subtract 2 from each side of the equation, thus

It will be seen that in this example is found on both sides of the equation, after we have cleared of fractions; accordingly it can be removed by subtraction, and so the equation remains a simple equation.

Here it is convenient to multiply by 4x+4, that is by 4x+1);

Multiply by 32+1; thus (3x+1) (4x+7)=12(x+1)2;

that is,

12x2+25x+7=12x2 + 24x + 12.

Subtract 12x2 from each side, and transpose; thus

Clear of fractions; thus (x-5) (x-6)=(-2) (x-3);

*45x-75 1.2 *3x-6

176. Solve 5x+

'6

•2

*9

To ensure accuracy it is advisable to express all the decimals as common fractions; thus

5х 10/45x 75 10 12

+

10 6 100

Simplifying,

that is,

= X 100 2 10

10/3x

9 10 10

Multiply by 12, 6x+9x-15=72−4x+8;

177. Equations may be proposed in which letters are used to represent known quantities; we shall continue to represent the unknown quantity by x, and any other letter will be supposed to represent a known quantity. We will solve three such equations.

Clear of fractions; thus

(x-a) (2x-b)2= (x —b) (2x− a)2;

that is, (-a) (4x2-4xb+b2)=(x−b) (4x2 - 4xa+a2).

181. Although the following equation does not strictly belong to the present chapter we give it as there will be no difficulty in following the steps of the solution, and it will serve as a model for similar examples. The equation resembles those already solved, in the circumstance that we obtain only a single value of the unknown quantity.

Solve

√x+ √(x−16)= 8.

By transposition,

(x-16)=8-√x;

square both sides; thus -16=(8-x)=64-16√x+x;

therefore

transpose,

therefore

therefore

-16 64-16x;

16/x64+16=80;

√x=5;

x=25.

15. x-3-(3-x) (x+1)=x (x-3)+8.

16. 3-x-2(x-1)(x+2)=(x-3) (5-2.x).

2x-1 3