Great part of the business of algebra consists in deducing, from given equations, the values of the unknown quantities which they contain. This is effected by making correspondent changes in both the members of the equation, so as to obtain, eventually, the unknown quantity on one side of the sign of equality, and known quantities on the other. SIMPLE EQUATIONS, Containing one unknown quantity. To facilitate the solution of equations generally, the following principles or modes of operation must be observed. MODE 1. 50. Any quantity may be transposed, that is, removed from one member of the equation to the other, by changing its sign.* EXAMPLES. Given 7x+5a=6x+3b to find x. By transposing 5a and 6x, we have 7x-6x=3b-5α, or x=3b-5a. 2. Given 8y-3-7y+5 to find y. Result, y=8. 3. Given 15x-6=14x+20, to find x. Result, x=26. 4. Given 20-13y=45-14y+4, to find y. Result, y=29. 5. Given 8x+15=15+9x-17, to find x. Result, x=17. * Hence, if any term be contained in both members of the equation with the same sign, it may be taken from both; and the signs of all the terms, may be changed, without destroying the equation. 6. Given 25x-b-24x+2b-c, to find x. Result, x=36-c. 7. Given 36—9x=7b—8x-4d, to find x. MODE 2. Result, x=4d-4b. 51. The co-efficient of an unknown quantity may be taken away, by dividing every term of the equation by it. EXAMPLES. Given ax+ba=ad, to find x. By dividing by a, we have x+b=d; hence by transposition, x=d-b. 2. Given 5x=25, to find x. Result, x=5. 3. Given 30x=180, to find x. Result, x=6. 4. Given 6x+17-44-3x, to find x. 5. Given 3ax-4ab-2ax-6ac, to find x. Result, x=3. Result, x=4b-6c. 6. Given 3x-10x=8x+x, to find x. Result, x=9. 7. Given 6y3—8y2=22y3—3y3, to find y. Result, y=31. MODE 3. 52. An equation may be cleared of fractions by multiplying all its terms by any common multiple of the denominators. Multiply by 30, and then we have 15x-10x+6x=55; whence 11x=55, and x=5. 53. When the unknown quantity is under a radical sign, such sign may be taken away, by first transposing the terms, so as to leave the radical quantity alone on one side of the equation, and then involving each member of the equation to the proper power. EXAMPLES. Given x+40=10-/x, to find x. Squaring both sides, x+40=100—20x+x. Squaring both members, x=9. 2. Given x+1—2—3, to find x. Result, x=24. 3. Given 5x+4+3=6, to find x. Result, x=7. 4. Given x16—8—/x, to find x. Result, x=25. 5. Given 4a+x=2✅b+x~x, to find x. 2 6. Given 4a2+x2='4b*+x, to find x. Result, x= 7. Given x+a+√2ax+x2=b, to find x. 54. When the member of an equation, which contains the unknown quantity, is a complete square, cube, &c. the equation may be reduced by extracting the square root, cube root, &c. of both its members. EXAMPLES. Given x3-6x+12x-8=343, to find x. Extracting the cube root, x-2=7, or x=9. 2. Given x2+4x+4=81, to find x. Result, x=7. Result, x-bα. Result, x=3. 4. Given +3ax2+3a2x+a3=b3, to find x. 5. Given 9x+17=98, to find x. 6. Given x-4x+6x2-4x+1=256, to find x. 4. Given a+x=√a2+x√4b2+x3, to find x. 5. Given 2+3x=√4+5x, to find x. α 7. Given ✔a+x+√a−x=✅✔✅ax, to find x. 13. Given x+2x2+1=4225, to find x. Result, x=8. * If four numbers be proportional, the product of the extremes is equal to that of the means; hence, an analogy is readily converted into an equation. Vide Article 62. |