. By substitution.
II. By comparison.
111. By combination.
CASE I.' 112. Elimination by substitution.
1. Given -[^'' + ^^ = ^^1, to find a and 2/.
OPERATION.
4a: + 5y=23 (1) 5a: + 4y= 22 (2) (3) 6x + 4('-?^)= 22 (4) 25X + 92 — 16a: = 110 (5) y = ^ = 3 0) ^= 2 (6) Transposing 4 a; in (1) and dividing by 5, we have (3), which gives an expression for the value of y. Substituting this value of y in (2...), we have (4), which contains but one unknown quantity ; i. e. y has been eliminated. Reducing (4) we obtain (6), or ar,= 2. Substituting this value of x in (3), we obtain (7), or y = 3. Hence, RULE.
Find an expression for the value of one of the unknown quantities in one of the equations, and substitute this value for the same unknown quantity in the other equation.
Note. — After eliminating, the resulting equation is reduced by the rule in Art. 102. The value of the unknown quantity thus found must be substituted in one of the equations containing the two unknown quantities, and this reduced by the rule in Art.
What to read after Bradbury's Elementary Algebra : Designed for the Use of High Schools And Academies?
You can find similar books in the "Read Also" column, or choose other free books by Bradbury, William Frothingham, 1829-1914 to read onlineMoreLess
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