bring all the terms which contain or its powers into the first member, and to make a positive. Transposing the terms from the second member which contain a, uniting like terms, and then changing all the signs; the Ans. is a 5x2-14x = - 14. 5. Given, 32x2+6x=x3-7x2+4x+30, to bring all the terms into the first member of the equation. Ans. 4a+5x+2x-30= 0. CASE II. 1. The unknown letter in a simple equation may be freed from any multiplier by dividing all the other terms of the equation by the multiplier; and it may be freed from any divisor by multiplying all the other terms of the equation by .the divisor, as is clear from Ax. II. 2. In like manner, the highest or any power of the unknown letter, in any equation, may be freed from any multiplier or divisor. 3. Hence any factor or divisor which is common to all the terms of any equation may be erased. EXAMPLES. 1. Given, 3x+5= 20, and 20, and 4 -72, to find x and y. Transposing 5 and 7, then dividing by 3 and multiplying 20-5 by 4, we get x = 2. Given, ax-bc, and 3 = c, and2 +ƒ = g, to find æ and y. (72) × 420. 3. Given, axbx2 + cx2-dx = -e, and 5y3 - 10y+ 40y500, to free x and y3 from their coefficients. d x2 and y3 from their divisors, and to make a positive. Ans. -2x=7, and y3 + aby2 — acy = ad. CASE III. Any equation may be freed from fractions by reducing all its terms to a common denominator, and then rejecting the common denominator, or by multiplying each term by the least common multiple of all the denominators, and then dividing the numerator of each fractional product by its denominator. The terms reduced to a common denominator, give the 20x 75 12x equation or rejecting the denominator we have the equation 20x - 75 = 12x, or 20x-12x 75 or x= Here we have 12 for the least common multiple of all the denominators; consequently, multiplying each term by 12 and reducing the fractional products, we get 10y + 28 = 9, the result to its most simple form, when the terms are brought into the first member and arranged according to the descending powers of x. Ans. 18x2+8x-7= 0. -2 Remark. Supposing Ax" + Aan-1 + A ̧¤¬2 + ....+ An-1+ An = 0 to stand for an equation which is freed from fractions, A, A1, A2, etc., being its coefficients, and n a positive whole number; then, if we multiply all the terms of ... . the equation by An-1, it can clearly be written in the form (Ax)” + A1(Ax)”−1 + AA1⁄2(Ax)”−2 + A3A ̧(Ax)”—3 + + A”-2An-1(A) + A”-1A2 = 0, which by representing Ax by y becomes y + A1y"-1 + AAy"-2 + A2Ayn-3 +.... + An-An-1Y + A"-1A, = 0, an equation in which the coefficient of y" is 1. Having found y from the preceding equation, since y is put for Ax, we shall clearly get x = value of x is found. x = - A ; consequently, the Thus, the terms of the equation 2x2-x-15=0, when multiplied by 2, give the equation (2x) - 2x-300, which by using y for 2x, becomes y-y-30 = 0, and since this equation is satisfied by putting 6 or 5 for y, it results that a 3 and 2.5 are the roots of the proposed equation. Similarly, if the terms of the equation 3x3 + 4x = 32 are multiplied by 32 = 9, and y put for 3x, the equation is reduced to y3 + 12y= 288; consequently, as this is satisfied by putting 6 for y, it results that = 2 must be a root of the proposed equation. = CASE IV. 1. If any equation contains the unknown letter under the form of a surd; then having simplified the equation as required, by the preceding cases, we may order the terms so that the part under the surd shall stand alone on one side of the equation; consequently, if we raise both members to the power denoted by the index of the surd, the surd will be removed. It is clear that we may (generally) proceed in a similar way to free an equation (in which the unknown letter is affected by two or more surds) from surds, by removing one surd at a time. 2. If the unknown letter in any equation is affected by any number of surds, we may put each surd part equal to a new letter, and free the equations thus obtained from surds; then substituting the letters for the surds in the proposed equation, we shall have as many equations as new letters and the unknown letter in the given equation; consequently, by the methods of elimination (to be given hereafter) we can always reduce these equations to one equation containing one unknown letter, which being solved, we easily get the roots of the given equation. EXAMPLES. 1. Given, 3 √x-2= 4, to find x. The equation, by transposition and division, becomes 2; consequently, by squaring, we get (√x)2 = 4, or √x = Freeing the equation from fractions and transposing, we have 4x-3= 19; consequently, by cubing, we have (4-3)= 193 = 6859, or 64(x-3)= 6859, which gives 3. Given, a + √bx = √ a2+ cx, to find x. Squaring both members, we get a2 + 2a √bx + bx: a2 + cx, or, erasing a2 from both members, and transposing, we get 2a vbx=(c-b); or, squaring, etc., we have = × b. Since 2a be(c-b) = 0 is satisfied by putting = 0, it is clear that = 0 is also a root of the proposed equation. It is clear if a is finite, while cb is an infinitesimal (or very small), that must be very great on account of the 2a c-b smallness of the divisor; consequently, when 2a is unlimc—b itedly great, a must evidently be unlimitedly greater, provided b is finite. 4. Given, V+3=2V, to show how to find. By transposition and cubing, we get x = (2 √x-3); consequently, putting a y2, we have y = (2y- 3)3 = 8y3 -36y+54y-27, or 8y3-37y+ 54y-27=0; consequently, solving this equation, we shall get y, and then from the equation ay we shall get the values of x, which will satisfy the given equation. Otherwise, squaring the members of the given equation, we have (Vx+3)= 4x; consequently, putting a = z3, we get (+3)=423, or 423 — 22 — 62 — 9 = 0, an equation of a form more simple than the equation in y. Solving the equation in 2, we get 2, and thence a is found from the equation 28. = CASE V. 1. In any equation, when the terms which contain the unknown letter are all brought to one side, and the terms which do not contain it are carried to the other side, if the side which contains the unknown letter is an exact power; then the equation will be simplified by extracting the root in both members of the equation. 2. If the unknown letter enters both members of an equation, and the terms can be so ordered that both members shall be an exact power, then the solution of the equation will be also simplified by extracting the root of both members. EXAMPLES. 1. Given, 22-3x2+6, to find x. The equation is easily reduced to x2 = 9; consequently, by extracting the square root of both members, we get x = 1/93, that is, we may put either 3 or 3 for a in the given equation, and it will be satisfied, or, which is the same, 3 and 3 are the roots of the proposed equation. Adding 1 to both members of the equation, it becomes x2+2x+1=8+ 1, or (x + 1) = 32; consequently, extracting the square roots, we have x+1=3, or x + 1 = 3, - 3, which give x = 2 and x = 4 for the roots x+1= of the proposed equation. 3. Given, 3a2 = 2x+1, to find x. Adding to both members, we get 4x2 = x2 + 2x + 1; consequently, extracting the square roots of both sides, we get 2x=x+1, or 2x = 1 3 x-1, which give x = 1, or x = |