CHAPTER XXIII.

HARDER EQUATIONS.

179. SOME of the equations in this chapter will serve as a useful exercise for revision of the methods already explained; but we also add others presenting more difficulty, the solution of which will often be facilitated by some special artifice.

The following examples worked in full will sufficiently illustrate the most useful methods.

Note. By a simple reduction many equations can be brought to the form in which the above equation is given. When this is the case, the necessary simplification is readily completed by multiplying across or "multiplying up," as it is sometimes called.

Example 2. Solve 8x+23 5x+2 2x+3

20

Multiplying by 20, we have

3x+4 5

- 1.

= 8x+12-20.

31 =

20(5x+2)

93x+124 = 20(5x+2),

180. When two or more fractions have the same denominator, they should be taken together and simplified.

This equation might be solved by at once clearing of fractions, but the work would be laborious. The solution will be much simplified by proceeding as follows.

The equation may be written in the form

(x-10)+2 (x −6)+2 (x −7)+2, (x-9)+2

+

x-7

x-9

X

+1+

х

10+

х

1

2

-

-

=1+

=

1

х 7

1

=

х

1

;

X- 10 X- 7 X- 9 X

Hence, since the numerators are equal, the denominators must be

181. IN the equations we have discussed hitherto the coefficients have been numerical quantities. When equations involve literal coefficients, these are supposed to be known, and will appear in the solution.

Example 1. Solve (x+a)(x+b) − c(a+c) = (x − c)(x+c)+ab.
Multiplying out, we have

Example 3. Solve the simultaneous equations :

ax - by = c

px+qy =r

To eliminate y, multiply (1) by q and (2) by b ;

(1),

(2).

We might obtain equations (1) or (2); ing x, as follows.

(aq+bp)x = cq+br;

by substituting this value of x in either of the but y is more conveniently found by eliminat

5. a(x+b)-b2 = a2 – b(a − x).

7. a(x-a)+b(x − b) + c(x − c) = 2(ab+be+ca).

6. c2x-d3

(b+1)(x+a) = (b-1)(x-α) d2x + c3.

10. x+(x-a) (x − b) + a2 + b2 = b+x2 - a(b − 1).

11.

x+a+b