 | C R. Lupton - 1879 - 194 pages
...26, /. Зж + 8 = 26 ; transposing, Зж =18, .'. ж = 6. SECOND METHOD. 82. By Substitution. —Find the value of one of the unknown quantities in terms of the other from either equation, and substitute this value in the other equation. Taking the same equation as... | |
 | Robert Potts - 1879 - 668 pages
...bo determined from these equations, a1z-\-bty = cl, а^с-\-1>^у = сг. First method : By finding the value of one of the unknown quantities in terms of the other and known quantities, from one equation, and substituting it in the other equation. Let the value of... | |
 | James Mackean - 1881 - 510 pages
...that may be employed. 158. I. When one of the equations is simple. Find from the simple equation a value of one of the unknown quantities in terms of the other, and substitute this value in the other equation. Illustrative Example. From [2], x = 2y + 3. Substitute... | |
 | Simon Newcomb - Algebra - 1882 - 302 pages
...?— = — ^— ; 2ж + « = 30. 8 — у 4 — ж' ' -; Elimination by Substitution. 159. RULE. Find the value of one of the unknown quantities in terms of the other from one equation, and substitute this value in the other equation. The latter will then have but one... | |
 | George Albert Wentworth - Algebra - 1885 - 300 pages
...г— *- (7) Reduce (7), 19у - - 19. лу — 1. Substitute the value of y in (1), 2x + 9 = 11. 183. Hence, to eliminate an unknown quantity by comparison,...last example, (3) be divided by (4), the resulting 2 IT 4- 9 1t equation, - = - — - — ™. would, when reduced, give the value of y. 3 This is the... | |
 | Webster Wells - Algebra - 1885 - 372 pages
...— 124 -ж= 8 Whence, a; = — 8 Substituting this value in (3) , у = ~40+44 _ 2 ¿à BULB. Find the value of one of the unknown quantities in terms of the other from one of the given equations, and substitute this value for thai quantity in the other equation.... | |
 | Webster Wells - 1885 - 368 pages
...274. When one equation is of the first degree. Equations of this kind may always be solved by finding the value of one of the unknown quantities in terms of the other from the simple equation, and substituting the result in the other equation. 1. Solve the equations... | |
 | Webster Wells - Algebra - 1885 - 370 pages
...274. When one equation is of theßrst degree. Equations of this kind may always be solved by finding the value of one of the unknown quantities in terms of the other from the simple equation, and substituting the result in the other equation. 1. Solve the equations... | |
 | George Albert Wentworth - Algebra - 1886 - 284 pages
...182. Hence, to eliminate an unknown quantity by substitution, From one of the equations obtain ¿he value of one of the unknown quantities in terms of the other. Ex. 66. Solve by substitn tion : 1. Зж — 4y = 2\ 8. Зr — 4y=181 y= 0/ .2. 1x — 5y = 241 9.... | |
 | George Albert Wentworth - Algebra - 1887 - 272 pages
...(7), 33 + 27y = 14 + 8y. 19y = -19. .-.y — L Substitute the value of y in (1), 2x + 9 = 11. 183. Hence, to eliminate an unknown quantity by comparison,...example, (3) be divided by (4), the resulting equation, - = • — t— 2, would, when reduced, give the value of y. This is the shortest method, and therefore... | |
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Find the value of one of the unknown quantities, in terms of the other and known quantities... 
