1 CASE III. Any equation may be cleared of fractions, by multiplying each of its terms, successively, by the denominators of those fractions, or by multiplying both sides by the product of all the denominators, or by any quantity that is a multiple of them. + =5, then, multiplying by 3, we have x+ =15; and this, multiplied by 4, gives 4x+3x=60; 3x 4 Thus, if 13 4 60 4 whence, by addition, 7x=60, or x= =8 And, if + = 10; then, multiplying by 12, (which 4 6 is a multiple of 4 and 6,) 3x+2x=120, or 5x=120, or It also appears, from this rule, that if the same number, or quantity, be found in each of the terms of an equation, either as a multiplier or divisor, it may be expunged from all of them, without altering the result. Thus, if ax=ab+ac; then, by cancelling, x=b+c. b a a CASE IV. If the unknown quantity, in any equation, be in the form of a surd, transpose the terms so that this may stand alone, on one side of the equation, and the remaining terms on the other (by Case 1); then involve each of the sides to such a power as corresponds with the index of 1 the surd, and the equation will be rendered free from any irrational expression. Thus, if x-2=3; then will √x=3+2=5, or, by squaring, x=52=25. And if 3x+4=5; then will 3x+4=25, or 3x= 25-4-21, or x= 21 Also, if 2x+3+4=8; then will 32x+3=8-4 =4, or 2x+3=43=64, and consequently 2x=64-3 If that side of the equation, which contains the unknown quantity, be a complete power, the equation may be reduced to a lower dimension, by extracting the root of the said power on both sides of the equation. Thus, if x2=81; then x=81=9; and if x3=27, then x=27=3. 33 Also, if 3x2-9=24; then 3x2=24+9=33, or x2= 3=11, and consequently x=√11. And, if x2+6x+9=27; then, since the left hand side of the equation is a complete square, we shall have, by extracting the roots, x+3=27=√9×3=3,3, or x=33-3. CASE VI. Any analogy, or proportion, may be converted into an equation, by making the product of the two extreme terms equal to that of the two means. Thus, if 3x: 16::5:6; then 3x×6=16×5, or 80 40 18x=80, or = 4 = 4 And, if 2x :a::b:c; then will 3ab; or, by division, x= 3ab Also, if 12-x: ::4: 1; then 12-x or 2x+x=12; and consequently x= 12 MISCELLANEOUS EXAMPLES. 1. Given 5x-15=2x+6, to find the value of x. Here 5x-2x=6+15, or 3x=6+15=21; and there 21 forex=7. 3 : 2. Given 40-6x-16=120-14x, to find the value of x. Here 14x-6x=120-40+16; or 8x = 136-40 3. Given 3x2-10x=8x+x2, to find the value of x. Here 3x-10=8+x, by dividing by x ; or 3x-x=8 +10=18, by transposition. And consequently 2x=18, or x=== 4. Given 6ax3-12abx2=3ax3+6ax2, to find the value of x. Here 2x-46=2+2, by dividing by 3ax2; or 2x-x =2+46; and therefore x=46+2. 5. Given x2+2x+1=16, to find the value of x. Here x+1=4, by extracting the square root of each side. And therefore, by transposition, x=4-1=3. 6. Given 5ax-36=2dx+c, to find the value of a. د Here 5ax-2dx=c+3b; or (5a-2d)x=c+3b; and therefore, by division, x = c+36 5a-2d 7. Given XX +-=10, to find the value of x. 234 6x 2x 2x Here x-+-=20 ; and 3x-2x+ = 60; or 4 12x-8x+6x=240; whence 10x=240, or x=24. 2x 13 3 Here x-3+=40-x+19; or 3x-9+2x=120 3 3x+57; whence 3x+2x+3x=120+57+9; that is 3x=186, or x=231. 2x 9. Given √3+5=7, to find the value of x. 2x Here =7-5=2; whence, by squaring, 3 =4, and 2x=12, or x=6. 10. Given x + √(a2+x3)= to find the value of x. √(a2+x2) Herex(a2+x2)+a2+x3 = 2a2; or x√(a2+x2) = a2 - x2,and x2 (a2+x2)=a4-2a2x2+x4; whence a2x2 + x=a4-2a2x2+x4, anda2x2= a2-2a2x2; therefore3a2x2 EXAMPLES FOR PRACTICE. 1. Given 3x-2+24=31, to find the value of x. Ans. x=3 2. Given 4-9y=14-11y, to find the value of y. Ans. y=5 3. Given x+18=3x-5, to find the value of x. Ans. x=11 2+3=11, to determine the value of x. Ans. x=6 5. Given 2x+1=5x-2, to find the value of x. X X 7 23 4 10' 6. Given t Ans.x= to determine the value of x. x-5 4 6 7 -, to find the value of x. Ans. x=3 x=36 8. Given 2+3x=4+5x, to find the value of x. 13 Ans. x=12 Ans. x= a to find the value 4 32 3 36 of x. Ans. x= 3a-26 12 |