Methods of Solving Simultaneous Linear Equations

There are different methods for solving simultaneous linear Equations:

I. Elimination of a variable

II. Substitution

III. Cross-multiplication

IV. Evaluation of proportional value of variables 

This topic is purely based upon numerical examples. So, let us solve some examples based upon solving linear equations in two variables.

I: Solved example on simultaneous linear Equations using elimination method:

Solve for ‘x’ and ‘y’:

3x + 2y = 18.  

4x + 5y = 25.  

Solution: 

3x + 2y = 18 ............. (i)

4x + 5y = 25 ............. (ii)

Let us multiply equation (i) by 4 on both sides and equation (ii) by 3 on both sides, so as to make coefficients of ‘x’ equal.

On multiplying, we get;

4(3x + 2y) = 4 ∙ 18

or, 12x + 8y = 72 ............. (iii)

and

3(4x + 5y) = 3 ∙ 25

or, 12x + 15y = 75 ............. (iv)

Subtracting (iii) from (iv), we get;

12x + 15y - (12x + 8y) = 75 - 72

or, 12x + 15y - 12x - 8y = 3

or, 7y = 3

or, y = $\frac{3}{7}$.

Substituting value of ‘y’ in equation (i), we get;

3x + 2($\frac{3}{7}$) = 18.

or, 3x + $\frac{6}{7}$ = 18

or, 3x = 18 – $\frac{6}{7}$

or, 3x = $\frac{120}{7}$.

or, x = $\frac{120}{21}$.

or, x = $\frac{40}{7}$.

Hence, x = $\frac{40}{7}$ and y = $\frac{3}{7}$.

II: Solved examples on simultaneous linear Equations using substitution method:

1.  Solve for ‘x’ and ‘y’:

            x + 3y = 9

           3x + 4y = 20

Solution:

x + 3y = 9 ............. (i)

3x + 4y = 20 ............. (ii)

Taking first equation in reference, i.e,

x + 3y = 9

x = 9 - 3y ............. (iii)

Substituting this value of ‘x’ from previous equation in 2nd equation, we get;

3x + 4y = 20.

or, 3(9 - 3y) + 4y = 20

or, 27 - 9y + 4y = 20

or, -5y = 20 - 27

or, -5y = -7

or, 5y = 7

or, y = $\frac{7}{5}$

Substituting this value of y into equation of x in (iii), we get;

x = 9 - 3y

or, x = 9 - 3$\frac{7}{5}$

or, x = 9 – $\frac{21}{5}$

or, x = $\frac{247}{5}$.

Hence, x = $\frac{247}{5}$ and y = $\frac{7}{5}$.

2. Solve for ‘x’ and ‘y’; 

x + y = 5

4x + y = 10

Solution:

x + y = 5 ............. (i)

4x + y = 10 ............. (ii)

From equation (i),  we get value of y as:

y = 5 - x

substituting this value of y in equation (ii), weget

4x + (5 - x) = 10

or, 4x + 5 - x = 10

or, 3x = 10 - 5

or, 3x = 5

x = $\frac{5}{3}$.

Substituting this value of x as $\frac{5}{3}$ in equation y = 5 - x , we get;

y = 5 - $\frac{5}{3}$

or, y = $\frac{10}{3}$.

Hence, x = $\frac{5}{3}$ and y = $\frac{10}{3}$.

III. Solved example on simultaneous linear Equations using Cross-multiplication method:

Solve for ‘x’ and ‘y’;

3x + 5y - 25 = 0.

5x + 3y – 35 = 0.

Solution:

Assume two linear equations be

A₁ x + B₁y + C₁ = 0, and

A₂x + B₂y + C₂ = 0.

The coefficients of x are: A₁ and  A_2.

The coefficients of y are: B₁ and B_2.

The constant terms are: C₁ and  C₂.

To solve the equations in a simplified way, we use following table:

$\frac{x}{B_{1}C_{2} - B_{2}C_{1}} = \frac{y}{C_{1}A_{2} - C_{2}A_{1}} = \frac{1}{A_{1}B_{2} - A_{2}B_{1}}$

In the given equations,

The coefficients of x are 3 and 5.

The coefficients of y are 5 and 3.

The constant terms are -25 and -35.

On substituting the respective values, we get

$\frac{x}{5 × (-35) - 3 × (-25)} = \frac{y}{(-25) × 5 - (-35) × 3} = \frac{1}{3 × 3 - 5 × 5}$.

or, $\frac{x}{- 175 + 75} = \frac{y}{-125 + 105} = \frac{1}{9 - 25}$.

or, $\frac{x}{-100} = \frac{y}{-20} = \frac{1}{-16}$.

On equating x term with constant term, we get;

x = $\frac{25}{4}$.

On equating y term with constant term, we get;

y = $\frac{5}{4}$.

IV: Solved example on simultaneous linear Equations using evaluation method:

Solve for x and y:

2x + 3y = 4; 3x - 5y = -2

Solution:

The given equations are 

2x + 3y = 4 .......... (1)

3x - 5y = -2.......... (2)

Multiplying equation (2) by 2, we get

6x - 10y = -4.......... (3)

Now we add equations (1) and (3) we get

8x - 7y = 0

or, 8x = 7y

or, $\frac{x}{7}$ = $\frac{y}{8}$ (= k)

Substituting x = 7k and y = 8k in equation (1) we get

2 ∙ 7k + 3 ∙ 8k = 4

or, 14k + 24k = 4

or, 38k = 4

or, K = $\frac{4}{38}$

or, k = $\frac{2}{19}$

Therefore, x = 7 ∙  $\frac{2}{19}$ and y = 8 ∙ $\frac{2}{19}$

Thus, x = $\frac{14}{19}$ and y = $\frac{16}{19}$

9th Grade Math

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