36. Simplify y2 1 2(x-1) 4(x-1) 4(x+1)(x-1)(x+1) 1 1 1 abxa (a−b) (x − a) * b (b− a) (x−b) ̄ ̄ ̄x (x−α) (x — b)° b) (x − a) + b (b − a) (b−a) (x−b) − − x (x − a) (x — t 38. If s=a+b+c+... to n terms, shew that -y2x2. ...)-n. 1 -3 1+ 40. Simplify α-x 41. Divide 3+/−3 (~-~o) + 4 (x + 1) by x+1. 42. If s=a+b+c+... to n terms, shew that a2+x a2-x2 a2+x2 p*+4p3q+6p2q2+4pq3+q* . p3+3p3q+3pq2+q3 2 p*-4p3q+6p2q3-4pq3+q 207. To determine several unknown quantities we must have as many independent equations as there are unknown quantities. Thus, if we had this equation given, x+y=6, we could determine no definite values of x and y, for or other values might be given to x and y, consistently with the equation. In fact we can find as many pairs of values of x and y as we please, which will satisfy the equation. We must have a second equation independent of the first, and then we may find a pair of values of x and y which will satisfy both equations. Thus if besides the equation x+y=6, we had another equation x-y=2, it is evident that the values of x and y which will satisfy both equations are x=4) since 4+2=6, and 4-2=2. Also, of all the pairs of values of x and y which will satisfy one of the equations, there is but one pair which will satisfy the other equation. We proceed to shew how this pair of values may be found. 208. Let the proposed equations be 2x+7y=34 5x+9y=51. Multiply the first equation by 5 and the second equation by 2, we then get 10x+35y=170 10x+18y=102. The coefficients of a are thus made alike in both equations. If we now subtract each member of the second equation from the corresponding member of the first equation, we shall get (Ax. II. page 58)* We have thus obtained the value of one of the unknown symbols. The value of the other may be found thus: Take one of the original equations, thus Hence the pair of values of x and y which satisfy the equations is 3 and 4. NOTE. The process of thus obtaining from two or more equations an equation from which one of the unknown quantities has disappeared is called Elimination. 209. We worked out the steps fully in the example given in the last article. We shall now work an example in the form in which the process is usually given. Ex. To solve the equations 3x+7y=67 5x+4y=58. Multiplying the first equation by 5 and the second by 3, Hence x=6 and y=7 are the values required. 210. In the examples given in the two preceding articles we made the coefficients of x alike. Sometimes it is more convenient to make the coefficients of y alike. Thus if we have to solve the equations Y 29x+2y=64 we leave the first equation as it stands, and multiply the second equation by 2, thus Hence x=2 and y=3 are the values required. |