 | C R. Lupton - 1879 - 194 pages
...and from (i), 3x + 2y = 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 - Algebra - 1879 - 672 pages
...may be determined from these equations, a¡x+bly = ol, a¿c+b¿y = cíl. First method : By finding the value of one of the unknown quantities in terms of the other and known quantities, from one equatiou, and substituting it in the other equation. Let the value of x be found from the... | |
 | 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 x be found from the... | |
 | Webster Wells - Algebra - 1879 - 468 pages
...uniting terms, 41 x = 82 Whence, ж =2. Substituting this value in (3), у = — - — = — 3. BULE. Find the value of one of the unknown quantities in terms of the other, from either of the given equations * and substitute this value for that quantity in the other equation.... | |
 | Charles Scott Venable - Algebra - 1880 - 168 pages
...equation. For this case we have the Rule. — From the equation of the first degree find the expression of one of the unknown quantities in terms of the other, and then substitute this expression in the second equation. Ex. 1. Given x 4 y = 10 (1) Ь0 find the values... | |
 | 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... | |
 | George Albert Wentworth - Algebra - 1881 - 400 pages
...11, .-. x=l. 191. Hence, to eliminate an unknown quantity by comparison, from each equation obtain the value of one of the unknown quantities in terms of the other. Form an equation from these equal values and reduce the equation. NOTE. If, in the last example, (3)... | |
 | 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... | |
 | Webster Wells - Algebra - 1885 - 370 pages
...— 124 — x= 8 "Whence, ж = — 8 Substituting this value in (3) , у = ~40+44 — 2 £ 136 RULE. 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 that quantity in the other equation.... | |
 | Webster Wells - Algebra - 1885 - 372 pages
...132 = — 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.... | |
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In one of the given equations obtain the value of one of the unknown quantities in terms of the other unknown quantity; Substitute this value in the other equation and solve. 
