Math Olympiad Test: Linear Educations in Two Variables- 2 - Class 9 MCQ

The given linear equation is 2x + 3y = 13
Now substituting the values of x and y from option in equation (i), we see
For (a) 2 × 4 + 3 × 2 = 8 + 6 = 14 ≠ 13
∴ (a) is not correct option.
Again 2 × 2 + 3 × 3 = 4 + 9 = 13 = 13
∴ (b) is required answer.

Given equations are
3x + 16y = 13 and x + y = p
These equations may have many set of solutions commons for different values of p.

(k, k) will satisfy x - 5y + 6 = 0
⇒ k² - 5k + 6 = 0
⇒ k² - 3k - 2k + 6 = 0
⇒ k(k -3) - 2(k - 3) = 0
⇒ (k -3) (k - 2) = 0
⇒ k - 2 = 0 or, k - 3 = 0
⇒ k = 2 or 3

(k³, 0) satisfies the equation, x - y + 8 = 0
⇒ k³ - (0) + 8 = 0
⇒ k³ = -8
⇒ k = (–8)^1/3 = -2

x + 3y - 3x - y + x - y = α - b
⇒ - x + y = α + b
⇒ y - x = α - 6
∴ (x, y) is satisfied by (b, α)

Let the point of intersection of lines be (a, b)
∴ 3α + 4b = 12 and 6α + 8b = 48
The above two equations have no solutions for (α, b)
∴ The graph will not intersect.

∴ The ordinate of every point on x-axis = 0
∴ The line 3x + 4y = 15 and the x-axis will intersect where value y of the line becomes zero
∴ 3x = 15
⇒ x = 5
∴ The point of intersection is (5, 0)

At y - axis, ordinate ≠ 0 abscissa = 0
∴ x = 0
⇒ 36y = 108
⇒ y = 3
∴ point of intersection = (0, 3)

The distance between the graphs = 3 - (-3)
= 3 + 3 = 6 units 

The equation can be written as,

∴ For different values of x, different values of y will exist.
∴ The above equation has many solutions.

Equation of line parallel to x-axis, will be of the form y = constant.
∴ Desired equation of line y = -4

Given 3x = 9 
⇒ x = 9/3 = 3
∴ Line is parallel to y-axis and passes through x = 3
∴ Point (3, 9) will lie on 3x = 9 

Income of A = 8x, Income of B = 7x
Expenditure of A = 19y
Expenditure of B = 16y
According to the question
8x -19y = 2500
⇒ 7x -16 y 2500
⇒ From these two equations,
We have,
x = 1500, y = 500
∴ Income of A = Rs 8 × 1500
= Rs 12000

Let the sum of ages of two sons be x, and their father’s age = y years
According to the question,
y = 3x
and y + 5 = 2(x + 10)
∴ 3x + 5 = 2x + 20
⇒ x = 15 years
and y = 45 years

We have ∠A + ∠B + ∠C = 180°
According to the question
∠C = 3, ∠B = 2 (180° - ∠C)
⇒ ∠C = 360 - 2∠C
⇒ 3∠C =  360°
⇒ ∠C = 120°