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x+2.(3x ---2)=17,

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.

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Ex. 2. Given
3x + y=60,

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;.

and by division, a= =16.

Whence, y=60-3.=60--3x16=60-48;

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rx+y=66—24, ,

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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 ;

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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.
And this value substituted in the second,

3(17 2y) y=2,
or 51-6y-y=2,
by transposition, &c, 7y=49,

.. by division, y=7, whence, x=17--2y=17--2X7=17-14,

.. y=3.

Ex. 6. Given

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ac +y =5,

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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.



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8 Ex. 7. Given

to find the values y

x .

Multiplying the first equation by 8,


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.

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6x-5Y=6, to

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

8 to find the values of wand y.

• Ans. «=16, and y=24.

20* .

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

= 4

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


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+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=

3 2


100—4 = 100-3y,


2 Multiplying by 2, 100-y=200—6y, by transposition, by-y=200-100,

or, 5y=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,

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