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9. In bx — cy — 0, and x3 — y3 — a3 ; what are the values of x and ya ?

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11. Given (x + 5) (y + 7) = (x + 1) (y — 9) + 112, and 2x + 10 = 3y+ 1; Ans. 5 and 3: which is x?

to find x and y.

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III. Let the given equations be so multiplied, or divided, by such numbers or quantities, as will make the terms which contain one of the unknown quantities the same in both equations.

Then by adding or subtracting the equations, according as the signs may require, there will result a new equation, with only one unknown quantity, as before: that is, add the two equations when the signs are unlike, but subtract them when the signs are alike, to cancel that common term.

The best multipliers generally are those of the selected term in the alternate equations; as in the example, ax + by = c2, and a,x + b,y= c,2, where the first equation being multiplied by a,, and the second by a, we get the co-efficients of * equal.

Again, it will often happen that a and a, have a common measure; and in this case, instead of a and a,, take the quotients of them by that common measure for the cross-multipliers. This will always, when it can be done, lessen the arithmetical labour. '

Let us take as a numerical example 4x + 6y= 2, and 10x — 3y = 59. Here in eliminating a, the co-efficients 4 and 10 have the common measure 2, and hence 5 and 2 are the multipliers. Hence we get 20x + 30y = 10, and 20x6y=118.

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Subtracting, we have 36y · 108, or y —— -3; and hence x = 5. Or again, multiply (2) by 2, and (1) by 1, then 4x+6y =2, and — 6y+ 20x = 118; and adding 24x = 120, or x = 5; and hence again y=— 3.

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4. In 3x — 4y = 38, and 4x + 3y = 9; x = 6, and y =

4x- 2y+3 18 -x+5y Ꮖ y

(+1)=(-2x+15) (+) { y

6. Given (x + 3) (y + '7) = (≈ + is) (y — •9) + 114, and 23 + 1 = 3 + 1;

to find x and y.


Ans. 51%, and y = 628.



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{2x−y + 15}

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find a and y3. Ans.

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a2 b2+a2 c2b2 c2



abc (ac


ac), and xy (ay + bx

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xy)=abc (x + y −c);

±√ ± √ ± ac, and y3 = ± √ ± √ ± bc.

In the examples given under the different rules for elimination, those have not always been chosen which are most simply solved by the rule under which they are given. This was purposely done for the sake of leading the student to examine the different questions by other of the rules, so as to afford him the means of judging in some degree from the appearance of any given equations, which rule will be most applicable to their solution. The improvement in his judgment will amply repay him for the trouble of solving each of them by all the methods.


THESE are in their nature and mode of solution precisely similar to those already treated of: they are, however, generally longer, and often the particular process to be employed is less obvious.

When it happens that the co-efficients of several of the equations are alike with the same or opposite signs, it will be found advantageous to add the equations together, and divide the sum by the number of them; then subtracting each equation from this quotient will give the value of each of the unknowns in succession. Sometimes when they are combined in factors, it will be advantageous to multiply or to divide only the others, and thus get equations in a simpler form. These, however, are special rules, and can only be acquired by observation and practice.

Ex. 1. Given x + y + z = 9, x + 2y + 3z = 16, and x + 3y + 4z = 21.
By the first method we have three values of x, viz. :

- 3y-4z.

x = 9—y — z, x = 16 — 2y — 3z, and x = 21 Equating the second value with each of the others, we have 9 y- z = 16 — 2y —3z, or y = 7 — 2z, and 162y3z21-3y- 4z, or y = 5—2.

Equating these values of y, we get 5—z=7

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Hence y 5-z=3, and x = =9—y―z=4.

Or, by the second method, we have from the first equation x=9—y — z ; and substituted in the two others gives

9+ y +22= 16, or y = 7—2z; and 9 + 2y + 3z=21. Substitute the former of these in the latter; then we get 23-z=21, or z = 2; and x and y will be found as in the last case.

Again, by the third method, the co-efficients of a are already equal; hence the equations are prepared for subtraction. Let (1) be taken from (2), and (2) from (3), then y + 2z = 7, and y + z = 5.

The coefficients of y are here equal, hence subtracting, z = 2; and hence ≈ and y may be found.

Ex. 2. Given x + y = 10, y + z = 23, and z + x = 19.

Add all three together and divide by 2: then we get

x + y + z = 26.

Subtract each of the given equations from this, and we find
x=3, y=7, and z = 16.

The student should also solve this by the other methods.

Ex. 3. Given xy= a2, yz = b2, and zx = c2.

Multiply all three together: then x2y2z2 = a2b2c2, or xyz =


Divide this by each of the given equations: then there will result

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This example is an instance of the remark on classification at p. 170; and would, in strict theory, like some that have gone before, be classed as a quadratic.

Ex. 4. There are given the three following equations for solution :

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a,,x + by + c11z = d12... (3)

Multiply (1) by a, and (2) by a, and subtract: then we get

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a,d2 - ad2

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Multiply (2) by a,, and (3) by a,, and subtract: then we have
(a,b, — a,b,) y + (a,c,
Multiply (4) by a,bab, and (5) by ab

- a,c,) = a, d2 — a,d2 .... (5)

ab, and subtract: then

resolving for z, and substituting in the values of x and y, we have

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This is the general solution for three unknowns, and by substituting any given numbers for the co-efficients in these, the corresponding values of x, y, z, would be obtained.

Ex. 5. Equations of the following forms are of very frequent occurrence in the subsequent parts of algebra, and hence it may be desirable to indicate the best mode of resolving them.





+ 13u + 12x + 11y + z = a1
+23u + 22x + 21y + z = ɑ2
+33u + 32x + 31y + z = αs
+43u + 42x + 41y + z = a1


where there are as many such equations as there are unknowns; and the second side of each equation given.

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Subtract the first from the second, the second from the third, the third from the fourth, and so on to the end. This will give n - 1 equations clear of z. Pursue the same course with these n - 1 equations, and we shall obtain n equations clear of z and y. Pursue again the same course, and n- 3 equations will be obtained clear of z, y, and x. Proceeding thus, we shall at last obtain a single equation involving only the letter to the left, whose value is thus found. Substitute this value in either of the two results obtained by the above process immediately before the last, and we obtain the value of the second letter. Substitute these two in either of the three next preceding results, and we shall get the value next unknown. We may thus obtain the whole very simply and conveniently.

The following example, adapted to four unknowns, may serve to illustrate the process and mode of writing the successive steps of the work.

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Whence u = 11; which substituted in either of the second differences (that with the least co-efficient will of course be most convenient) gives x = 31 · би — — 35; and these values of u and x in the first difference gives y = 22 7u22 + 105 77 50; and lastly, z = 2 — y — x


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2. If

3. If x

(x x +
x +

u+ x + y +2= 1 8u+4x+2y+2= 5 27u+9x+3y+2=14 64u+16x+4y+z=30

Ans. u=},x=}, y=1, z=0.


· x + y + z = 18

x + 3y+2z = 38 1x + by + 1/2 = 10

+ 12 = 27

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; then x4, y = 6, z = 8.

- 20 ; then x = 1, y = 12, z = 60. · x + ‡y + z = 16

4. When

u = 2

y = 2, x-2= 3, and y + z = 9; then x = 7, y = 5, z = 4.

x (x + y + z) = 45)

y (x + y + z) = 75
z (x + y + z) = 105,

; then x=3, y= ± 5, and z = ± 7.

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6. Given x + y + z = a, my = nx, and pz = qx; to find x, y, z.

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9. Given x (y + z) = a2, y (x + z) = b2, and z (x + y) = c2; to find the unknowns.

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11. Given xy = a (x + y), xz = b (x + z), and yz = c (y + z); to find the reciprocals of x, y, z.

́u + x + y + z = 4

12. Given

ux + uy + uz + xy + xz + yz = 6
uxy + uxz + uyz + xyz= 4

; to find u, x, y, z.

A COLLECTION OF QUESTIONS PRODUCING SIMPLE EQUATIONS. QUEST. 1. To find two numbers, such, that their sum shall be 10, and their difference 6.

Let x denote the greater number, and y the less *.

Then the first and second conditions are x + y = 10, and x-y=6, x = 10 -y; whence x = 8, y = 2.

QUEST. 2. Divide 1007 among A, B, C, so that A may have 201 more than B, and B 107 more than C.

Let A's share=x, B's=y, and C's = z.

Then x + y + z = 100, x = y + 20, and y =z + 10.

From which x= 50, y = 30, and z = 20.

QUEST. 3. A prize of 5001 is to be divided between two persons, so that their shares may be in proportion as 7 to 8; required the share of each.

Put x and y for the two shares; then the conditions are

7:8::x:y, or 7y = 8x, and x + y = 500; hence x = 233} and y = 266}. QUEST. 4. What fraction is that, to the numerator of which if 1 be added, the value will be ; but if 1 be added to the denominator, its value will be }? x + 1

Denote the fraction by then


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= 1, and

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QUEST. 5. A labourer engaged to serve for 30 days on these conditions: that for every day he worked, he was to receive 20d, but for every day he played, or was absent, he was to forfeit 10d. Now at the end of the time he had to receive just 20 shillings, or 240 pence. It is required to find how many days he worked, and how many he was idle?

Let x be the days worked, and y the days idled.
Then 20% are the pence earned, and 10y the forfeits;

Hence, by the question x + y = 30, and 20x — 10y= 240;
Whence x 18, the days worked; and y = 12, the days idled.

QUEST. 6. Out of a cask of wine which had leaked away 4, 30 gallons were drawn, and then, being gauged, it appeared to be half full: how much did it hold?

Suppose it held a gallons; then it leaked 4 gallons.


Hence there had been 4x + 30 gallons taken away, and by the question, x = x+30; and x = 120, the gallons it held.

In these solutions, as many unknown letters are always used as there are unknown numbers to be found, purposely for exercise in the modes of reducing the equations: avoiding the short ways of notation, which, though they may give neater solutions, afford less exercise in practising several rules in reducing equations. It is also considered unnecessary to carry out the solutions to their completion, as the steps are so familiar to the student from the exercise in reduction which has preceded. The examples, indeed, are given principally with a view to practice in the translation of the verbal conditions of a question into the symbolical language of algebra.

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