CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Notes Set 05

Algebraic Methods of Solving a Pair of Linear Equations

For finding the solutions of a pair of linear equations, the following algebraic methods are used: 1. Substitution method, 2. Elimination method.

Substitution Method

In this method, we first solve one equation to get the value of one variable (y) in terms of the other variable (x). Then, in the second equation, this value of y is substituted. The second equation becomes a linear equation in x only, which may be solved for x. Then we can find the value of y by putting the value of x in the first equation. Solving a system of linear equations using this method is known as the method of elimination by substitution.

A pair of linear equations in two variables x and y can be solved by the substitution method using the following steps, illustrated for \( 2x - 3y = 7 \) ...(i) and \( x + y = 1 \) ...(ii).
Step 1. Express y in terms of x (or x in terms of y) by choosing one of the two equations. Using the second equation, \( x + y = 1 \Rightarrow y = 1 - x \).
Step 2. The value of y obtained in Step 1 is substituted in the other equation to get a linear equation in x: \( 2x - 3(1 - x) = 7 \).
Step 3. Solve the linear equation obtained in Step 2: \( 2x - 3 + 3x = 7 \Rightarrow 5x = 10 \Rightarrow x = 2 \).
Step 4. The value of x obtained is substituted in the relation obtained in Step 1 to find y: \( y = 1 - x = 1 - 2 = -1 \).
Therefore the solution of the given pair of equations is \( x = 2 \) and \( y = -1 \).

Note. This method is known as the method of elimination by substitution. The reason is that y is eliminated from the second equation (‘elimination’) because the value of y is substituted in the second equation (‘substitution’).

Solved Examples on the Substitution Method

Example 1. Solve the given equations for x and y: \( 4x + 3y = 24 \), \( 3y - 2x = 6 \).
Answer:
The given equations are \( 4x + 3y = 24 \) ...(i) and \( 3y - 2x = 6 \) ...(ii).
From (i), \( y = \frac{24 - 4x}{3} \) ...(iii).
Substituting in (ii): \( 3\left(\frac{24 - 4x}{3}\right) - 2x = 6 \Rightarrow 24 - 4x - 2x = 6 \Rightarrow -6x = -18 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (iii): \( y = \frac{24 - 12}{3} = \frac{12}{3} = 4 \).
Therefore, \( x = 3 \) and \( y = 4 \).

Example 2. Solve the following pair of linear equations by the substitution method: \( \sqrt{2}x + \sqrt{3}y = 0 \) and \( \sqrt{3}x - \sqrt{8}y = 0 \).
Answer:
The given equations are \( \sqrt{2}x + \sqrt{3}y = 0 \) ...(i) and \( \sqrt{3}x - \sqrt{8}y = 0 \) ...(ii).
From (i), \( y = \frac{-\sqrt{2}x}{\sqrt{3}} \) ...(iii).
Substituting in (ii): \( \sqrt{3}x - \sqrt{8}\left(\frac{-\sqrt{2}x}{\sqrt{3}}\right) = 0 \Rightarrow \sqrt{3}x + \frac{4x}{\sqrt{3}} = 0 \Rightarrow 3x + 4x = 0 \Rightarrow 7x = 0 \Rightarrow x = 0 \).
Substituting \( x = 0 \) in (iii): \( y = \frac{-\sqrt{2} \times 0}{\sqrt{3}} = 0 \).
Therefore the solution is \( x = 0 \) and \( y = 0 \).

Example 3. Rajan combines 2 solutions to form a mixture that is 40% acid. One solution was 20% acid and the other was 50% acid. If Rajan had 90 millilitres of the mixture, how much of each solution was used to create the mixture?
Answer:
Let the volume of solution A (20% acid) be x mL and the volume of solution B (50% acid) be y mL.
Volume of A + volume of B = volume of mixture: \( x + y = 90 \) ...(i).
Acid in A + acid in B = acid in mixture: \( 0.2x + 0.5y = 0.4 \times 90 = 36 \) ...(ii).
Multiplying (ii) by 10: \( 2x + 5y = 360 \) ...(iv). From (i), \( x = 90 - y \) ...(iii).
Putting (iii) in (iv): \( 2(90 - y) + 5y = 360 \Rightarrow 180 + 3y = 360 \Rightarrow y = 60 \). Then \( x = 90 - 60 = 30 \).
Hence 30 mL of solution A and 60 mL of solution B were used to form the mixture.

Example 4. One day, the National Museum of India, New Delhi admitted 321 adults and children and collected Rs. 1590. The price of admission is Rs. 6 for an adult and Rs. 4 for a child. How many adults and how many children were admitted to the museum on the day?
Answer:
Let x and y be the number of adults and children respectively. Total number of people = 321, so \( x + y = 321 \) ...(i). Total amount collected = Rs. 1590, so \( 6x + 4y = 1590 \) ...(ii).
From (i), \( x = 321 - y \) ...(iii). Putting in (ii): \( 6(321 - y) + 4y = 1590 \Rightarrow 1926 - 2y = 1590 \Rightarrow y = 168 \).
Then \( x = 321 - 168 = 153 \).
Hence 153 adults and 168 children were admitted to the museum that day.

Elimination Method

A pair of linear equations in two variables x and y can be solved by the elimination method using the following steps, illustrated for \( 2x + 3y = 19 \) ...(i) and \( 5x + 4y = 37 \) ...(ii).
Step 1. Multiply the given equations by suitable numbers so that the coefficient of one of the variables becomes numerically equal. To eliminate y, multiply the first equation by 4 and the second by 3: \( 8x + 12y = 76 \) ...(iii) and \( 15x + 12y = 111 \) ...(iv).
Step 2. If the numerically equal coefficients are opposite in sign, add the new equations; otherwise subtract them. Here the signs are the same, so subtract (iii) from (iv): \( 7x = 35 \).
Step 3. Solve the linear equation in one variable obtained in Step 2: \( x = 5 \).
Step 4. Substitute this value in any of the two equations to find the other variable. From (ii): \( 5(5) + 4y = 37 \Rightarrow 4y = 12 \Rightarrow y = 3 \).
Therefore the solution of the given pair of equations is \( x = 5 \) and \( y = 3 \).

Solved Examples on the Elimination Method

Example 1. Use the elimination method to solve the following pair of equations: \( x + y = 5 \) and \( 2x - 3y = 4 \).
Answer:
The given equations are \( x + y = 5 \) ...(1) and \( 2x - 3y = 4 \) ...(2).
Multiplying (1) by 3 and (2) by 1 and adding: \( 3x + 3y = 15 \) and \( 2x - 3y = 4 \) give \( 5x = 19 \Rightarrow x = \frac{19}{5} \).
Putting \( x = \frac{19}{5} \) in (1): \( \frac{19}{5} + y = 5 \Rightarrow y = 5 - \frac{19}{5} = \frac{6}{5} \).
Therefore, \( x = \frac{19}{5} \) and \( y = \frac{6}{5} \).

Example 2. Solve the following pair of linear equations by the elimination method: \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \).
Answer:
The given equations are \( 3x + 4y = 10 \) ...(1) and \( 2x - 2y = 2 \) ...(2).
Multiplying (2) by 2: \( 4x - 4y = 4 \) ...(3). Adding (1) and (3): \( 7x = 14 \Rightarrow x = 2 \).
Putting \( x = 2 \) in (2): \( 4 - 2y = 2 \Rightarrow y = 1 \).
Hence \( x = 2 \) and \( y = 1 \).

Example 3. A gold crown, suspected of containing some silver, was found to have a mass of 714 grams and a volume of 46 cubic centimetres. The density of gold is about 19 grams per cubic centimetre. The density of silver is about 10.5 grams per cubic centimetre. What percent of the crown is silver?
Answer:
Let the volume of gold be G cm³ and the volume of silver be S cm³. Total volume = 46 cm³ and total mass = 714 g.
\( G + S = 46 \) ...(i) and \( 19G + 10.5S = 714 \) ...(ii).
Multiplying (i) by 19: \( 19G + 19S = 874 \) ...(iii). Subtracting (ii) from (iii): \( 8.5S = 160 \Rightarrow S = \frac{160}{8.5} \approx 18.8 \).
Hence the volume of silver is nearly 19 cm³. As the crown has a volume of 46 cm³, the crown is \( \frac{19}{46} \times 100 \approx 41\% \) silver by volume.

Equations of the Form ax + by = c and bx + ay = d, where a ≠ b

To solve \( ax + by = c \) ...(i) and \( bx + ay = d \) ...(ii) where \( a \neq b \), follow these steps.
Step 1. Add (i) and (ii) to get \( (a + b)x + (a + b)y = c + d \), that is, \( x + y = \frac{c + d}{a + b} \).
Step 2. Subtract (ii) from (i) to get \( (a - b)x - (a - b)y = c - d \), that is, \( x - y = \frac{c - d}{a - b} \).
Step 3. Solve the two equations obtained in Step 1 and Step 2 to get x and y.

Example 1. Solve the following equation for x and y: \( 47x + 31y = 63 \), \( 31x + 47y = 15 \).
Answer:
The given equations are \( 47x + 31y = 63 \) ...(i) and \( 31x + 47y = 15 \) ...(ii).
Adding (i) and (ii): \( 78x + 78y = 78 \Rightarrow x + y = 1 \) ...(iii).
Subtracting (ii) from (i): \( 16x - 16y = 48 \Rightarrow x - y = 3 \) ...(iv).
Adding (iii) and (iv): \( 2x = 4 \Rightarrow x = 2 \). Putting in (iii): \( 2 + y = 1 \Rightarrow y = -1 \).
Therefore, \( x = 2 \) and \( y = -1 \).

Word Problems on Fractions and Costs

Example 1. It is given that the denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator. Determine the fraction.
Answer:
Let the numerator be x and the denominator be y, so the fraction is \( \frac{x}{y} \).
Given: \( y = 2x + 4 \), i.e., \( 2x - y + 4 = 0 \) ...(i). Also \( y - 6 = 12(x - 6) \Rightarrow 12x - y - 66 = 0 \) ...(ii).
Subtracting (i) from (ii): \( 10x - 70 = 0 \Rightarrow x = 7 \). Putting in (i): \( 14 - y + 4 = 0 \Rightarrow y = 18 \).
Therefore the required fraction is \( \frac{7}{18} \).

Example 2. 37 pens and 53 pencils together cost Rs. 320, while 53 pens and 37 pencils together cost Rs. 400. Find the cost of a pen and that of a pencil.
Answer:
Let the cost of a pen be Rs. x and the cost of a pencil be Rs. y. Then \( 37x + 53y = 320 \) ...(i) and \( 53x + 37y = 400 \) ...(ii).
Adding: \( 90x + 90y = 720 \Rightarrow x + y = 8 \) ...(iii). Subtracting (i) from (ii): \( 16x - 16y = 80 \Rightarrow x - y = 5 \) ...(iv).
Adding (iii) and (iv): \( 2x = 13 \Rightarrow x = 6.5 \). From (iii): \( y = 8 - 6.5 = 1.5 \).
Therefore the cost of one pen is Rs. 6.50 and the cost of one pencil is Rs. 1.50.

Problems Based on Time, Distance and Speed

The following formulae are used to solve problems based on time, distance and speed: Distance = Speed × Time, Time = \( \frac{\text{Distance}}{\text{Speed}} \) and Speed = \( \frac{\text{Distance}}{\text{Time}} \).

Some basic definitions.
1. Stream: the moving water in a river is called a stream.
2. Upstream: the boat is moving in the direction opposite to the stream.
3. Downstream: the boat is moving along the direction of the flow of the stream.
4. Still water: when the water is still and not moving, as in a pond, it is called still water.
If the speed of a boat in still water is x km/h and the speed of the stream (current) is y km/h, then speed upstream = \( (x - y) \) km/h and speed downstream = \( (x + y) \) km/h.

Example 1. A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream.
Answer:
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h. Speed upstream = \( (x - y) \) km/h and speed downstream = \( (x + y) \) km/h.
Time = Distance ÷ Speed, so \( \frac{32}{x - y} + \frac{36}{x + y} = 7 \) ...(i) and \( \frac{40}{x - y} + \frac{48}{x + y} = 9 \) ...(ii).
Put \( \frac{1}{x - y} = u \) and \( \frac{1}{x + y} = v \): \( 32u + 36v = 7 \) ...(iii) and \( 40u + 48v = 9 \) ...(iv).
Multiplying (iii) by 5 and (iv) by 4: \( 160u + 180v = 35 \) and \( 160u + 192v = 36 \). Subtracting: \( 12v = 1 \Rightarrow v = \frac{1}{12} \). Then \( 32u = 7 - 3 = 4 \Rightarrow u = \frac{1}{8} \).
So \( x - y = 8 \) ...(v) and \( x + y = 12 \) ...(vi). Solving, \( x = 10 \) and \( y = 2 \).
Therefore the speed of the boat in still water is 10 km/h and the speed of the stream is 2 km/h.

Summary

1. An equation of the form \( ax + by + c = 0 \), where a, b, c are real numbers and \( a, b \neq 0 \), is called a linear equation in two variables x and y.
2. Two linear equations in the same two variables are called a pair of linear equations in two variables. The general form is \( a_1x + b_1y + c_1 = 0 \), \( a_2x + b_2y + c_2 = 0 \), where \( a_1, a_2, b_1, b_2, c_1, c_2 \) are real numbers with \( a_1^2 + b_1^2 \neq 0 \) and \( a_2^2 + b_2^2 \neq 0 \).
3. A pair of linear equations can be solved by the graphical method and by the algebraic methods (substitution and elimination). The graph of a linear equation in two variables is always a straight line.

4. Substitution method: express one variable in terms of the other from one of the equations and substitute this in the other equation.
5. Elimination method: remove one of the variables by making its coefficients numerically equal and adding or subtracting the equations.