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 = \(\frac{7}{5}\).

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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

Ax + B1y + C= 0, and

A2x + B2y + C= 0.

The coefficients of x are: Aand  A2.

The coefficients of y are: B1 and B2.

The constant terms are: C1 and  C2.

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

Simultaneous Linear Equations

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

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 = \(\frac{16}{19}\)

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9th Grade Math

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