ICSE Class 8 Maths Algebra Chapter 21 Simultaneous Linear Equations

Simultaneous Linear Equations

Simultaneous Linear Equation

Indeterminate

Truth Table

Elimination of Variable

Mr Ahuja goes to a 'Fixed Rate' fruit-market to buy bananas and apples. Two fruit-sellers call out to him and since he knows both of them well, he feels obliged to buy from both fruit-sellers. He buys 12 bananas and 8 apples from one and pays him Rs 73. Can you tell how much each fruit costs?

Now, we have 2 variables in the equation 12x + 8y = 73. For every new value of x, there will be a new value of y. Here, x represents the cost of 1 banana and y represents the cost of 1 apple. Thus, a linear equation with two variables is indeterminate. At best we can construct a truth table calculating the values of y for different values of x, as shown below:

Now if he buys 6 bananas and 9 apples from the second fruit-seller and pays him Rs 69, we have another indeterminate linear equation with two variables as 6x + 9y = 69. A truth table for this equation would be:

Now if both equations are considered together, we notice one pair of solutions that match. When x = 1.75, in both the equations the value of y is 6.5.

Thus, the bananas were for Rs 1.75 and apples were Rs 6.50 each.

Two or more equations which have only one set of values of the variable involved as a common solution are known as simultaneous equations.

The solution of the simultaneous equations 12x + 8y = 73; 6x + 9y = 69 is x = 1.75 and y = 6.5. No other set of values for x and y will satisfy both the equations simultaneously.

Solving Simultaneous Equations

Case I: Eliminating one variable by adding or subtracting equations

Step 1: The numerical coefficient of one variable is made the same in the given simultaneous equations.

Step 2: The equations are added or subtracted to eliminate the above variable.

Step 3: The resultant equation is solved for the other variable.

Step 4: The value of the other variable is substituted in any one of the original simultaneous equations to find the value of the first variable.

Example 1: Solve x + 4y = 20; x - 2y = 2

Let us select the variable x as its numerical coefficient is the same in both equations.

Now, as the same quantity can be added to or subtracted from both sides of an equation,

\[x + 4y - (x - 2y) = 20 - 2\]

(subtracting equations)

\[\Rightarrow x + 4y - x + 2y = 18\]

(eliminating variable x)

\[\Rightarrow 6y = 18\]

\[\Rightarrow y = 3\]

\[x + (4 \times 3) = 20\]

(substituting the value of y in any one equation)

\[\Rightarrow x + 12 = 20 \Rightarrow x = 8\]

Thus, the solution of x + 4y = 20; x - 2y = 2 is x = 8 and y = 3.

Example 2: Let us now solve the equations (for fruit bought) in the example at the beginning of the chapter.

To solve 12x + 8y = 73; 6x + 9y = 69, let us make the numerical coefficients of x the same in both the equations.

\[(6x \times 2) + (9y \times 2) = 69 \times 2\]

(multiplying LHS and RHS by 2)

\[\Rightarrow 12x + 18y = 138\]

\[\Rightarrow 12x + 18y - (12x + 8y) = 138 - 73\]

(subtracting equations)

\[\Rightarrow 12x + 18y - 12x - 8y = 65\]

\[\Rightarrow 10y = 65\]

\[\Rightarrow y = 6.5\]

\[12x + (8 \times 6.5) = 73\]

(substituting the value of y in any one equation)

\[\Rightarrow 12x + 52 = 73\]

\[\Rightarrow x = \frac{73 - 52}{12} \Rightarrow x = 1.75\]

Thus, the solution of 12x + 8y = 73; 6x + 9y = 69 is x = 1.75 and y = 6.5.

Example 3: Solve:

\[\frac{x + 3}{3} + \frac{y - 2}{4} = 6; \frac{4x + 1}{5} - \frac{3y - 2}{10} = 1\]

In order to simplify the fractional terms of the equation, we multiply both sides of the equation by the LCM of the denominators.

\[\frac{4}{12}(x + 3) + \frac{3}{12}(y - 2) = 6 \times 12\]

and

\[\frac{2}{10}(4x + 1) - \frac{1}{10}(3y - 2) = 1 \times 10\]

\[\Rightarrow 4x + 12 + 3(y - 2) = 72\]

and

\[8x + 2 - (3y - 2) = 10\]

\[\Rightarrow 4x + 12 + 3y - 6 = 72\]

and

\[8x + 2 - 3y + 2 = 10\]

\[\Rightarrow 4x + 3y = 72 - 6\]

and

\[8x - 3y = 10 - 4\]

\[\Rightarrow 4x + 3y = 66\]

and

\[8x - 3y = 6\]

\[\Rightarrow 4x + 3y + 8x - 3y = 66 + 6\]

(adding equations to eliminate 3y)

\[\Rightarrow 12x = 72 \Rightarrow x = 6\]

\[(4 \times 6) + 3y = 66\]

(substituting the value of x in any one equation)

\[\Rightarrow 3y = 66 - 24\]

\[\Rightarrow y = \frac{42}{3} \Rightarrow y = 14\]

Thus, the solution of \[\frac{x + 3}{3} + \frac{y - 2}{4} = 6; \frac{4x + 1}{5} - \frac{3y - 2}{10} = 1\] is x = 6 and y = 14.

Example 4: Solve \[\frac{2}{x} + \frac{6}{y} = 2; \frac{9}{x} - \frac{9}{y} = 3\]

Let \[\frac{1}{x} = a\] and \[\frac{1}{y} = b\], then the simultaneous equations become 2a + 6b = 2; 9a - 9b = 3

Let us make the numerical coefficient of b the same in both equations.

\[(2a \times 3) + (6b \times 3) = 2 \times 3;\]

\[(9a \times 2) - (9b \times 2) = 3 \times 2\]

\[\Rightarrow 6a + 18b = 6;\]

\[18a - 18b = 6\]

\[\Rightarrow 6a + 18b + 18a - 18b = 6 + 6\]

(adding equations to eliminate 18b)

\[\Rightarrow 24a = 12 \Rightarrow a = \frac{1}{2}\]

Now

\[a = \frac{1}{2} = \frac{1}{x} \Rightarrow x = 2\]

\[\frac{2}{2} + \frac{6}{y} = 2\]

(substituting the value of x in any one equation)

\[\Rightarrow 1 + \frac{6}{y} = 2 \Rightarrow \frac{6}{y} = 2 - 1\]

\[\Rightarrow \frac{6}{y} = 1 \Rightarrow 6 = y\]

Thus, the solution of \[\frac{2}{x} + \frac{6}{y} = 2; \frac{9}{x} - \frac{9}{y} = 3\] is x = 2 and y = 6.

Case II: Eliminating one variable by substitution

Step 1: Make one variable the subject of the formula or bring it to the LHS of the equation.

Step 2: Substitute its value (the expression on the RHS) in the other equation.

Step 3: Solve the equation obtained to find the value of the other variable.

Step 4: Substitute the value of the other variable in any one equation to find the value of the first variable.

Example 5: Solve x - 7 = y; 3x - 4y = 8

Select the first equation as it would be easier to express the value of the variable x.

\[x - 7 = y\]

\[\Rightarrow x = y + 7\]

Substitute this value of x in the second equation,

\[3(y + 7) - 4y = 8\]

\[\Rightarrow 3y + 21 - 4y = 8\]

\[\Rightarrow -y = 8 - 21 = -13\]

\[\Rightarrow y = 13\]

(multiplying both sides by -1)

\[x - 7 = 13\]

(substituting the value of y in any one equation)

\[\Rightarrow x = 13 + 7 \Rightarrow x = 20\]

Thus, the solution of x - 7 = y; 3x - 4y = 8 is x = 20 and y = 13.

Example 6: The example at the beginning of the chapter, solving the equations for fruit bought, can also be solved using this method.

12x + 8y = 73; 6x + 9y = 69

\[\Rightarrow 12x = 73 - 8y \Rightarrow x = \frac{73 - 8y}{12}\]

Substituting the value of x in the second equation,

\[6\left(\frac{73 - 8y}{12}\right) + 9y = 69\]

\[\Rightarrow \frac{438}{12} - 4y + 9y = 69\]

\[\Rightarrow 5y = 69 - \frac{438}{12}\]

\[\Rightarrow 5y = \frac{65}{2} \Rightarrow y = \frac{65}{10} = 6.5\]

\[\Rightarrow 12x + (8 \times 6.5) = 73\]

\[\Rightarrow 12x = 73 - 52 \Rightarrow x = \frac{21}{12} = 1.75\]

Thus, the solution of 12x + 8y = 73; 6x + 9y = 69 is x = 1.75 and y = 6.5.

Observe that the formula for x in the above example is not as simple as in the previous example. Thus, the substitution method is used only in equations where one variable can be made the subject in a 'simple' formula.

Try this!

Solve 7x + 8y = 9; x - 2y = 0

Teacher's Note

Simultaneous equations help us solve real-world problems involving two unknowns, such as calculating prices of items or determining ages. Understanding both elimination and substitution methods equips students with flexible problem-solving strategies applicable in physics, economics, and daily financial planning.

Exercise 21.1

1. Solve the following simultaneous equations by eliminating one variable either by adding or by subtracting the equations.

(i) x + 3y = 8; x - 4y = 1

(ii) x + 2y = 22; x - 3y = 2

(iii) x + 3y = 20; x - 6y = -7

(iv) x = 33 - 3y; x = 5y - 15

(v) 3x - 4y = 9; 3x + 5y = 36

(vi) 5x - 2y = 5; 8x - 2y = 14

(vii) 2x + 6y = 30; 6x - 2y = 50

(viii) 5x - 6y = 18; 2x - y = 17

(ix) 2y - 4x = 2; 8x - 3y = 7

(x) 6x + 2y = 46; 6y + 2x = 42

(xi) 5x = 67 - 2y; 3y = 3x + 6

(xii) 9x = 62 + 8y; 3y = 52 - 2x

(xiii) 5y - 4x = 32; 8y + 2x = 26

(xiv) 2x - 3y = 29; 5x + 2y = 25

(xv) 3x + 7y = -43; 7x - 2y = -27

(xvi) 4x - 3y = -24; 11x - 4y = -15

(xvii) \[\frac{x}{4} + 3y = 18; y - \frac{x}{3} = 1\]

(xviii) \[2x + \frac{2y}{3} = 26; 3x - \frac{5y}{6} = 6\]

(xix) \[\frac{x}{2} + \frac{y}{5} = 5; \frac{y}{2} - \frac{x}{3} = 3\]

(xx) \[\frac{x}{6} + \frac{y}{7} = 6; \frac{x}{3} - \frac{y}{2} = 1\]

2. Solve the following simultaneous equations by eliminating one variable by substitution.

(i) x - 2y = 5; 2x + 7y = 32

(ii) x + 12y = 15; 5x - 2y = 13

(iii) 3x + y = 22; 6x - 3y = 9

(iv) 4x + y = 16; 12x - 2y = 8

3. Solve the following simultaneous equations.

(i) 6x + 9y = 6; 12y - 6x = 1

(ii) 15x + 6y = 7; 9y - 5x = 5

(iii) 30x - 35y = 13; 10x + 7y = 7

(iv) 21y + 5x = -1; 20x + 28y = -12

(v) \[\frac{x + 4}{3} + \frac{y - 1}{5} = 5; \frac{x - 2}{4} - \frac{y - 3}{6} = 1\]

(vi) \[\frac{x + 2}{7} + \frac{y - 1}{2} = 6; \frac{x - 3}{3} - \frac{y + 1}{5} = 1\]

(vii) \[\frac{12}{x} - \frac{6}{y} = 1; \frac{6}{x} + \frac{10}{y} = 7\]

(viii) \[\frac{12}{x} - \frac{6}{y} = 6; \frac{16}{x} + \frac{4}{y} = 2\]

(ix) \[\frac{2}{x} + \frac{3}{y} = 18; \frac{6}{x} - \frac{2}{y} = 10\]

(x) \[\frac{6}{x} + \frac{4}{y} = 16; \frac{3}{x} - \frac{6}{y} = -4\]

Teacher's Note

Providing varied problem types helps students recognize when elimination is more efficient than substitution. Practicing with fractions and variables in denominators builds algebraic flexibility necessary for higher mathematics.

Solving Word Problems Using Simultaneous Equations

Word problems involving two unknown quantities can be easily solved by formulating a pair of simultaneous equations.

Example 7: Mr. Karkare bought \[2\frac{1}{2}\] kg of mangoes and \[3\frac{1}{4}\] kg of apples from a fruit-seller for Rs 226.25 while Rekha bought \[\frac{3}{4}\] kg of mangoes and \[1\frac{1}{2}\] kg of apples for Rs 123.50. At what price was the fruit-seller selling each kg of mangoes and apples?

Let the price of 1 kg of mangoes be Rs x and the price of 1 kg of apples be Rs y.

Given \[\frac{5x}{2} + \frac{13y}{4} = 226.25; \frac{7x}{4} + \frac{3y}{2} = 123.50\]

\[\Rightarrow \frac{5x \times 4}{2} + \frac{13y \times 4}{4} = 226.25 \times 4;\]

\[\frac{7x \times 4}{4} + \frac{3y \times 4}{2} = 123.50 \times 4\]

\[\Rightarrow 10x + 13y = 905; \quad 7x + 6y = 494\]

\[\Rightarrow (10x \times 7) + (13y \times 7) = 905 \times 7;\]

\[(7x \times 10) + (6y \times 10) = 494 \times 10\]

\[\Rightarrow 70x + 91y = 6335; \quad 70x + 60y = 4940\]

\[\Rightarrow 70x + 91y - (70x + 60y) = 6335 - 4940\]

\[\Rightarrow 70x + 91y - 70x - 60y = 1395\]

\[\Rightarrow 31y = 1395 \Rightarrow y = \frac{1395}{31} = 45\]

\[10x + (13 \times 45) = 905\]

(substituting value of y in any one equation)

\[\Rightarrow 10x = 905 - 585 \Rightarrow x = \frac{320}{10} = 32\]

Thus, mangoes were being sold for Rs 32 per kg while apples cost Rs 45 per kg.

Example 8: Renu's age is double the age of her cousin Roy. 8 years hence, her age will be 1.5 times Roy's age. Find the present ages of the cousins.

Let Roy present age = x and Renu's age = y

8 years hence, Roy's age = x + 8 and Renu's age = y + 8

\[y = 2x; \quad y + 8 = 1.5(x + 8)\]

Substituting the value of y in the second equation,

\[2x + 8 = 1.5x + 12\]

\[\Rightarrow 2x - 1.5x = 12 - 8\]

\[\Rightarrow 0.5x = 4 \quad \Rightarrow x = 8\]

Substituting the value of x in y = 2x,

\[y = 2 \times 8 = 16\]

Thus, Renu is 16 years old while her cousin Roy is 8 years old.

Example 9: If Sheetal gives 15 berries to Suresh, they will have the same number of berries, but if Suresh gives 15 berries to Sheetal, she will have 4 times as many berries as him. How many berries do Sheetal and Suresh have?

Let Suresh have x berries and Sheetal have y berries. When Sheetal gives 15 berries to Suresh,

\[x + 15 = y - 15\]

When Suresh gives 15 berries to Sheetal,

\[4(x - 15) = y + 15\]

Thus

\[x + 15 = y - 15; \quad 4x - 60 = y + 15\]

\[\Rightarrow x = y - 30; \quad 4x - y = 75\]

Substituting the value of x in the second equation

\[4(y - 30) - y = 75 \quad \Rightarrow \quad 4y - 120 - y = 75\]

\[\Rightarrow 3y = 75 + 120 \quad \Rightarrow \quad y = \frac{195}{3} = 65\]

Substituting the value of y in x = y - 30, we get

\[x = 65 - 30 = 35\]

Thus, Suresh has 35 berries while Sheetal has 65 berries.

Example 10: If 1 is added to the numerator of a fraction and subtracted from its denominator, the fraction becomes 1, but if 2 is subtracted from the numerator and 1 is added to the denominator the fraction becomes \[\frac{1}{2}\]. Find the fraction.

Let the numerator and the denominator of the fraction be x and y. When 1 is added to the numerator and subtracted from the denominator respectively,

\[\frac{x + 1}{y - 1} = 1\]

\[\Rightarrow x + 1 = y - 1 \Rightarrow x = y - 2\]

When 2 is subtracted from the numerator and 1 is added to the denominator,

\[\frac{x - 2}{y + 1} = \frac{1}{2} \Rightarrow 2(x - 2) = y + 1\]

\[\Rightarrow 2x - 4 = y + 1 \Rightarrow 2x - y = 5\]

Substituting x = y - 2 in the above equation, we get

\[2(y - 2) - y = 5\]

\[\Rightarrow 2y - 4 - y = 5 \quad \Rightarrow \quad y = 5 + 4 = 9\]

Substituting the value of y in x = y - 2, we get

\[x = 9 - 2 = 7\]

Thus, the fraction is \[\frac{7}{9}\].

Example 11: A 2-digit number is 7 times the sum of its digits. If 27 is subtracted from the number, its digits are reversed. Find the number.

Let the digit in the ones place be x and the digit in the tens place be y. Then the 2-digit number is 10y + x. Given that the number is 7 times the sum of its digits,

Or \[10y + x = 7(y + x)\]

\[\Rightarrow 10y + x = 7y + 7x\]

\[\Rightarrow 10y - 7y = 7x - x\]

\[\Rightarrow 3y = 6x \Rightarrow y = 2x\]

When 27 is subtracted from 10y + x, its digits are reversed or the new number has x in the tens place and y in the ones place.

Or \[10y + x - 27 = 10x + y\]

\[\Rightarrow 10y - y = 10x - x + 27\]

\[\Rightarrow 9y = 9x + 27\]

Substituting y = 2x in the above equation, we get

\[9 \times 2x = 9x + 27\]

\[\Rightarrow 18x - 9x = 27\]

\[\Rightarrow 9x = 27 \Rightarrow x = \frac{27}{9} = 3\]

Substituting the value of x in y = 2x, we get

\[y = 2 \times 3 = 6\]

Thus, the number is 63.

Example 12: A boat first travels 36 km upstream and 30 km downstream in \[8\frac{1}{2}\] hours. Then it travels for \[3\frac{1}{2}\] hours going 24 km downstream and 21 km upstream. Find the speed of the boat in still water and the speed of the