CHAPTER III

LITERAL EQUATIONS IN ONE UNKNOWN NUMBER

188. It is of great importance to determine the unknown numbers of an equation in terms of general numbers, i. e., in terms of numbers expressed by letters. In the preceding chapter the known numbers have been represented by numerals. Known quantities when represented by letters are usually represented by the first letters of the alphabet, a, b, c, l, m, n, a, B, y, etc.; the unknown quantities by the letters, y, z, Thus, in the equation,

a and b are the known numbers and x is the unknown number.

It is purely an arbitrary agreement to represent the known numbers by a, b, c, and the unknown numbers by x, y, z, The number a might be supposed to be unknown and x and b to be known numbers. Then the solution of the equation would give

a = b―x.

If b were the unknown number and a and x were known we would have as our solution

b = x+a.

189. A Numerical Equation is one in which all the known numbers are numerals.

A Literal Equation is one in which some or all of the numbers are represented by letters.

E. g.,

3 ax + a2 = b2-2bx; ax + by = c; 3x+5b = 19.

190. The principles of equivalent equations hold when the equations are literal.

EXAMPLE 1. Solve the equation ax + b cx+d.

=

EXAMPLE 2. Solve (a + x) (b + x) = (m + x) (n + x).

By multiplication, ab + (a+b) x + x2 = mn + (m + n) x + x2 by transposition, (a+b) x (m + n) x = mn

ab

By reducing terms in the first member to common denominator,

Clear fractions and get (be a2 — ab + ac) (x—b) = (x − a) (x —c)b

or bx-ax-abx+acx-b2x+a(a+b-c) b=bx-(a+c) bx+abc.

Transpose and get

abx+bcx-ax-abx+acx-b2x=-ab (a+b-c)+abc;

(-a2-b2+ac+bc)x=-ab(a+b-2c)

Or, by clearing fractions at first, the result is obtained,

x=

-bc'

(a+b) (x-a) (x—b)=a (x—b) (x—c)+b(x—a) (x—c)

(a+b) x2 — (a+b)2x+ab (a+b)=ax2-a (b+c)x+ab c
+bx-b (a+c)x+abe.

[b (a+c)—(a+b)2+a (b+c)] x=ab (2c—a—b);

ab (a+b2 c) a2 + b2- beac

Ans.

Ans.

13. (ab) (cx) + (b −c) (a− x) + (c− a) (b−x)=a-x.

14. (a-x)b(a— c — x) (x — b) = x (a — x).

15. m (a+b-x)=n (a+b− x).

16. (ab) (a-c+x) + (a+b) (a+c-x)=2a2.

17. (m+x) (a+b−x) + (a− x) (b − x) = a (m+b).

18. (ax-1) (bx-1) (ex-1)+1= ax + bx + cx.

19. (ax) (b+x) (c+x) − (a− x) (b − x) (cx)=2(x+abc).

20. (ab) (ac) (a + x) + (a+b) (a + c) (a− x) = 0.

21. 5a3cx+ac2x-5 abc2-3 a3c3 = 5 a2bcx+bc2x −3 a2bc3 — 5 a2c2.

22. 2 a b c + ab2x-2 ab3c — abc2d — 3 a3x = (b3 — 3 a2b) x — b2c2d.

25.

a2x
+ de

=

- bx-ac.

(3bc — ad)x ___ 5 a(2 b − a) ̧

2 ab(a - b)

--

a2 - b2

a2c
+x2.

b

+

0.

ab

a+

=

b(x - b)
2a+b

x

+

+

m

n

cx

d

+

(bn+dm)x(bg+dp) =

(mx − p) (nx − q)

49.

+

2.

X- m

x-n

61. (x− a)3 +(x − b)3 + (x − c )3 — 3(x — a) (x — b) (x — c).