Elimination by the method of the Greatest Common Divisor. Hence, divide one of these first members by the other, and proceed, as in arts. 60, &c., to find their greatest common divisor; each successive remainder may be placed equal to zero. But a remainder will at last be obtained, which does not contain the quantity to be eliminated; and the equation, formed from placing this remainder equal to zero, is the equation from which this quantity is eliminated. By eliminating, in this way, the unknown quantity from either of the equations which contain it, taken with each of the others, a number of equations is formed one less than that of the given equations, and containing one less unknown quantity; and to which this process of elimination may be again applied until one equation is finally obtained. 156. Scholium. It sometimes happens, that the first members have a common divisor which contain the given unknown quantity; and, in this case, the process cannot be continued beyond this divisor. But as the given first members are multiples of their common divisor, they must be rendered equal to zero by those values of the unknown quantities which render the common divisor equal to zero; that is, the two given equations are satisfied by such values of the unknown quantities; so that, though they are in appearance distinct equations, they are, in reality, equivalent to but one equation, that is, to the equation formed by placing their common divisor equal to zero. 157. Scholium. Care must be taken that no factor be suppressed which may be equal to zero. Examples of Elimination by the method of the Greatest Common Divisor. 158. EXAMPLES. 1. Obtain one equation with one unknown quantity from the two equations x3 + y x2 - y3 +5=0, x3 + y2 x-5=0, by the elimination of x. Solution. 1st Rem. Divide the first members as follows: x2 1 Divide the preceding divisor by this remainder after multiplying by y to render the first term divisible. Divide the preceding divisor by this remainder after multiplying by (3 y3-10) to render the first term divisible. y2x-y310 y x2 (3 y3—10) (—4 y5+25 y2) x −9 yo+150 y6—700 y3+1000 (3 y3—10) (—4 y5+25 y2) x — 4 yo+ 85 y6—375 y3 · whence the required equation is -5y965 y6-325 y3+1000 = 0. Examples of Elimination by the method of the Greatest Common Divisor. 2. Obtain one equation with one unknown quantity from the two equations 23+ y3 = a, by the elimination of x. Ans. (y3 — a) 5 — (ÿ5 — b)3 — 0. 3. Obtain one equation with one unknown quantity from the two equations x2 + y2 = 2, x2 + x3 y + x2 y2+ x y3 + y2 = 1, by the elimination of x. Ans. y8-4 y 14 y1 — 20 y2+9=0. 4. Obtain one equation with one unknown quantity from the two equations x2 + xy + y2 = 1,, x3 + y3 = 0, by the elimination of x. Ans. 4 y6 — 6 y1 +3 y2 — 1 = 0. 5. Obtain one equation with one unknown quantity, from the two equations x3 + y x2 + x + y = 4, x3 + x2 + y x = 3, by the elimination of x. Ans. Either y-10, or y2-3y+21=0. 6. Obtain one equation with one unknown quantity from the three equations Examples of Elimination by the method of the Greatest Common Divisor. 7. Obtain one equation with one unknown quantity from the three equations x + y + z = a, x2 + y2 + z2 = b, x y + x z + y z = c, by the elimination of x and y. Ans. These three equations involve an impossibility unless and in case this equation is satisfied by the given values of a, b, and c, the three given equations are equivalent to but two, one of them being superfluous, and, by the elimination of x, they give the indeterminate equation with two unknown quantities y2 + y z + z2 — ay—az+c=0. 8. Obtain one equation with one unknown quantity from the three equations Ans. 28-826+16x+x-10=0. 9. Obtain one equation with one unknown quantity from the four equations x+y+z+u = a, x y + xx+xu+yz+yu+zu=b, x y z + xyu+xzu+yzu= c, xyzu= e, by the elimination of x, y, and z. = Ans. u4-a u3+bu2-cu+c=0. which, being substituted in the first of the given equations, produces X 3. 11. Solve the two equations x2 y1—8 y2x2+16x2=90 xy+60 (x—y2)—720 (y-1), ~ (y2-4y+4) x 12 5 3 12. Solve the three equations x y + z = 5, x y z +z2 15, x y2+ x2 y2x+2x=8. Ans. x= 2, y = 1, z=3. 159. Problem. To solve two equations of the first degree by Elimination by Addition and Subtraction. Solution. The given equations may, as in art. 146, be reduced to the forms The process of the preceding article, being applied to these equations in order to eliminate x, will be found to be the same as to Multiply the first equation by A' the coefficient of x in the second, multiply the second by A the coefficient of x in the first, and subtract the first of these products from the second. |