Introduction

Substitution method is one of the ways to tackle and solve a problem. Although this method is predominantly used in solution to system of equations, it is also a method used in calculus. Substitution method is in fact a subtopic in differential calculus and integral calculus. In this topic let us study a number of examples so that the concept of substitution method becomes very clear to us.

Let us first start with the application of substitution method in solving system of two equations. Substitution method is one of the ways to tackle and solve a problem. Although this method is predominantly used in solution to system of equations, it is also a method used in calculus. Substitution method is in fact a subtopic in differential calculus and integral calculus. In this topic let us study a number of examples so that the concept of substitution method becomes very clear to us.

Let us consider the set of simultaneous equations x – 2y = 2 and 2x + 3y = 25.

Looking at both the equations, it is apparent that ‘x’ can be easily be expressed in terms of ‘y’ by isolating ‘x’ in the first equation. There is no harm even if ‘x’ or ‘y’ can be isolated from the second equation and used for subsequent steps as you would end up getting the same final answer but the steps would be cumbersome. It is a great technique to correctly identify the correct option so that the subsequent steps become easier. One can acquire this technique with a little practice. So, from the first equation we will start and follow the steps shown below.

The first equation is, x – 2y = 2.

By isolating ‘x’, we get x = 2 + 2y

Now consider the second equation 2x + 3y = 25.

‘Substituting’ x = 2 + 2y in the second equation, 2(2+ 2y) + 3y = 25

By simplification, 4 + 4y + 3y = 25 or, 4 + 7y = 25 or, 7y = 21, which gives the solution as y = 3.

Now consider the substitution x = 2 + 2y.

Evaluating 2 + 2y for y = 3 (as we solved y = 3), x = 2 + 2*3 = 8

Thus, the solution to the system is x = 8 and y = 3.

It may not be the case always that the coefficient of one of the variable turns out to 1. In such a case, care should be taken with the fraction we might face.

Suppose the first equation was give as 3x – 2y = 18 in the same example, the steps would be similar excepting that isolation of ‘x’ would involve a step of division.

That is, 3x – 2y = 18 or, 3x = 18 + 2y or, x = (18 + 2y)/(3)

‘Substituting’ x = (18 + 2y)/(3), in the second equation, 2(18 + 2y)/(3)+ 3y = 25

At this stage, we need to do one more step of multiplying by 3 throughout to eliminate the fraction. That is, 2(18 + 2y) + 9y = 75

By simplification, 36 + 4y + 9y = 75 or, 36 + 13y = 75 or, 13y = 39, which gives the solution as y = 3.

Now consider the substitution x = (18 + 2y)/(3).

Evaluating (18 + 2y)/(3) for y = 3 (as we solved y = 3), x = (18 + 2*3)/3 = 24/3 = 8.

Thus, the solution to the system is x = 8 and y = 3.

Let us now see how substitution method is helpful in case we have the system of equations of the reciprocals of the variables.

Let us consider the set of equations (3/x) + (2/y) = 7/2 and (4/x) – (1/y) = 1. The first substitution should be such that the reciprocals are eliminated. That can be done by substitutions like, a = 1/x and b = 1/y. So, the given equations turn into, 3a + 2b = 7/2 and 4a – b = 1.

Now, to eliminate the fraction let us multiply the first equation throughout by 2.

Hence it becomes as, 6a + 4b = 7.

Now from the second equation we isolate ‘b’ as b = 4a – 1.

Substituting this in 6a + 4b = 7, 6a + 4(4a – 1) = 7 or, 6a + 16a – 4 = 7 or, 22a – 4 = 7 or, 22a = 11 or a = 1/2.

Since b = 4a – 1, when a = 1/2, b = 1.

Now reversing the substitution, 1/2 = 1/x or x = 2 and 1 = 1/y or y = 1.

Hence, the solution set to the system of equation is x = 2 and y = 1.

In this section let us see how substitution method works in calculus. Suppose we need to find the derivative of the function y = [ln (sin (x)].

As per chain rule of differentiation, dy/dx = dy/du*du/dx.

This implies that let us substitute u = sin (x)So, the function now becomes as y = ln (u) and hence dy/du = 1/u.

Since u = sin (x), du/dx = cos (x).

Therefore, the derivative of [ln (sin (x)] is ln (u)* cos (x).

Reversing the substitution, dy/dx = [ln (sin(x)]cos (x).

Now let us consider the integration of [dx/√(1 – x

^{2})]

The term √(1 – x

^{2}) reminds us of Pythagorean theorem and in turn the trigonometry identity cos u = √(1 – sin

^{2}(u)).

Therefore, substituting x = sin (u), the term √(1 – x

^{2}) become as cos (u).

Now, since x = sin (u), dx = cos (u) du.

Therefore, the given integral [dx/√(1 – x

^{2})] becomes as, integral of [cos (u) du] /[cos (u)] = integral of du, = u + ‘c’, where ‘c’ is the constant of integration.

Since we assumed x = sin (u), we can express ‘u’ as sin-1 (x).

Therefore, the integration of [dx/√(1 – x

^{2})] is sin-1 (x)Thus, we had seen how substitution method helps from the examples we worked out. It may be worth noting that substitution method is also helpful in higher algebraic topics like solution of even power trinomials, logarithmic functions, exponential functions etc.