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or x+6x-4=17; .. by transposition, 7x=17+4=21,
21 by division, x= ...=3;
7? and .. y=3x-2=3X3--2=9--2=7.
Ex. 2. Given
to find the 5x+10=78+y, S values of a and
y. From the first equation, y=60-3x;
Let the value of y be substituted in the second equation, and it becomes,
5x+10=78+(60-32) Then, by transposition, 8x=78+60-10;.
to find the va
3 Ex. 3. Given
lues of x and -Y -62– 2x,
3 Mult. the 1st equation by 3, then x+y=198-by: ..
(1); 2nd by 3, then x-y=186-63 .. (2),
(2; From equation (1), we have x=198-7y,
70-y=186 ; By substituting the above value of x, in the last equation, it becomes
7(198—7y)-y=186, or, 1386-49y-y=186 ;
by transposition, -50y=186--1336=-1200, by changing the signs, 50y=1200,
1200 .. by division, y= =24.
50 Whence, v=198-7y=198-7X24=198-168,
Sx+2y=80, Ex. 4. Given
to find the values ut y=60, of x and y. From the second equation, x=60-y:
= By substituting this value of x in the 1st equation, we have,
60-y+2y=80, by transposition, y=80-60,
..y=20. And x=60-y=(by substitution) 60-_20,
Ex. 5. Given
to find the va
3x — y= 2, lues of x and y.
From the 1st equation, x=17--2y.
3(17 — 2y) — y=2,
.. by division, y=7, whence, x=17--2y=17--2X7=17-14,
Ex. 6. Given
ac +y =5,
to find the values
of x and
yo From the first equation, s=5-y,
squaring both sides, xo=(5-3).
And by substituting this value for to in the sea cond equation, it becomes,
(5-y) - Yo=4,
by reduction, 25+10y=5, by transposition, 10y=20,
.. by division, y=2. Whence, a=54-y=5--2=3.
, of and y.
8 Ex. 7. Given
to find the values y
s'. by transposition, x=1552-64y. And substituting this value for x, in the second equation, it becomes,
+8(1552—64y)=131, by reduction, y+99328-4096y=1048, by transposition, 4095y=98280,
98280 by division, y
..y=24. Whence x=1552-64y=1552-64 X 24,
or x=1552-1536 ;
=16. The value of y might be found from the second equation, in terms of x and the known quantities ; which value of y substituted for it in the first, an equation would arise involving only x, the value of which might be found; and therefore the value of y also may be obtained by substitution.
5x+6y Ex. 8. Given
-=27, and 3
4 the values of w and
Ans. x=9, and y=6. Ex. 9. Given 15y +45x=300, and +-15y=36, to find the values of x and y.
Ans. x=6, and y=2. Ex. 10. Given 3x+y=60, and 5x+10=78+y, to find the values of w and y.
Ans. x=16, and y=12. Ex. 11. Given 10x --3y=38, and 3x-y=11, to find the values of x and
Ans. x=5, and y=4. Ex. 12. Given x+y=193-6y, and x-y=186 --6c, to find the values of x and y.
Ans. =30, and y=24. .
y y = ,
8 to find the values of wand y.
• Ans. «=16, and y=24.
3. 2 values of x and y
Ans. x=6, and y=12. Es. 15. Given 4x+y=34, and 4y +x=16, to find the values of x and y.
Ans. x=8, and y=2. Ex. 16. Given 3x +2y=54, and x:y::4:3, to find the values of x and
Ans. r=12, and y=9. Ex. 17. Given
3 23, to find the values of cand y.
Ans. x=4, and y=3,
Ex. 13. Given + y = 26, and +0x = 131,
Ex. 14. Giyen +%=7, and +9 =8, to find the
+6y=21, and 9+ 6 + 5x =
249. Find the value of the same unknown quantiiy in terms of the other and known quantities, in each of the equations; then, let the two values, thus found, be put equal to each other; an equation arises involving only one unknown quantity; the value of which may be found, and therefore, that of the other unknown quantity, as in the preceding rules.
This rule depends upon the well known axiom, (Art 47); and the two preceding methods are founded on principles which are equally simple and obvious.
s x+3y=100, Ex. 1. Given
to find the va12 + y=100, 5 lues of x and y. From the first equation, x=100-3y,
= and from the second, x=
100—4 = 100-3y,
2 Multiplying by 2, 100-y=200—6y, by transposition, by-y=200-100,
... by division, y=20. whence, x=100 - 3y=100—3 X 20;
.. x=40. Here, two values of y might have been found, which would have given an equation involving only x; and from the solution of this new equation, a value of , and therefore of y, might be found,