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LITERAL SIMULTANEOUS EQUATIONS. 163. Example 1. Solve ax+by=C

(1), a'x+b'y=c

(2). The notation here first used is one that the student will frequently meet with in the course of his reading. In the first equation we choose certain letters as the coefficients of x and y, and we choose corresponding letters with accents to denote corresponding quantities in the second equation. There is no necessary connection between the values of a and a', and they are as different as a and b; but it is often convenient to use the same letter thus slightly varied to mark some common meaning of such letters, and thereby assist the memory. Thus a, a' have a common property as being coefficients of x; 6, bl as being coefficients of y.

Sometimes instead of accents letters are used with a suffix, such as di, A2, A3 ; 61, 62, 63, &c.

To return to the equations ax+by=C

.(1), alc+b'y=c

(2). Multiply (1) by b' and (2) by b. Thus

ab'x+-bb'y=b'c,

a'bx+bb'y=bc'; by subtraction,

(ab' -alb)x=b'c-bc';
b'c-bc

. (3).

ab' - alb As previously explained in Art. 157, we might obtain y by substituting this value of x in either of the equations (1) or (2); but y is more conveniently found by eliminating x, as follows: Multiplying (1) by a' and (2) by a, we have

aa's+a'by=a'c,

aa'x + ab'y=ac' ; by subtraction

(a'b ab')y=a'c- ac' ;

a'c-ac .. Y=

a'b - ab or, changing signs in the terms of the denominator so as to have the same denominator as in (3),

ac! - a'c

b'c-bc
y=
ab' -a'l'

ab' - ab

and x =

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From (1) by clearing of fractions, we have

(C-5)-a(c-)+y(c-a-b(c-a)=(c-a) (c-b), 2(0–6)+y(c-a)=ac- ab + bc-ab+c2-ac-6c+ab, c(cb)+(^-a)=co-ab .

(3). Again, from (2), we have (a-b)+a(a-b)+cy-cara(a-6) 2 (ab)+cy=ac.

(4). Multiply (3) by c and (4) by c-a and subtract, x{c(c-b)-(c-a)(2-6)}=c3 – abc-ac(c-a), (c2 — ac+a? ab)=C(c2 - ab-ac+a%);

.: =; and therefore from (4)

y=b. H. A.

10

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(x+y) (m2 +22)=2 (m3 +73) +ml (x+y). 19. bx+cy=a+b 1 1

+ cy a-6

ta 20. (a - b)x+(a+)y=2(a2- 62)

ax by=a+ 62.

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CHAPTER XX.

PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS.

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(2)

164. In the Examples discussed in the last chapter we have seen that it is essential to have as many equations as there are unknown quantities to determine. Consequently in the solution of problems which give rise to simultaneous equations, it will always be necessary that the statement of the question should contain as many independent conditions as there are quantities to be determined.

Example 1. Find two numbers whose difference is 11, and one-fifth of whose sum is 9.

Let » be the greater number, y the less;
Then
X-Y=11.

(1), Also

*

5

x+y=45. By addition 2x=56 ; and by subtraction 2y=34. The numbers are therefore 28 and 17.

Example 2. If 15 lbs. of tea and 10 lbs. of coffee together cost $15.50, and 25 lbs. of tea and 13 lbs. of coffee together cost $ 24.55, find the price of each per pound. Suppose a pound of tea to cost x cents

and a pound of coffee to cost y cents. Then from the question, we have 15x+10y=1550

.(1), 25x+13y=2455.

(2). Multiplying (1) by 5 and (2) by 3, we have

75x + 50y=7750,

75x+39y=7365. Subtracting,

1ly=385,

y=35. And from (1) 15x+350=1550. Whence

15x=1200 ;

.: x=80. •. the cost of a pound of tea is 80 cents, and

the cost of a pound of coffee is 35 cents.

cents ;

or

and apples cost

or

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Example 3. A person spent $6.80 in buying oranges at the rate of 3 for 10 cents, and apples at 15 cents a dozen ; if he had bought five times as many oranges and a quarter of the number of apples he would have spent $ 25.45. How many of each did he buy ? Let be the number of oranges and y the number of apples.

10x 3 oranges cost

cents, 3

157
y apples cost

12
10x 15y
+ =680

(1),
3 12
10

50x Again, 5x oranges cost 5x x

3'

cents,

3 y

y

15 157 Х

cents ;
4 12' 48

50x 15y
+ =2545.

(2) 3 48 Multiply (1) by 5 and subtract (2) from the result;

75 15 then

y=855; 12 48

2854

48

.: y=144 ; and from (1)

X=150.
Thus there were 150 oranges and 144 apples.

Example 4. If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes equal to ; and, if the numerator and denominator are each diminished by 1, it becomes equal to į: find the fraction.

Let be the numerator of the fraction, y the denominator; then the fraction is

Y
From the first supposition,
X+2 5

..(1),

8 from the second, -1 1

(2).

2 These equations give x=8, y=15.

8 Thus the fraction is

15

or

=855;

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y+1

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y-1

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