Substitution Method

Students can learn about solving equations by substitution method here. The topic is dealt under algebra. The students can learn about the caclulation process in detail with elaborate explanation and solved examples.

Solving simultaneous equation by substitution method is as explained in below numerical:

Example for substitution method :

Solve 2x - 9y = 0 …(i)

x - 18y = 27 … (ii)

Suggested Answer for first simultaneous equation equation question solving using substitution method.

From (i)

2x - 9y = 0

2x = 9y

... x = $\frac{9y}{2}$ ... (iii)

Substituting this value of x in (ii), we get,

($\frac{9y}{2}$) - 18y = 27.

9y - 36y = 54

- 27y = 54

y = -2

Substitute this value of y in (iii):

x = $\frac{9}{2}$ ( - 2)

= - 9

The solution is x = -9 and y = -2.

Here is an example of solving simulatenous equation using the substitution method:

Example :

Solve 43x + 31y = 241 …(i)

31x + 43y = 277 …(ii)

Suggested Answer :

By adding (i) and (ii), we get

74x + 74y = 518

x + y = 7 …(iii)

By subtracting (ii) from (i)

12x - 12y = -36

or x - y = -3 …(iv)

By adding (iii) and (iv) 2x = 4

x = 2

Substituting x = 2 in (iii), we get:

(2) + y = 7

y = 5

The solution is x = 2 and y = 5.

Below you could see the example of substitution method

Example :

Solve: $\frac{5}{x}$ - $\frac{2}{y}$ = 2 ... (i)

$\frac{2}{x}$ + $\frac{3}{y}$ = 16 ..(ii)

Suggested Answer :

Let $\frac{1}{x}$ = a and $\frac{1}{y}$ = b

(i) and (ii) can be written as

5a - 2b = 2 …(iii)

2a + 3b = 16 …(iv)

Multiplying (iii) by 3 and (iv) by 2, we get

15a - 6b = 6 …(v)

4a + 6b = 32 …(vi)

Adding (v) and (vi), we get 19a = 38 a = 2

Substituting a = 2 in (iv), we get

2(2) + 3b = 16

3b = 12

b = 4

Re-substituting

a = 2 = $\frac{1}{x}$

∴ x = $\frac{1}{2}$

b = 4 = $\frac{1}{y}$

∴ y = $\frac{1}{4}$

The solution is x = $\frac{1}{2}$ and y = $\frac{1}{4}$ .

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