Common School Algebra

as B; but if B had lent A $5, each would have had the How much money had each?

same sum.

11. Ten boxes of raisins and 3 barrels of flour weigh 850 lbs.; also, 12 boxes of raisins and 7 barrels of flour weigh 1700 lbs. Required the weight of a box of raisins, also that of a barrel of flour.

12. What fraction is that, to the numerator of which if 1 be added, the value of the fraction will be, but if 3 be subtracted from the denominator, the value of the fraction will be ğ?

ART. 75. 1. A boy bought an apple, an orange, and a peach, for 6 cents; 2 apples, 3 oranges, and 4 peaches, for 19 cents; 5 apples, 2 oranges, and 7 peaches, for 25 cents. What did he pay for one of each?

Let x = the price of an apple, y = that of an orange, and z=that of a peach, in cents. Then,

(1) x+ y + z= 6;

(2) 2x+3y+4x=19;

(3) 5x+2y+7z=25.

In this problem, we have three independent equations, containing three unknown quantities. Our first step, in the solution, is to obtain from these original equations two others, which shall contain only two unknown quantities. Let us eliminate x, that is, find two equations which shall not contain x.

First method. Bring down the 2d, and multiply the 1st by 2, so as to make the coefficient of x the same as it is in 2d;

(2) 2x+3y+4x+19;

(4) 2x+2y+2x=12. Subtract 4th from 2d,
(5) y+2x=7. { Multiply 1st by 5, and bring

(6) 5x+5y+5x=30;

down 3d,

(3) 5x+2y+7z=25. Subtract 3d from 6th,
(7) 3y-2x=5. Bring down 5th,

The 5th and 7th contain only the two unknown quantities ý and z. These equations may, therefore, be solved as we solved similar equations in the preceding section. Since the coefficients of z are alike in the 5th and 7th, by adding these equations, we have

4y=12; .. y=3 cents, price of an orange.

Substitute 3 for y in the 5th, which contains only y and z, 3+2x=7,..z=2 cents, price of a peach. Substitute 3 for y, and 2 for z, in the 1st, x+3+2=6,.. x=1 cent, price of an apple. Second method. Resume the original equations, (1) x + y + z= 6;

(2) 2x+3y+4x=19;

(3) 5x+2y+7x=25.

Find the value of x from each of the equations, as if y and z were known.

Put equal to each other the values of x in the 4th and

5th; also the values of x in the 4th and 6th,

Clear the 7th and 8th of fractions; then transpose the terms containing y into the first members, all the other terms into the second, and reduce,

and

(9) y= 7-2z; Change the signs in 10th, (10) -3y=-5-2z. § find the value of y, 5+22 Equalize the values of y in 9th

(11)

5+2x=7—2%,..5+2z=21—6 %, ..

z=2 cents, price of a peach.
Substitute 2 for z in 9th,

y=7-4=3 cents, price of an orange.
Substitute 3 for y, and 2 for z, in 4th,
x=6—3—2=1 cent, price of an apple.

Third method. Take the original equations, (1) x + y + z = 6;

(2) 2x+3y+4x=19;

(3) 5x+2y+7z=25.

Find the value of x from

the 1st, as if y and z were known.

-y-z. Substitute this value in 2d and 3d,

(4) x=6

(5) 12-2y-2x+3y+4z=19;)

(6) 30-5y-5x+2y+7x=25. Transpose the known terms, and reduce in the 5th and 6th,

Find the value of z in 7th,

7-y

Substitute this value in 8th,

-3y+7-y=-5,..-4y=-12,.. 4 y 12, and

y=3 cents, price of an orange.

Substitute 3 for y in 9th,

7-3

2 cents, price of a peach.

Substitute 3 for y, and 2 for z, in 4th,

x=6—3—2=1 cent, price of an apple.

of his of

2. A farmer found that the number of his sheep exceeded by 20 that of his cows and horses together; that his horses and of his cows equalled in number sheep; and that of his cows, of his horses, and his sheep, made 12. Required the number of each. Suppose he had z sheep, y cows, and z horses. Then (1) x=y+z+20;