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19. Find two numbers such that the one may be n times as great as the other, and their sum equal to b.
20. A man agreed to work a days on these conditions : for each day he worked he was to receive c cents, and for each day he was idle he was to forfeit d cents. At the end of a days he received m cents. How many days was he idle ?
21. A sum of money is divided among three persons; the first receives a dollars more than a third of the whole sum ; the second receives b dollars more than a half of what remains; and the third receives c dollars, the amount which is left : find the original
22. Out of a certain sum a man paid $96; he loaned half of the remainder, and then spent one-fifth of what he had left. After these deductions he still had one-tenth of the original sum. How much had he at first ?
23. A man moves 12 miles in an hour and a half rowing with the tide, and requires four hours to return, rowing against a tide one-quarter as strong: find the velocity of the stronger tide.
24. A man moves a miles in b hours, rowing with the tide, but requires c hours to return, rowing against a tide d times as strong as the first : find the velocity of the stronger tide.
25. A has a certain sum of money from which he gives to B $4 and one-sixth of what remains ; he then gives to C $5 and onefifth of what remains, and finds that he has given to each an equal amount. How many dollars had A, and how many dollars did B receive?
26. The fore-wheel of a carriage is a feet, and the hind-wheel is b feet in circumference. What is the distance passed over when the fore-wheel has made c revolutions more than the hind-wheel ?
27. In a certain weight of gunpowder the nitre composed 10 pounds more than two-thirds of the weight, the sulphur 41 pounds less than one-sixth, and the charcoal 54 pounds less than one-fifth of the nitre. What was the weight of the gunpowder ?
28. Two-thirds of A's money is equal to B's, and three-fourths of B's is equal to C's; together they have $650. What amount has each ?
,153. CONSIDER the equation 2x + 5y=23, which contains two unknown quantities. From this we get
23-2.x that is, y=
5 From this it appears that for every value we choose to give to x there will be one corresponding value of y. Thus we shall be able to find as many pairs of values as we please which satisfy the given equation. Such an equation is called indeterminate.
21 For instance, if x=1, then from (1) y=
27 Again, if x= -2, then
and so on.
5 But if also we have a second equation of the same kind expressing a different relation between x and y, such as
24-3.x we have from this Y=
4 If now we seek values of x and y which satisfy both equations, the values of y in (1) and (2) must be identical.
23 - 2.x 24 - 3.0 Therefore
4 Multiplying up, 92 - 8x=120 – 15x;
.. y=3, and
-4. S Thus, if both equations are to be satisfied by the same values of x and y, there is only one solution possible.
154. DEFINITION. When two or more equations are satisfied by the same values of the unknown quantities they are called simultaneous equations.
155, In the example already worked we have used the method of solution which best illustrates the meaning of the term simultaneous equation; but in practice it will be found that this is rarely the readiest mode of solution. It must be borne in mind that since the two equations are simultaneously true, any equation formed by combining them will be satisfied by the values of x and y which satisfy the original equations. Our object will always be to obtain an equation which involves one only of the unknown quantities.
156. The process by which we cause either of the unknown quantities to disappear is called elimination. It may be effected in different ways, but three methods are in general use: (1) by Addition or Subtraction ; (2) by Substitution; and (3) by Comparison.
ELIMINATION BY ADDITION OR SUBTRACTION.
157. Example 1. Solve 7+2y=47
(1), 5x – 4y= 1.
(2). Here it will be more convenient to eliminate y.
Multiplying (1) by 2, 14x+4y=94, and from (2)
5x – 4y=1; adding,
... x=5. To find y, substitute this value of x in either of the given equations. Thus from (1)
x=5. In this solution we eliminated y by addition.
Example 2. Solve 3x + 7y=27
(2). To eliminate a we multiply (1) by 5 and (2) by 3, so as to make the coefficients of x in both equations equal. This gives
15x + 6y=48; subtracting,
29y=87; .: y=3.
To find x, substitute this value of y in either of the given equations. Thus from (1)
y=3. In this solution we eliminated x by subtraction.
RULE. Multiply, when necessary, in such a manner as to make the coefficients of the unknown quantity to be eliminated equal in both equations. Add the resulting equations if these coefficients are unlike in sign; subtract if like in sign.
ELIMINATION BY SUBSTITUTION.
158. Example. Solve 2x - 5y= 1..
7x+3y=24 Transposing – 5y in (1), and dividing by 2, we obtain
2 Substituting this value of x in (2) gives
7 (5+2) +3y=24.
.. y=1. This value substituted in either (1) or (2) gives
I=3. RULE. From one of the equations, find the value of the unknown quantity to be eliminated in terms of the other and known quantities; then substitute this value for that quantity in the other equation, and reduce.
Placing these values of x equal to each other, we have
X=8. RULE. From each equation find the value of the unknown quantity to be eliminated in terms of the other and known quantities; then form an equation with these values and reduce.
EXAMPLES XIX. a. Solve the equations : 1. 3x+4y=10, 2. x+2y=13, 3. 4.x+y=29, 4.x+y= 9. 3.x+y=14.
30+3y=11. 4. 2.30 — y= 9,
5. 5c+6=17, 6. 2x+y=10, 3x – 7y=19.
6.0 +5y=16. 7x+8y=53. 7. 8x - Y=34, 8. 15x+7y=29, 9. 14x – 3y=39, x+8y=53.
9x+15=39. 6.x+17y=35. 10. 23x – 23y=33, 11. 35x+17y=86, 12. 15x+779=92,
63. - 25y=101. 56x – 13y=17 55x – 33y=22. 13. 58-7y=0, 14. 21x – 50y=60, 15. 393 — by=99, 7x+54=74
28x – 27y=199. 52x – 15y=80. 16. 5x=7y – 21, 17. 6 – 50=18, 18. 8x=5y, 21x – 9y=75.
12x – 9y=0.
13x=8y +1. 19. 3x=7y,
20. 19x+17y=0, 21. 93.0+ 15y=123, 12y=5x – 1,
2.c - y=53. 15x +93y=201.
160. We add a few cases in which, before proceeding to solve, it will be necessary to simplify the equations. Example 1. Solve 5 (x+2y) – (3x+11y)=14
(1), 7x – Oy - 3 (x – 4y)=38
(2). From (1)
5x+10y - 3x – 11y=14;
(3). From (2)
7x – 9y – 3x +12y=38;
. . 4.x + 3y=38 From (3)
6x – 3y=42 and hence we may find x=8, and y=2.