223. These are the simplest cases that arise, but they are specially important since the solution in a large number of other cases is dependent upon them. As a rule our object is to solve the proposed equations symmetrically, by finding the values of x+y and x-y. From the foregoing examples it will be seen that we can always do this as soon as we have obtained the product of the unknowns, and either their sum or their difference. (1), Example 1. Solve (2). Whence Multiply (2) by 2; then by addition and subtraction we have x2+2xy + y2=144, x+y=±12, x2+ y2=74 x-y=± 2. We have now four cases to consider; namely, x+y=12,) x+y= 12,) x+y=-12, x-y=- 2. x-y= 2. From which the values of x are 7, 5, -5, -7; and the corresponding values of y are 5, 7, -7, −5. x-y= 2. Example 2. Solve By subtracting (1) from the square of (2) we have 2xy=-104; x2+y2=185 .*. ..(3). xy= 52.. Equations (2) and (3) can now be solved by the method of Art. 222, Example 1; and the solution is EXAMPLES XXVIII. a. 2. x+y=51, x+y=-12, x-y= 5. x-y=8, 8. x-y=-8, 2. 3. x+y=74, 6. xy=1075, 9. x-y=-22, (1), (2). By division, from (2) by subtraction, xy=-1914, x+y=-65. 14. x2+y2-170, 15. 2+y2-65, xy=13. xy=28. 17. x+y=15, x2+ y2=125. 20. x2+y2=185, x-y=3. 23. x-y=3, 26. 1 1 x y = 2, x3-y3=999 x-y= 3 x2+xy+ y2=333 x2 - 2xy + y2= 9; 3xy=324, xy=108 224. Any pair of equations of the form x2±pxy+ y2=a2. 18. 21. x= ±7, ±2,) y= ±2, ±7.S 24. Dividing (1) by (2) From (2) and (3) by addition, x2+y2=53; by subtraction, xy=14; whence 27. 29. x2+pxy+y2=p+2, x=12, or From (2) and (4) y= 9, or Example 2. Solve x+x2y2+y1=2613........ x2+xy+y2= 67.... x2-xy+y2= 39... x-y=4, where p is any numerical quantity, can be reduced to one of the cases already considered; for by squaring (2) and combining with (1), an equation to find xy is obtained; the solution can then be completed by the aid of equation (2). Example 1. Solve 9,) 12.) .(1), .(2), (1), (2). (3); (4). .(1), .(2). .(3). [Art. 223, Ex. 1.] and or 1 + 1 x 12 By division, +==1}. HOMOGENEOUS EQUATIONS Of the Same Degree. 225. The following method of solution may always be used when the equations are of the same degree and homogeneous. Example. Solve 20. 1 Put y=mx, and substitute in both equations. Thus x2+xy+2y2=74. 2x2+2xy + y2=73.. 1+m+2m2 74 = ; .. 73+73m+146m2=148+148m+74m2; 1 ყვ 64 × 74 74 y =91, =74; =1. =64; and substitute in either (3) or (4). x2 22 ( 1 - 3 + 161) 5 50 .. x2 .. x=8; 5 (ii) Take m=ğ; then from (3), x2 (1+5+50) = =74, x2= 74×9 =9; 5 ... x=±3; •'. y=mx=3x=±5. .(1), .(2). (3), ..(4). The student will notice that, having found the values of x, we obtained those of y from the equation y = mx, using, in each case, the value of m employed in obtaining those particular values of x. 226. When one of the equations is of the first degree and the other of a higher degree, we may from the simple equation find the value of one of the unknowns in terms of the other, and substitute in the second equation. Example. Solve From (1) we have and substituting in (2), 3x-4y=5 3x2-xy-3y2-21. and by substituting in (1), whence and Thus, x= 3(5+4y)2 y(5+4y) -3y2=21; 3 9 9y2+105y-114=0, .. (y-1) (3y+38)=0; Put then from (2) we obtain Substituting in (1), 5+4y. ; SYMMETRICAL EQUATIONS. 227. The following method of solution may always be used when the given equations are symmetrical, that is, when the unknown quantities in each equation may be interchanged without destroying the equality. The same method may generally be employed with advantage where the given equations are symmetrical except with respect to the signs of the terms. Example. Solve x+y=82. x=u+v, and y=u-v; x=u+v=3, −1, 1± √−10; y=u-v=1, -3, −1±√−10. (1), (2). · (1), . (2). NOTE. We may assume x+y=2u and x-y=2v, u and v being any unknown quantities, whence we obtain x=u+v, and y=u-v, the values used in the above. |