Putting these two values of x2 equal to each other, ±4, or ±7, (263). The From this Equation y will be found 4. Find the value of x and y in the Equations By transposition in the first Equation, x=14-4y. Substituting this value of x in the second Equation, 56-18y+y211; hence y=3, or 15. We shall now easily find x=2, or -46. 5. Find the values of x and y in the Equations x+2y=7, and x2+3xy—y2=23. Transposing 2y in the first Equation, x=7—2y. Hence 22-49-28y+4y2, and 3xy=21y-6y2. Substituting these values of x2 and 3xy in the second Equation, and adding similar terms in the first member of the resulting Equation, 49-7y-3y2-23; hence y=2, or -41. Substituting these values of y, successively, in the first Equation, we shall find x=3, or 15%. 6. Find the values of x and y in the Equations y x+ 2/2 =11, and xy+2y2=120. Clearing the first Equation of its fraction, 2x+y= 22; hence y=22-2x. Then xy=22x-2x2, and 2y2=968—176x+8x2. Substituting these values of xy and 2y2 in the second Equation, transposing, and uniting similar terms, 6x2-154x=- from which x will be found equal to 8, or 173. The value of y will then be found to be 6, or -13. 7. Find the values of x and y in the Equations Hence 10x=10y-20, and 3xy-3y2-6y. Substituting these values of 10x and 3xy in the third Equation, and reducing, we find 17y-3y2-20; hence y=4, or 13. Substituting these values of y, successively in the first Equation, we find x=2, or —1. 8. Find the values of x and y in the Equations X x-y =4, and y =1. Clearing these Equations of fractions, and reducing each resulting Equation, we have x+y=8, and xy-y-2x=2. Transposing y in the third Equation, x=8-y; hence xy=8y-y2. Substituting these values of x and xy in the fourth Equation, and reducing, 9y-y2 = 18; which will give y=6, or 3; and we shall now find x=2, or 5. 9. Find the values of x and y in the Equations 5(x+y)=13(x−y), and x+y2=52. Developing both sides of the first Equation, and reducing, From this Equation we shall find y=4, or −64; and the value of x will then be found to be 9, or — 14. Solutions by Means of an Auxiliary Unknown Quantity. 10. Find the values of x and y in the Equations x2+xy 54, and 2xy+y2=45. Assuming x=vy, and substituting this value for x in each of the given Equations, we have v2y2+vy2=54, and 2vy2+y2=45. From the first of these Equations, Clearing this Equation of fractions, 108v+54 45v2+45v; hence Substituting these values of v, successively, for v in the fourth Equation, we have The first of these Equations will give y= ±3. The second of the two will give y=±√-225=±15√/−1. 3 =9-1. 5 11. Find the values of x and y in the Equations xy=28, and x2+y2=65. Substituting vy for x in each of these Equations, we have Substituting the first of these values for v in the third Equation, we shall find y±4. Substituting the second value for v in the same Equation, we shall find y±7. Then since x=vy, we shall have x=±4× 74 =±7; or x=±7 12. Find the values of x and У in the Equations x2+xy=12, and xy—2y2=1. Substituting vy for x in each of these Equations, v2y2+vy2=12, and vy2-2y2=1. From the first of these Equations, we shall find Substituting these values, successively, in the fourth Equation we find 13. Find the values of x and y in the Equations 4x2-2xy=12, and 2y2+3xy=8. Assuming y=vx, and substituting this value for x in the given Equations, 4x2-2vx2-12, and 2v2x2+3vx2=8. Finding the values of 22 from these Equations, and putting them equal to each other, |