75. Consider the equation 7x=14. It is required to find what numerical value x must have to Again, in the equation 7x-2x-x=23+15-10, by collecting terms, we have 4x=28. .. x= 7. Transposition of Terms. 76. To solve 3x-8=x+12. This case differs from the preceding in that the unknown quantity occurs on both sides of the equation. We can, however, transpose any term from one side to the other by simply changing its sign. This we proceed to show. Subtract x from both sides of the equation, and we get 3x-x-8=12, Adding 8 to both sides, we have 3x-x=12+8, [Axiom 2]. [Axiom 1]. Thus we see that +x has been removed from one side, and appears as -x on the other; and -8 has been removed from one side and appears as +8 on the other. Hence we may enunciate the following rule: Rule. Any term may be transposed from one side of the equation to the other by changing its sign. It appears from this that we may change the sign of every term in an equation; for this is equivalent to transposing all the terms, and then making the right and left hand members change places. or Example. Take the equation -3x-12= x-24. - x+24= 3x+12, which is the original equation with the sign of every term changed. Here it will be convenient to begin by clearing the equation of fractional coefficients. This can always be done by multiplying both sides of the equation by the least common multiple of the denominators. [Axiom 3.] 78. We can now give a general rule for solving any simple equation with one unknown quantity. Rule. First, if necessary, clear of fractions; then transpose all the terms containing the unknown quantity to one side of the equation, and the known quantities to the other. Collect the terms on each side; divide both sides by the coefficient of the unknown quantity, and the value required is obtained. Example 1. Solve 5(x-3) -7(6 − x) + 3 = 24 − 3(8 − x). Removing brackets, 5x-15-42+7x+3=24 −24+3x ; transposing, 5x+7x-3x=24-24+15+42-3; Example 2. Solve (x+1)(2x − 1) − 5x = (2x − 3)(x − 5)+47. 2x2+x-1-5x=2x2-13x+15+47. Erasing the term 21⁄22 on each side, and transposing, 79. It is extremely useful for the beginner to acquire the habit of verifying, that is, proving the truth of his results; the habit of applying such tests tends to make the student self-reliant and confident in his own accuracy. In the case of simple equations we have only to show that when we substitute the value of x in the two sides of the equation we obtain the same result. Example. To show that x=7 satisfies the equation (x+1)(2x-1)- 5x = (2x-3)(x − 5)+47. When x=7, the left side (x+1)(2x − 1) − 5x = (7+1)(14-1)-35 = (8 x 13) - 35 = 69. The right side (2x-3)(x-5)+47 = (14 − 3)(7 − 5) + 47 = (11 × 2) +47 = 69. Thus, since these two results are the same, x=7 satisfies the equation. 17. 9. 4x=18. 10. 12x=42. 13. 6x=26. 14. 0=11x. 15. 1=11x. 0=-2x. 18. 6x=3. 19. 5=15x. 11. 306x. 33. 34. 5x-17+3x-5=6x-7-8x+115. 7x-21-4x+13+2x=41-5x-7+6x. 35. 15-7x-9x-28 +14x-17=21-3x+13-9x+8x. 36. 5x-6x+30-7x=2x+10-7x+5x - 20. 37. 5(x-3)=4(x − 2). 39. 3-7(x-1) = 5 – 4x. 38. 11(5-4)=7(5-6x). 40. 5-4(x-3)= x -2(x-1). 48. (x+2)(x+3) + (x − 3) (x − 2) − 2x(x + 1) = 0. 49. 50. = 1 52. Show that x= 51. Show that x = 5 satisfies the equation 2(x+1)(x+3) + 8 = (2x + 1)(x+5). 54. Show that x = 4 satisfies the equation (3x+1)(2x − 7) = 6(x − 3)2 + 7. 80. We shall now give some equations of greater difficulty. Example 1. Solve 5x-(4x-7)(3x-5)= 6-3(4x-9)(x − 1). Simplifying, we have 5x − (12x2 - 41x+35) = 6 − 3(4x2 – 13x+9); and by removing brackets 5x-12x2+41x-356-12x2 + 39x − 27. Erase the term - 12x2 on each side and transpose; Note. Since the sign before a bracket affects every term within it, in the first line of work we do not remove the brackets until we have formed the products. Multiply by 88, the least common multiple of the denominators; removing brackets, transposing, 352-11(x-9)=4x-44; 352-11x+99=4x-44; - 11x-4x=-44-352-99; collecting terms and changing signs, 15x=495; .. x = 33. Note. In this equation X- 9 is regarded as a single term with the the minus sign before it. In fact it is equivalent to vinculum or line between the numerator and denominator having the same effect as a bracket. [Art. 58.] In certain cases it will be found more convenient not to multiply throughout by the L.C.M. of the denominator, but to clear of fractions in two or more steps. Now clear of fractions by multiplying by 5 × 7 × 4 or 140; .. 72x-108+45x +405 = 280x - 2800; 81. To solve equations whose coefficients are decimals, we may express the decimals as common fractions, and proceed as before; but it is often found more simple to work entirely in decimals. Example. Solve 375x-1.875 = 12x+1·185. 2. 15-3x=(2x + 1)(2x − 1) − (2x − 1)(2x+3). 2. 21-x(2x+1)+2(x − 4)(x+2) = 0. 4. 3(x+5)-3(2x − 1) = 32 - 4(x-5)2+4x2. 5. 3x2--7x-(x+2)(x − 2) = (x + 1)(x − 1 ) + (x − 3)(x+3). 6. (x-6)(2x-9)-(11-2x)(7 − x)=5x-4-7(x-2). |