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6. A cask which held 146 gallons was filled with a mixture of brandy, wine, and water. In it there were 15 gallons of wine more than there were of brandy, and as much water as both wine and brandy. What quantity was there of each ?

Let the number of gallons of brandy, ..x+15= number of galions of wine, and 2x+15= number of gallons of water. :.:+x+15+2x+15–146, .. by transposition, 4.c=116, and

29. .. there were 29, 44, and 73 gallons respectively of brandy, wine, and water.

7. A person employed 4 workmen; to the first of whom he gave 2 shillings more than to the second ; to the second 3 shillings more than to the third ; and to the third 4 shillings more than to the fourth. Their wages amounted to 32 shillings. What did each receive ?

Let x= the sum*received by the fourth, i.x+4=

third, 3+7=

second, and +9=

first. 2+3+4+3+7+*+9=32, and, by transposition, 4.2=12,

consequently x=3. .. they received 12, 10, 7, and 3 shillings respectively.

8. A father, taking his 4 sons sons; the second received 10d., to school, divided a certain sum the third 14d., the fourth 25d., amongst them. Now the third the fifth 28d., and the sixth 33d. had 9 shillings more than the less than the first. Now the sum youngest ; the second 12 shil-distributed was 10d. more than lings more than the third ; and treble of what the first received. the eldest 18 shillings more than What did each receive ? the second ; and the whole sum Let x= what the first rec'd., was 6 shillings more than 7 times..x—10— 6

second, the sum which the youngest re

-14=

third, ceived. How much had each ?

X-25=

fourth, Suppose the youngest re

X-28

fifth, ceived 2 shillings,

sixth, then the third received x+ 9 The sum of which =62_110 the second,

x+21=3x+10 by supposition... by and the eldest,

x+39 transposition, 3r=120 and = :: 2+x+9++21+x+39=40. :. they received 40, 30, 26, 7x+6; ... by transposition, 63–15, 12, 7 pence respectively. 3z, and ... 213r. Consequent3xX Consequent- 10. It is required to divide the ly they received 21, 30, 42, and number 99 into five such parts, 60 shillings respectively. that the first may exceed the sec

9. A sum of money was to be ond by 3; be less than the third divided amongst six poor per-by 10 ; greater than the fourth

X-33=

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by 9; and less than the fifth by 11. What two numbers are 16.

those whose sum is 59, and difLet = the first part, ference 17 ? ..x— 3 second,

Let the less, x+10= third,

... 3+17= the greater,
9 fourth,

and . . x++17=59 ;
x+16=
fifth.

by transposition, 2x=42, .. 2+2-3+*+10+149+

and x=21, the less, 2+16=99, or 5x+14=99.

.. the greater =38. .. by transposition, 5x=85, and

-17. .. the parts are 17, 14, 27, 8, and 33.

12. What number is that, the treble of which increased by 12 shall as much exceed 54 as that treble is below 144 ?

Let x= the number. .:: 3x+12454144_3x, :: 6x=186, and x=31.

13. Two persons began to play with equal sums of money: the first lost 14 dollars, the other won 24 dollars, and then the second had twice as many dollars as the first. What sum had each at first ?

Let the sum ; Then x—14 and +-24

= the sums each had after playing; :: 2.--285x+24; : 52.

14. At a certain election 943 men voted, and the candidate chosen had a majority of 65. How many voted for each ?

Let x=the number of votes the unsuccessful candidate had;

... 34-65= the number the successful one had. ..:-+-+65 =943; 2x=878, and x=439. ... the numbers were 439 and 504.

RULE III. Any equation may be cleared of fractions by multiplying each of its terms, successively, by the denominators of those fractions, or by multiplying both sides by the products of all the denominators, or by any quantity that is divisible by each of the denominators.

3.2 if

, x+ 4

4 15; and this, multiplied by 4, gives 4x +30=60; whence, by ad

60 4 dition, 72=60, or

..

Thus, ir į +1 = 5, then, multiplying by 3, we have za

7-8

2 And, if + =10; then, multiplying by 12, (which is a 4 6

multiple of 4 and 6.) 32 +2.120, or 5x=120, or

5

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-24.

с

a

a

а 2

Ans. 2196

=19

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За

It also appears, from this rule, that if the same number, or quantity, be found in each of the terms of an equation, either as a multiplier or divisor, it may be expunged from all of them, without altering the result. Thus, if ax=abtac; then, by cancelling, tot.

6 And if-+ ; then, 2+b=c, or X-.

Зх 1. Given + 24 to find 4

X. 2 Multiplying by 4, (which is a multiple of 4 and 2,) 6x=+96,

96 or 63—96, or 5.1=96, and by division 195, Ans.

,

5 2. Given g ++= 62 to find x.

Ans. X=60. 3 5

2 Here + + 62; then, multiplying by 3, we have at

3 5 2 3.x

152 5 + =186, and multiplying by 5, gives 5x+3x+ 2 2

2 930; and this, multiplied by 2, gives 10x+6x+15=1860, or

1860 312-1860, and, by division,

=60, Ans.

31 -3 3. Given + 20

to find x. Ans. X=9. 2

2 Multiplying by 6, (which is a multiple of 2, 3, and 2,) 32—9+

72 2x=120—3x-57, and, by transposition and division, ==9,

=

8 Ans.

2+2 4. Given

Ans.

=13. 2 3

4 Multiplying by 12, we get 6x+6+4x+8=192—3.2--9, or 132 109; and, by division, x=y=13, Ans. ata

2x

ato 5. Given + +

to find 2. 6

d

a cb+aco?-acd Ans. x

acd tabd2cbd 2x ato

; then, multiplying by b, xtra 26x

26cx abc+cbe abtes

, mult. by c, cztcatbze + +

= + d

d and, by multiplying by a and d successively, we have adcx+a'cd tabdz2bcdatabc+acb", or, by transposition, adcz+abdi

a'bc+acb_a'cd Roodz ta'bctacba'd, and, by division, there

'

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x+19

2+1

+ 16-3 to find z.

69 1

2

с

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Here *ta +

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to

a

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a

adci-abd2bcd A.

а

Rule IV. If the unknown quantity be a surd, transpose the rest of the terms, and let the surd quantity stand alone on one side of the equation; then take away the radical sign from it, and raise the other side of the equation to the power denoted by the index of the surd.

Thus, if Nx_2=3; then will XX3+2=5, or, by squaring, X=5=25.

And if „(3x+4)=5; then will 3x+4=25, or 32=25—4= 21, or x=4=7.

Also, if (2x+3)+48; then will (2x+3)=8-44, or 2x+3=4%64, and .22—64—3—61, or 21–304. 1. Given 2/2+3=9 to find z.

Ans. x=9. Here 2wx+3=9; then, 2x x=9—3—6 by transposition, and 4r=36 by squaring, or t==9, Ans. 2. Given ✓ (x+1)^2=3 to find x.

Ans. x=24. Here ✓(x+1)^2=3; then » (x+1)=3+2=5, or ætl= 25 by squaring, and consequently x==25—1=24, Ans. 3. Given ♡ (3x+4)+3=6 to find x.

Ans. N=75 Here (3.c+4)+3=6; then (3x+4)=6—3=3, or 3x+4 =27 by cubing, and 3x=27–4=23, or 2=43=7}, Ans. 4. Given ✓(4+x)=4-^x to find X.

Ans. =24 . Here (4+x)=4Wx; then 4+x=16_8Xx+x by squaring, and, by trans. 8/12, or 2X=3; and 4t=9 by squaring, or x==24, Ans. 5. Given ✓ (4a++x)=N(464+24) to find X.

Here „(4a2+2)=(464+X*); then, by squaring 4a+'= ✓(484+24), and squaring again, 16a++Baʼzi+x=464+2*, by

74_4a"

74_4a4 transposition and division 2

and ..x=NG

-), An. 2a*

2a? Rule V. If that side of the equation which contains the unknown quantity be a complete power, the equation may be reduced to a lower dimension, by extracting the root of the said power from both sides of the equation.

Thus, if m=81; then x=81=9; and if x=27, then = 327–3. Also, if 3.x2–9=24; then 32=24+9=33, or = =11, and ... x=N11. And, if x2 +6x+9=27; then, since the left hand side of the equation is a complete square, we shall have, by extracting the roots, 2+3=127=Ň (9X3)=3X3, or X=3N3–3. 1. Given 9x4—6=30 to find x.

Ans. 2. Then 9x=30+6=36, or r=864, and ...=N42, Ans. 2. Given *+9–36 to find z.

Ans. 3.

ر

=

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4

a

2

Then 36-9-27, or 273, Ans.

=

81 3. Given +2+1=3 to find Z. 3

Ans. 24. 4 By extracting the square of both sides 2+1==41, or x= 4-=4, Ans. 4. Given zo +-axt = to find z. 32 2

Ans. x=

2 By extracting the square root of both sides 2+ =b, or comAns.

. 5. Given +14c+49=121 to find 2.

Ans. 4. By extracting the square root of both sides of the equation et 7=ll, or 211–754, Ans.

RULE VI. Any proportion may be converted into an equation by making the products of the extremes and means equal to each other.

Thus, if 3x : 16::5:6; then 3x X6=16X5, or 18x80, or : =ff==4.

2cx And if : a::8:6; then will :c

=ab, or 2cx=3ab; or, by 3

3 3ab division, 2c

4.0 Also, if 12–7:59:4:1; then 12–

;

=2.x, or 2x+te 12 , and consequently ==4. 3

20ab 1. Given -::a::5bc: cd to find X. 4

3d

20abc BzXcd=aX 5bc, or 3cda20abc, and consequently a

3cd

2x

Ans. I

20ab

3d Ans.

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:

:

2. Given 10—2 : 4x::3:1 to find 2.

Ans. 231. Then 10—2=2x by mult. ext. and means, and 3x=10, or = 40=3}, Ans. 3. Given 8+82 : 4x : : 8: 2 to find 2.

Ans. xl. Here 8782: 4x : : 8:2; then 16+16x32x by mult. ext. and means, and 322—16x=16, or 16x=16; .. X=1=1, Ans. 4. Given x: 6-2 :: 2:4 to find x.

Ans. 22. Here 2 :6_* :: 2:4 ; then 4x=12—2x by mult. ext. and means, and 4x+2x=12, or 6x=12; h=2, Ans.

a 5. Given 42 : a :: 9Nx: 9 to find X.

Ans.

16

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