5. Given x2 + 2x + 1 16, to find the value of x. Here x+1: 4, by extracting the square root of each side. And therefore, by transposition, x 4 1 3. 6. Given 5ax 3b= 2dx+c, to find the value of x. Here 5ax 2dx =c+3b; or (5a — 2d) x = c + 36; and 2x 9. Given ✔ +5=7, to find the value of x. 2x 3 Here ✓7-5=2; whence, by squaring, 4, and 2x 12, or x = = 6. 10. Given x +✓ (a2 + x2) 2a2 to find the value + ∞2 aa 4 2a2; or x √ (a2 +x2) 2a2x2 + x2; whence a2x2 a2x2 a2 ✓ 3, the answer required. 9 3 3a2x2 = a*, or x2 a2 - 2a2x2; therefore EXAMPLES FOR PRACTICE. 1. Given 3x − 2 + 24 = 31, to find the value of x. 8. Given 2 + 3x = √ (4 + 5x), to find the value of x. (4+ 12. Given (a2 + x2) = 1/ (b1 + x), to find the value of x. b4 a4 Ans. xv 2a2 13. Given v (a + x) + √ (a − x) = √ ax, to find the value of x. 4a2 Ans. x= 15. Given a +≈ = √ [a2 + x √ (b2 + x2)], to find the value of x. 2 v 16. Given (x2 + 3a2) — √(x2 - 3a2) = xva, to find the value of x. 17. Given √ (a + x) + √ (a — x) = b, to find the value of ⇓ Ans. x= √ (4a — b2). b 2 18. Given (a + x) + V (α-x)=b, to find the value of x √ (c2 — 4) 21. Given v(a2+ ax) = a −√ (a2 — ax), to find the value of a. = 22. Given √(a2 —- x2) + ∞ √ (a2 — 1) — a3 √ (1 — x3), to find the value of x. xv 23 Given √(x+a) = C √(x+b), to find the value of x. 2c α to find the value a) 2 3. a + x Of the resolution of simple equations, containing two unknown quantities. When there are two unknown quantities, and two independent simple equations involving them, they may be reduced to one, by any of the three following rules : RULE 1.-Observe which of the unknown quantities is the least involved, and find its value in each of the equations, by the methods already explained; then let the two values, thus found, be put equal to each other, and there will arise a new equation with only one unknown quantity in it, the value of which may be found as before.* * This rule depends upon the well-known axiom, that things which are equal to the same thing, are equal to each other; and the two following methods are founded on principles which are equally simple and obvious. 1. Given EXAMPLES. 2x + 3y = 23 5x-2y=10} to find the values of x and y. Here, from the first equation, x = 23 3y 2 And consequently, 42-2y= 72 — Or, by multiplication, 84-4y = 144 9y; 2' 1. Given 4x + y = 34, and 4y + x = 16, to find the values of x and y. Ans. x= 8, y = 2. 2. Given 2x+3y=16, and 3x-2y= 11, to find the values of x and y. Ans. x = 5, y = 2. Ans. x= 24, and У = 18. RULE 2.-Find the values of either of the unknown quantities in that equation in which it is the least involved; then substitute this value in the place of its equal in the other equation, and there will arise a new equation with only one unknown quantity in it; the value of which may be found as before. У = From the first equation, a 13-y; which value, being substituted for x, in the second, y — y = 3, or 2y = 13 — 3 = 10. Gives 13 y-y |