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Class 10th Maths
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Online Class For 10th Standard Students (CBSE) (English Medium)
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Frequently Asked Questions on Chapter 3 – Pair of Linear Equations in Two Variables

Given the linear equation 2x + 3y – 8 = 0?

To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy the below condition; a1/a2=b1/b2=c1/c2
Thus, another equation could be 4x + 6y – 16 = 0, such that; a1/a2=2/4=1/2,b1/b2=3/6=1/2,c1/c2=-8/-16=1/2 Clearly, you can see another equation satisfies the condition.

Solve the following pair of linear equations by the substitution method x + y = 14 x – y = 4?

Given, x + y = 14 and x – y = 4 are the two equations. From 1st equation, we get, x = 14 – y Now, substitute the value of x in second equation to get, (14 – y) – y = 4 14 – 2y = 4 2y = 10 Or y = 5 By the value of y, we can now find the exact value of x; ∵ x = 14 – y ∴ x = 14 – 5 Or x = 9 Hence, x = 9 and y = 5.

Solve the following pair of linear equations by the elimination method and the substitution method 3x + 4y = 10 and 2x – 2y = 2?

By the method of elimination. 3x + 4y = 10……………………….(i) 2x – 2y = 2 ………………………. (ii) When the equation (i) and (ii) is multiplied by 2, we get: 4x – 4y = 4 ………………………..(iii) When the Equation (i) and (iii) are added, we get: 7x = 14 x = 2 ……………………………….(iv) Substituting equation (iv) in (i) we get, 6 + 4y = 10 4y = 4 y = 1
Hence, x = 2 and y = 1
By the Method of Substitution
From equation (ii) we get, x = 1 + y……………………………… (v) Substituting equation (v) in equation (i) we get, 3(1 + y) + 4y = 10 7y = 7 y = 1 When y = 1 is substituted in equation (v) we get, A = 1 + 1 = 2
Therefore, A = 2 and B = 1

Solve the following pairs of equations by reducing them to a pair of linear equations 1/2x+1/3y=2 1/3x+1/2y=13/6?

Let us assume 1/a=m and 1/b=n, then the equation will change as follows. m/2+n/3=2 3m+2n-12=0 ………………………….(1) m/3+n/2=13/6 2m+3n-13=0 ………………………….(2) Now, using cross-multiplication method, we get, m/(-26-(-36) )=n/(-24-(-39) )=1/(9-4) m/10=q/15=1/5 m/10=1/5 and q/15=15 So, m = 2 and n = 3 1/a=2 and 1/b=3 a=1/2 and b=1/3

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Let the number of rows be A and the number of students in a row be B. Total number of students = Number of rows x Number of students in a row =AB Using the information,that is given,
First Condition:
Total number of students = (A – 1) ( B + 3) Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3 Or 3A – B – 3 = 0 Or 3A – Y = 3 – – – – – – – – – – – – – (1)
Second condition:
Total Number of students = (A + 2 ) ( B – 3 ) Or AB = AB + 2B – 3A – 6 Or 3A – 2B = -6 – – – – – – – – – (2) When equation (2) is subtracted from (1) (3A – B) – (3A – 2B) = 3 – (-6) -B + 2B = 3 + 6B = 9 By using the equation (1) we get, 3A – 9 =3 3A = 9+3 = 12 A = 4 Number of rows, A = 4 Number of students in a row, B = 9 Number of total students in a class => AB => 4 x 9 = 36

By using the substitution method solve the following pair of linear equations x + y = 14 and x – y = 4.

x + y = 14 and x – y = 4 are the two equations. From 1st equation, we get, x = 14 – y Now, substitute the value of x in second equation we get, (14 – y) – y = 4 14 – 2y = 4 2y = 10 Or y = 5 By the value of y, we can now find the value of x, x = 14 – y x = 14 – 5 Or x = 9 Hence value of x = 9 and y = 5.

Form the pair of linear equations and find their solution by using the substitution method. Find the numbers if the difference between two numbers is 26 and one number is three times the other.

Let the two numbers be x and y respectively, such that y > x. So according to the question, y = 3x ……………… (1) y – x = 26 …………..(2) Now by substituting the value of (1) in (2), we get 3x – x = 26 x = 13 ……………. (3) Substituting equation (3) in (1), we get y = 39 Hence, the numbers are 13 and 39.

Solve the pair of linear equations by the elimination method and the substitution method: 3x + 4y = 10 and 2x – 2y = 2

By using the method of elimination. 3x + 4y = 10……………………….(i) 2x – 2y = 2 ………………………. (ii) When the equation (i) and (ii) is multiplied by 2, we get: 4x – 4y = 4 ………………………..(iii) When the Equation (i) and (iii) are added, we get: 7x = 14 x = 2 ……………………………….(iv) Substituting equation (iv) in (i) we get, 6 + 4y = 10 4y = 4 y = 1 Hence, x = 2 and y = 1 By the method of Substitution From equation (ii) we get, x = 1 + y……………………………… (v) Substituting equation (v) in equation (i) we get, 3(1 + y) + 4y = 10 7y = 7 y = 1 When y = 1 is substituted in equation (v) we get, A = 1 + 1 = 2 Hence, A = 2 and B = 1

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