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y – 1
Transpose the fractions to the left side of the equation and the integral forms to the right.
8 – X – y
3 As x and y must be positive integers, the right side of the equation has an integral value, hence the left side must also be equal to an integer.
Let m be some integer. Then
3 Solving for y, y
y 3m + 1. Substitute this value of y in the original equation.
3x + 4(3m + 1) = 25 Solve for x, 3x 25 12m
- 4 =
21 12m .. c = 7 – 4m
3m +1. Assign values to m to find positive integral values of x and y, when m = = 0, x =
1 1, x
4 when m > 1, x becomes negative and when m < 0, y becomes negative; hence we have only two sets of positive integral value for x
7, y 3, y
2. Solve in positive integers, 5x + 8y = 34.
5 Зу 4 Transpose,
=-6 x - y
5 3y 4
must be integral.
5 Let Зу
5 Then 3y
5m + 4
3 This value of y is not an integral form because the coefficient
3y - 4 of y in the fraction is not unity.
Зу 4 If the fraction is multiplied by an integer, it will still be
5 equal to an integer; hence multiply the fraction by a number that will make the coefficient of y one more than a multiple of the denominator.
Multiply the fraction by 2,
3 Hence, to be integral, the fraction must be equal to some
Y – 3
-5 Then y
3 = 5m and
5m + 3
5x 34 – 40m 24
3 + 5m When m 0, x = 3.
Any value of m > 0 will make x negative and any value of m < 0 will make y negative. Therefore, the only positive integral values 2, y
and y 2, y
are x =
Solve in positive integers: 1. 4x + 5y = 26.
6. 2x + y
3. 2. 5x + 2y = 39.
7. 7x + 5y
21. 3. 3x + 5y = 50.
8. x + 2y 9. 4. 2x + 3y 43.
9. 3x + 7y 16. 5. 3x + 7y 50.
10. 2.c + 5y = 13. Solve in least positive integers: 11. 5x 2y 15.
14. 7x 2y 7y = 8.
39. 13. 3x 4y 17.
26. 4x + 3y + 2z = 25 5x + 2y 32 160 17.
52 45. 19. A man spent $28 in buying shirts, some at $2 and some at $3. How many of each kind did he buy?
20. A coal dealer sold 10 tons of pea coal, 12 tons of nut coal and 5 tons of egg coal for $245. The next day, at the same price, he sold 8 tons of pea, 5 tons of nut, and 10 tons of egg coal for $204. Find the price of each.
21. In how many ways can a man pay a bill of $53 using $2 bills and $5 bills?
22. Find the smallest number that when divided by 4 or divided by 7, leaves a remainder of 3.
183. Graphical Solution of Indeterminate Equations. Any indeterminate equation of the first degree may be represented graphically by a straight line.
The positive integral values of x and y are then determined by noting at what points the line crosses the
intersections of the ruled lines of the graph paper, in the first quadrant, since the values of x and y are both positive in the first quadrant only.
Thus, the equation of Problem 2, yields the graph shown in Fig. 1. When x = 3, the line cuts the corner of one of the squares. Since this is the only point in the first quadrant, the equation 5x + 8y = 34 has only one set of positive integral values of x and y.
Note.—The success of this solution depends entirely upon the accuracy of the drawing. The line should be accurately determined by locating two points some distance apart.
The equations of Exercise 134 furnish ample material for the application of this method.
REVIEW EXERCISE XXV
y + 42
1. In how many ways may $2.75 be paid with quarters and half dollars?
2. Find the smallest number that, when divided by 5 or divided by 7, gives 4 for a remainder. 3. Solve in positive integers, 4x + 7y
= 94. 4. A farmer spends $752 in buying horses and cows; if each horse cost $37 and each cow $23, how many of each kind does he buy?
5. Find two fractions whose denominators are 9 and 5 respectively, and whose sum shall be equal to 438. 6. Solve in positive integers 5x
2x + 3y 52 -8. 7. A party of 20 people, consisting of men, women, and children, pay a hotel bill of $67. Each man pays $5, each woman $4 and each child $1.50. How many of each were there?
8. Solve in positive integers, 8x + 5y = 107.
9. A woman expended 93 cents for 14 yards of cloth, some at 5, some at 7, and the rest at 10 cents a yard. How many whole yards of each did she buy?
10. What is the least number that will contain 25 with a remainder of 1, and 33 with a remainder of 2?
DISCUSSION OF EQUATIONS
184. Character of the Roots of a Quadratic Equation. Every quadratic equation may be reduced to the form
ax2 + bx + c = 0 in which a is positive, while b and c may be positive or negative.
If the roots of this general form are denoted by rıand r2 then, as previously shown, - b + 12 4ac
- b-V62 ri 2a
2a These values of the two roots show that the character of these roots, as real or imaginary, rational or irrational, equal or unequal, depends upon the value of ✓ 12 – 4ac.
1. To determine whether the roots are real or imaginary.
If the quantity under the radical sign is positive, the roots are real, if negative, the roots are imaginary. Hence, when b2 - 4ac is positive, the roots are real,
when b2 - 4ac is negative, the roots are imaginary. 2. To determine whether the roots are rational or irrational. If the roots are imaginary, they are also irrational.
If the roots are real, they may be rational or irrational, according to whether the quantity under the radical sign is a perfect square or not.
Hence, when 62 4ac is a perfect square or equal to zero, the roots are rational; otherwise, they are irrational.
3. To determine whether the roots are equal or unequal.
It is evident that the two roots rı and r. differ from each other in that the radical is added to -b in the one and subtracted from –b in the other. If the radical vanishes, the difference between the two roots vanishes and the roots are equal.