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and symbols and by converting it to an algebraical equation.

The quantity which is to be found out in the problem is to be represented by x, then the conditions of the

problem to be expressed in terms of x. Thus an equation in x is obtained. Finally, solving this equation the

value of x is found.

If there are more than one or two variables or unknown quantities in the given algebraic problem then the

quantities are represented by x and y etc. Thus the simultaneous equations in x and y are obtained.

multiplied by 4 and the fourth divide by 2, the results will be all equal.

Then according to the problem,

A + b + c + d = 60 … (1)

A – 3 = b + 11 = 4 c = d / 2 … (2)

Now from (2)

A – 3 = b + 11 = 4 c = d / 2 = k (suppose)

Therefore

A – 3 = k, a = k + 3

B + 11 = k, b = k – 11

4 c = k, c = k / 4

D / 2 = k, d = 2 k

Putting in (1)

(k+3)+ (k-11) +k/4+2k=60

17k=272

K=16