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Examples for Practice. 5. 3.1 - 5 2x + 9 required the value of x. Ans. I = 14. 6.° 2.0 + 3 1 +17

Ans. = 14. 7. 5.r 4 -4.0 + 25

Ans, I = 29. - 9 Or- 3.

Ans. = 6. 9. 4.1 + 2a= 3.7 + 9b

Ans.

96--2a. (62.) Rule Il. If the unknozun quantity has a coefficient, then its value may be found by dividing both sides of the equation by that coefficient. Note: This rule is founded on the 5th Axiom.

Examples. 1. Let 4r=28, then, by dividing each side of the equation by 4, 47 28

28 we have

;
but

and <7; therefore, r=7.

4 2. If 6r=20-4r, then will x=2. 3. If 12x+20=6r+44, then will r=4. 4. If 25x+10=20x+50, then will x=8.

Examples for Practice. 5. 10r=150...

Ans, x=15. 6. 30r +8=68.

Ans. I= 2. 7. 1őr +14=12r +54

Ans. X=10. 8. gr-3=4r+22.....

Ans. r= 5. 9. 17r-4r+9=3x +39..

Ans. x= 3. 10. 12.r+6r+18=9r+36.

Ans, x= 2. (63.) Rule III. An equation may be cleared of fractions by multiplying each side of it by the denominators of the fractions in succession. Or, . all the denominators may be taken away in one operation, by multiplying each term by their least common multiple. Note.-The foundation of this rule is found in Axiom 3d.

Examples. = 12; then, by multiplying each side of the equation 6x

6r by 6, we shall have = 12 6; but

=x, and 12 x672; 6

6 therefore, x = 72.

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2. If

+

=7, then will x = 10. 2 5

21 2.2 For, multiply by 2, then +

= 14. 2 5

10r Again, multiply by 5, and 5x +

= 70. 5 70

10, Or, 5x+27=70; that is, 7x=70, or <=;

7 which is the answer, a3 above,

+

3

3. If +

13, then will $ = 12.

4
Here the least common multiple of 2, 3, and 4, is 12 ;
Whence, by multiplying each term by 12,
We have 6x + 47+3.5=156;
That is, 13x=156;. x=156+13=12.

(64.) Note.- In the solution of the following questions, the learner must observe the following rules :

Ist. To clear the equation of fractions by Rule 3d.

2d. To collect the unknown quantities on one side of the equation, and the known on the other, by the First Rule; and,

3d. To find the value of the unknown quantity by dividing each side of the equation, by its co-efficient, as in Rule 2d.

Examples for Practice.

4. Let 25 + = 22 ; it is required to find o ?

x Ans. x24.

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(65.) QUESTIONS WHICH INVOLVE EQUATIONS CONTAINING ONLY ONE UNKNOWN QUANTITY,

Note. — These questions are designed to shew the method of app!ying the foregoing rules, previously to our entering upon the management of equations with two naknown quantities.

QUESTION 1. There are two numbers whose difference is 12, and their sum 20; what are the numbers ?

Obs. As their difference is 12, the greater number must evidently exceed the less pumber by 12.

Solu. Let x = the less number; then x+12 will be the greater

But by the equation, the sum of the two numbers is 20; or, the greater + the less number = 20. Hence, by addition, (2 + 12) + x = 20. That is, 2x + 12

= 20. therefore, 2x

20 12 z: 8.

8 and ..

4 the less No.

2 Hence the greater number=.+12=4+12 = 16.

QUESTION 2. There are two numbers whose difference is 9, and if three times the greater be added to five times the less, the sum will be 35; what are those two numbers ?

Solu. Let x = the less ; then x+9 = the greater number.
And 3 times the greater = 3 X X + 9 = 3.2 + 27.

5 times the tess 5 X X But, by the conditions of the question, 3 times the greater + 5 times the less number = 35.

Hence, by addition, (36 + 27) + 5x = 35;
That is, 8x + 27 = 35, or 8x = 35 – 27 :

= 8;
8
::==== the less number ;

and +9=1+9=10 the greater number. QUESTION 3. What number is that to which if we add 10, three-fifths of the sum shall be 66 ? Solu. Let x=the number sought, then x+10 = the number + 10. 3

3x +30
;
5
5

5

5 3 But, by the question, gths of + 10 = 66.

3x + 30 Hence, by substitution,

= 66.

5 Multiplying by 5, we have 3x + 30

:. 36 =330-30=300, and x = -100=the number required.

3

= 5r.

Now @ths of 3+ 10 =

xx +10=3X x + 10

330;

QUESTION 4. What is that number which being multiplied by 6, the product increased by 36, and that sum divided by 18, the quotient shall be 20 ? Solu. Let = the number sought.

Then Ox = the number multiplied by 6.
Also Or + 36= the product increased by 38,

6.0 + 36
And

= that sum divided by 18 18

6.0 + 36 Hence, by the question,

=20.

18
And 6: +36 = 360, multiplying by 18.
or Or = 360 36 324 ;
324

54, the number required.

6 QUESTION 5. A post is one-fifth in the earth, three-sevenths in water, and 13 feet out of the water; what is the length of the post ? Solu. Let x = the length of the post ;

Then = the part of it in the earth ;

3.2
Also = the part of it in the water.

7

And 13 = the part of it out of the water ; But the part in earth + part in water t part out of water the

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whole post.

3.2 Therefore

5
+

+ 13 = %.
7

15.0
Multiply by 5, then * + + 65 * 5x.

7
by 7, and 7 x + 15.3 + 455

35% ;
or

455 = 35-7- 15.4 =13%; 455 Hence x = = 35 feet the length of the post.

13 QUESTION 6. Having paid away one-fourth and one-seventh of my money, I had eighty-five pounds left in my purse; how much money was in it previously to these disbursements : Solu. Let s= the money in purse at first; then

is the money paid away

Now, by the question,

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2380 Therefore, r= =140

The money at first in purse. 17

Examples for practice. 7. The difference of two numbers is 10, and if to their sum 15 be added, the whole will be 43 ; it is required to find those two numbers ?

Ans. 9 and 19. 8. The difference of two numbers is 14, and if 9 times the less be. subtracted from 6 times the greater, the remainder will be 33 ; what are those numbers ?

Ans. 17 and 31. 9. If to a certain number I add 20, and from two-thirds of this sum I take away 12, the remainder will be 10; what is that number?

Ans. 13. 10. What number is that, of which if I add one-third, one-fourth, and two-sevenths together, the sum shall be 73 ?

Ans. 84. 11. What number is that the third part of which exceeds its fifth by 72 ?

Ans. 540. 12. Two merchants, A and B, lay out equal sums of money in trade ? A gains 1201., and B loses 80l.; and now A.'s money is treble

. of B.'s :- Pray what sum had each at first ?

Ans. 180.

(66.) The equations which we have hitherto considered contained but one unknown quantity; but, in the solutions of Algebraic Problems, equations frequently arise containing two or more unknown quantities; as, ax + by = c, ar + ly + cx= d, &c. From the first of these it is evident, by the rules already laid down, that

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In this equation the value of x, which is found in terms of y; but y is an unknown quantity, consequently I still remains undetermined; whence it is manifest, that from the equation ar + by = 0, neither x nor y can be found. But if in addition to ax + by =c, we have another equation, as,

f - ey dr + ey=f, from which x may be found as before =

d ly fsey we should then have

.

d From this last equation, y may be found by the foregoing rules; and

by when y is found

becomes known.

a

с —

From what has been said, it appears that in order to determine the values of two unknown quantities, two equations must be given. In

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