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a-cz = d', then ax+by = d, an equation, the solution of which has been determined in Art. 2.

But as ď may admit of more values than one, there may be several equations involving x and y, corresponding to the number of values of d. And for every value of : there will be corresponding values of x and y. The given equation may be of the three following forms:

ar+by+cz =d, ax+by-cz =d, ax-by-cz =d,

or ax+by =d-iz, ax cz = d-by, ax_by = d+c%. In equations of the first form, the number of solutions in positive integers is limited.

In equations of the second and third form, the number of solutions in positive integers is unlimited.* 7. To solve the general independent equations

ayx+by+z=, Ayx+by+cam= d,, involving three unknown quantities, limiting the solution to positive integers.

First eliminate , between the two given equations, and the result is (a,02-0,2x+(6,c7-bc)y=c,d, -da, an equation of the form ax+by =c, according as b,c,-6,C is positive or negative.

The equations ax+by=c and ax_by = c have been considered in Art. 2, and when the values of x and y have been found, the corresponding values of % may be determined by the substitution of the values of x and y in either of the given equations.

If the elimination of from the two given equations give an equation of the form ar+by = 0, the number of solutions in positive integers is limited; but if it gives an equation of the form ax_by = 0, the number of solutions in positive integers is unlimited.

* Ex. Solve the equation 5x + 6y + 20z=187 in positive integers.

Here 5x+6y=187 - 20z. But x and y cannot be each less than unity, and z must not be greater than 9.

Hence z may be 1, 2, 3, 4, 5, 6, 7, 8, and these 8 values of y will give 8 different equations for determining the values of x and y corresponding to the 8 values of z.

When these 8 values of z are successively substituted in the given equation, the values of x and y determined in each case, will be those which correspond to the several values of z.

In the first case, when z=1, 5x+6y=167, and this equation being solved, x=31—6p, from which it appears that p must not be greater than 5.

And y=5p+2, from which it also appears that p must not be less than 0.
Hence this equation 5x+6y=167 admits of six solutions, when z=1.
p=

0, 1, 2, 3, 4, 5. x=31 – 6p-31, 25, 19, 13, 7, 1. y=52+2= 2, 7, 12, 17, 22, 27.

1, 1, 1, 1, 1, 1, The rest of the solutions can be found in the same manner, when z is 2, 3, 4, 5, 6, 7, or 8, respectively.

EXERCISES.

EQUATIONS OF THE FIRST DEGREE.

I. Express generally in words at length, and solve each of the following equations: 1. x= 14- 6x

7. 5x –9= 7x+15. 2. 2x + x = 46+2x.

8. x+7= 3x-9.
3. 2x +11 = 78—14.

9. 3.x—7=4 - 5x.
4. 3x +14 = 9x+22. 10. 2x -3x - 2 = 0.
5. 8x^5 = 13–7x. 11. 5x—7-2x = 10x+8 - 2x.
6. 3.0- 7 = 5x +17. 12. x-10+8 = 70-3.

II. Solve the following equations :1. 375x+.5 = 2.25 +8. 2. •5x – 2 = .252 +.23-1. 3. -5x + 60-8=75x+25. 4. •15x +1:575 -.875x = .0625. 5. •15x+1.2-875x+:375 = .0625x. 6. 1.20-3(-180—05) = 4x+8.9. 7. •84x - 7.6-2.23 - 10 = 0.

III. Solve the following equations, and verify the correctness of the results in each case :

1. (x-3)-3(2-5)+5(x-7)= 0. 2. 2(x-1)+3(2-2)+4(x-3) = 0. 3. (x+4)(x+2)+(x+3)(x-3) = 2(x+3)(x+2). 4. (4-3)-(3-2)(x+1)=(x-3)(1+x)+(3-X). 5. (4x+2)x+(4x+1.x = (2x+1)(4x-2). 6. (1-5)(2-3)-(3-5)-(x+7)(x-2) = 0. 7. 24.0-1248 - x) = 504. 8. (x-4)+3(x-1)(x-2) = 3(-2x-4). 9. (x-1)(x-2)=x^. 10. (x-9)(x-7)(x-1)=(x-2)(2— 4)(x–11). 11. (x-3)(x-4)= x? - 5x – 8. 12. (x+1)(x+2)(x+3) = (x-1)(x-2)(x-3)+3(4x — 1)(x+1).

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X. 1. Find the two numbers whose sum shall be 100, and whose difference shall be 10.

2. What number is that whose third part exceeds the seventh part by 20?

3. What value of x will make the difference between (2x+4)(3x+6) and (3x-2)(2x—8) equal to 96 ?

4. What number is that to which, if 8 be added, one fourth of the sum is equal to 29 ?

5. The difference between a certain number and its defect from 17, is equal to 3 times its defect from 21; find the number.

6. Divide 240 into two parts, such that one-fourth of one of them added to one-tenth of the other shall be equal to 36.

7. Divide the number 144 into three such parts, that one half of the first, one-third of the second, and one-fourth of the third, shall be equal to one another.

8. Find a number which, when trebled and diminished by 3, exceeds the sum of its fifth and seventh parts by 90.

9. Find a number such that if of it be subtracted from 20, and

of the remainder from of the original number, 12 times the second remainder shall be half of the original number.

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