Ch 3: Linear Equations (Part 1)
Based on the latest CBSE rationalized syllabus. Master Graphical, Substitution, and Elimination Methods.
Exercise 3.1 Solutions
Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls.
Solution:
Let the number of boys be x and girls be y.
Equation 1: x + y = 10 (Total students)
Equation 2: y = x + 4 ⇒ -x + y = 4
Plotting these on a graph, the lines intersect at the point (3, 7).
Therefore, Number of boys = 3, Number of girls = 7.
Q2. On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the lines intersect at a point, are parallel or coincident:
(i) 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0
a1/a2 = 5/7, b1/b2 = -4/6 = -2/3.
Since a1/a2 ≠ b1/b2, the lines intersect at a single point (Consistent).
(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
a1/a2 = 9/18 = 1/2, b1/b2 = 3/6 = 1/2, c1/c2 = 12/24 = 1/2.
Since a1/a2 = b1/b2 = c1/c2, the lines are coincident (Consistent, infinite solutions).
Exercise 3.2 (Substitution & Elimination)
Q1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14 ; x - y = 4
From eq 2: x = y + 4. Substitute this in eq 1:
(y + 4) + y = 14 ⇒ 2y = 10 ⇒ y = 5.
Put y=5 in x = y + 4 ⇒ x = 5 + 4 = 9.
Q3. Form the pair of linear equations for the following problems and find their solution by elimination method.
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Let fraction be x/y.
Condition 1: (x + 1) / (y - 1) = 1 ⇒ x - y = -2 (Eq 1)
Condition 2: x / (y + 1) = 1/2 ⇒ 2x - y = 1 (Eq 2)
Subtract Eq 1 from Eq 2:
(2x - y) - (x - y) = 1 - (-2)
x = 3
Put x=3 in Eq 1: 3 - y = -2 ⇒ y = 5.
Therefore, the fraction is 3/5.
Important PYQs
Q. For what value of 'k', will the following pair of linear equations have infinitely many solutions? kx + 3y - (k-3) = 0 ; 12x + ky - k = 0
Solution:
For infinitely many solutions: a1/a2 = b1/b2 = c1/c2
k/12 = 3/k = -(k-3)/(-k)
Taking first two parts: k² = 36 ⇒ k = ±6
Taking last two parts: 3/k = (k-3)/k ⇒ 3k = k² - 3k ⇒ k² - 6k = 0 ⇒ k(k - 6) = 0 ⇒ k = 0 or k = 6.
The common value satisfying both is k = 6.