12. Given, V3 ± ∞ = √9 ± x, to find x. Ans. x 0 or 5. 13. Given, 15x+7= 149 +36x, to find æ. 15. "Given, va + x + Va-x=b, to find x." Cubing both members, we get 2a + 3[va + x + Va-x] × Va2x2 = 2a+3b va2x2=b3; consequently, we easily 16. “Given, √(a + x)2 + √ (a − x)2 = 3√a2 — x2, to find a." Proceeding as in the last question, we easily get (a + x)2 +(ax)2+91 (a2 — a2)3 23 -x); consequently, x = = 2(a2 + x2) + 9(a2 — x2) = 27(a2 tion can easily be put under the form b 26 α x 3c a2 b 3c-d 5a(26-a); consequently, = 3c-d 5a(2b — a) 3c-d Remarks.-1st. We have taken this and the preceding example from the Translation of Hirsch's Collection of Examples, Formulæ, etc., by Rev. J. A. Ross, A. M. (See the edition of the work published in 1827, vol. 1, page 131, examples 38 and 39). 2d. We have introduced the examples for the purpose of showing the great superiority which particular processes sometimes have over the most general. Because any expression of the form A - B is equivalent to (√A + √B) × (√Ā – √B), we easily get x= 20. Given, 18 + α 4x2 50 = 11 Ans. x = 9. · (3x − 1)2 + (4x + 1)2, to find æ. -- Ans. x 2. 22. Given, 4:1:: 4x8x + 3 √x: x2-xVx, to - find x. 23. Given, 2 : 1 :: Va + x + Va − x : √ a + x − to find x. 24. Given, + 3 : x2 − 5 :: x: 4, to find x. The proportion gives the cubic equation - 4— 5x — 120, which shows that a must be greater than 5.3 and less than 5.4; since by putting 5. 3 for x in the equation, we get the negative result -1.983, and by putting 5.4 for æ, we get the positive result 1.824. 25. Given, 15x+6x+x-1= 0, to reduce the equation to a proper form for solution. Multiply each of the terms of the equation by 3 x = x, and it becomes - 15+ 6x + x − x3 = 0; consequently, putting xy, and ordering the equation according to the descending powers of y, we get y1 — y3 + by — 15 = 0, an equation of the fourth degree, whose solution gives y, and thence a can be found from x = y. The equation has two real and two imaginary roots, one of the real roots lying between 2, and the other between 1 and 2. - 3 and SOLUTION OF TWO OR MORE SEPARATE (SIMPLE) EQUATIONS, CONTAINING AS MANY UNKNOWN LETTERS AS THERE ARE EQUATIONS. 1. When two or more equations exist together (as supposed), they are sometimes called a system of two or more equations; and it is plain that their solution consists in reducing them to a single equation, or to an equation which contains only one unknown letter, whose solution will clearly give that of all the equations. 2. When a system of equations is reduced to a single equation, the unknown letters not contained in it are said to be eliminated, and the process used is called the method of elimination. 3. We will now proceed to give the common methods of elimination, together with some others, which will often be found useful in practice. I. ELIMINATION BY SUBSTITUTION. RULE. Find the value of one of the unknown letters in one of the equations, and substitute or put the value for the letter in each of the other equations; then we shall obtain a new system of equations, which will be one less in number than the given equations. Proceeding in a similar way with the equations thus obtained, their solution will be reduced to that of a system of equations one less in number (than the equations from which they have been deduced); and in like manner we may clearly continue the reduction of the number of equations, until a single equation is obtained, from whose solution we can get that of each of the given equations. EXAMPLES. 1. Given, 3x + 2y = 12 and 4x + 3y = 17, to find x and y. From the first equation we get y = and putting 12 3x " 2 this value for y in the second equation, we have the single for x and y in the given equations, we get the identical equations 6+ 6 = 12 and 8+ 9 = 17; consequently, the values of x and y are correct. 2. Given, 4x-13y = 11 and 5x-18y = 12, to find a and y. Ans. 6 and y = 1. 3. Given, ax + by =c and a'x + b'y = c', to find x and y. we shall have x=0 to do, since the given equations become by c and by c'. If ab' - a'b=0, while the numerators of the expressions for x and y are not equal to 0, it is clear that x and y will be unlimitedly great. If b'e be' 0 and ac' a'c 0, then x and y will be re =c and ax + by = c, the equations are reducible to ax + by which are equivalent to ax + bye, it is clear that the preceding values of x and y show them to be really indeterminates. In like manner it may be shown that a and Y will be indeterminates if b'c - bc' = 0, and ab' — a’b = 0. 4. Given, + y = 11 and y2+x=7, to reduce the equations to a single equation. From the second we get x=7-y, which, put for x in the first, gives the single equation (7 — y3)2 + y = 11, or y* — 14y2 + y + 380, one of whose roots is 2. Putting 2 for y in x=7-y, we get x = 3; consequently, 3 and 2 are values of x and y which satisfy the given equations. 5. Given, x2 + y2 = 13 and xy=6, to reduce the equations to a single equation. 6 From the second equation we get y = which, put for y 2 in the first, gives the single equation a2 + (9)2 = 13, or a 13x2 +36=0, which is satisfied by putting x = 3, which gives y = 2. 1 1 5 1 1 3 6. Given,+== + 6' x x, y, and z. y 1 1 7 = to find 12'. = and + From the first equation we get = in the second, gives the two equations 1 1 У 2 12 1 1 3 = + y we have x = 2. If we put 2, 3, and 4 for x, y, 1145 6 and 2, in the given equations, they will be satisfied, or become identical equations; consequently, the values of x, y, and z are correct. II. ELIMINATION BY COMPARISON. RULE. Find the value of one of the unknown letters in each of the given equations; then put any one of these values equal |