Linear Equations In Two Variables

(i) -2x+3y =12

⇒ -2x + 3y – 12 = 0

⇒ a = - 2, b = 3, c = -12

(ii) x--5 =0

⇒ 2x – y – 10 = 0

⇒ a = 2, b = -1, c = -10

(iii) 2x+3y=

⇒ 2x + 3y – = 0

⇒ a = 2, b = 3, c =

(iv) 3x = -7y

⇒ 3x + 7y = 0

⇒ a = 3, b = 7 and c = 0

(v) 2x+3 = 0

⇒ a = 2, b = 0 and c = 3

(vi) y – 5 = 0

⇒ a = 0, b = 1, c = - 5

(vii) 4 = 3x

⇒ 3x – 4 = 0

⇒ a = 3, b = 0, c = -4

(viii) y = x/2

⇒ x – 2y = 0

⇒ a = 1, b = -2, c = 0

(i) 2x= -3

⇒ 2x + 0y + 3 = 0

(ii) y = 3

⇒ 0x + y – 3 = 0

(iii) 5x = 7/2

⇒ 10x + 0y – 7 = 0

(iv) y = 3x/2

⇒ 3x – 2y + 0 = 0

Let the cost of one fountain pen be Rs. ‘x’ and the cost of one ball pen is Rs. ‘y’

Given, cost of ball pen is Rs. 5 less than half of the cost of fountain pen.

⇒ y = () – 5

⇒ 2y = x – 10

⇒ x – 2y – 10 = 0

(i) 3x + 4y = 7

At, x = 1

3 + 4y = 7

⇒ y = 1

Thus, x = 1, y = 1 is a solution

At, x = 0,

0 + 4y = 7

⇒ y = 7/4

Thus, x = 0, y = 7/4 is a solution.

(ii) x = 6y

At, y = 0

⇒ x = 0

Thus, x = 0, y = 0 is a solution.

At y = 1,

⇒ x = 6

Thus, x = 6, y = 1 is a solution

(iii) x+πy =4

At x = 0,

πy = 4

⇒ y = 4/π

Thus, x = 0, y = 4/π is a solution

At y = 0,

⇒ x + 0 = 4

⇒ x = 4

Thus, x = 4, y = 0 is a solution

(iv) x-y =4

At x = 0,

⇒ 0 – y = 4

⇒ y = -4

Thus, x = 0, y = 4 is a solution

At x = 3,

⇒ 2 – y = 4

⇒ y = -2

Thus, x = 3, y = -2 is a solution

(i) 5x – 2y = 10

At x = 0,

⇒ 0 – 2y = 10

⇒ y = -5

Thus, x = 0, y = -5 is a solution

At y = 0 ,

⇒ 5x = 10

⇒ x = 2

Thus, x = 2, y = 0 is a solution

(ii) -4x+3y =12

At x = 0,

0 + 3y = 12

⇒ y = 4

Thus, x = 0 and y = 4 is a solution

At y = 0,

-4x + 0 = 12

⇒ x = -3

Thus, x = -3 and y = 0 is a solution

(iii) 2x+3y =24

At x = 0,

⇒ 0 + 3y = 24

⇒ y = 8

Thus, (0, 8) is a solution

At y = 0

(i) (3, 0)

⇒ 2 × 3 – 0 = 6

⇒ 6 = 6

Thus (3, 0) is a solution

(ii) (0, 6)

⇒ 2 × 0 – 6 = 6

⇒ -6 = 6

This is not true, thus (0, 6) is not a solution

(iii) (2, -2)

⇒ 2 × 2 + 2 = 6

⇒ 6 = 6

Thus, (2, -2) is a solution

(iv) (√3, 0)

⇒ 2√3 – 0 = 6

⇒ 2√3 = 6

This is not true, thus (2√3, 0) is not a solution

(v) (1/2, - 5)

⇒ (2/2) – (-5) = 6

⇒ 6 = 6

Thus (1/2, -5) is a solution.

3x + 4y = k

If x = -1, y = 2 is a solution of the equation, then

⇒ 3 × -1 + 4 × 2 = k

⇒ k = 5

x + 4y – 7 = 0

For, x=-λ and y =to be a solution

⇒ -λ + 4 × (5/2) – 7 = 0

⇒ λ = 10 – 7 = 3

Given, x = 2α+1 and y = α-1 is a solution of the equation 2x-3y+5=0

⇒ 2 × (2α + 1) – 3(α – 1) + 5 = 0

⇒ 4α +_2 – 3α + 3 + 5 = 0

⇒ α = -10

Given, x=1 and y=6 is a solution of the equation 8x-ay+a²=0

⇒ 8 × 1 – a × 6 + a² = 0

⇒ a² – 6a + 8 = 0

⇒ a² – 4a – 2a + 8 = 0

⇒ a(a – 4) – 2(a – 4) = 0

⇒ (a – 2)(a – 4) = 0

⇒ a = 2, 4

(i) x + y = 4

It passes through (0, 4) and (4, 0)

(ii) x – y = 2

It passes through (0, -2) and (2, 0)

(iii) –x + y = 6

It passes through (0, 6) and (-6, 0)

(iv) y = 2x

It passes through (0, 0)

(v) 3x + 5y = 15

It passes through (5, 0) and (0, 3)

(vi) x/2 – y/3 = 2

It passes through (4, 0) and (0, - 6)

(vii) (x – 2)/3 = y – 3

⇒ x – 3y + 7 = 0

It passes through (-7, 0) and (0, 7/3)

(viii) 2y = -x + 1

It passes through (0, 1/2) and (1, 0)

Equation of line passing through a point (a, b)

(x – a) = m(y – b)

Where ‘m’ is an integer

There are infinite lines passing through a point.

Equation of two lines passing through (3, 12)

Taking m = 1, 2

⇒ x – 3 = y – 12

⇒ x – y + 9 = 0

And, (x – 3) = 2(y – 12)

⇒ x - 3 = 2y – 24

⇒ x – 2y + 21 = 0

Given, a three-wheeler scooter charges Rs. 15 for first kilometre and Rs. 8 each for every subsequent

kilometre. For a distance of x km, an amount of Rs. y is paid.

⇒ 15 × 1 + (x – 1) × 8 = y

⇒ 8x – y + 7 = 0

Given, lending library has a fixed charge for the first three days and an additional charge for each day

thereafter. Aarushi paid Rs. 27 for a book kept for seven days. If fixed charges are Rs. x and per day charges

are Rs. y.

⇒ 3 × x + (7 – 3) × y = 27

⇒ 3x + 4y = 27

Given, a number is 27 more than the number obtained by reversing its digits.

Number is 10y + x

Reverse of the number is 10x + y

⇒ 10y + x = 10x + y + 27

⇒ 9x – 9y + 27 = 0

⇒ x – y + 3 = 0

Given, sum of a two digit number and the number obtained by reversing the order of its digits is 121

Number is 10y + x

Reverse of the number is 10x + y

⇒ 10y + x + 10x + y = 121

⇒ 11x + 11y = 121

⇒ x + y = 11

Thus, the line passing through (-1, 3) and (3, 5) passes through the point (1, 4).

From the graph, the line passes through (1, - 1) and (-1, 1)

(i) y = x

⇒ 1 = -1 which is not true

(ii) x + y = 0

⇒ 1 – 1 = 0

⇒ 0 = 0

Also, - 1 + 1 = 0

⇒ 0 = 0

Thus x + y = 0 is a equation

(iii) y = 2x

⇒ -1 = 2 which is not true

(iv) 2 + 3y = 7x

⇒ 2 – 3 = 7

⇒ -1 = 7 which is not true

The line passes through (-1, 3) and (2, 0)

(i) y = x + 2

⇒ 3 = -1 + 2

⇒ 3 = 1 which is not true

(ii) y = x – 2

⇒ 3 = -1 – 2

⇒ 3 = -3 which is not true

(iii) y = -x + 2

⇒ 3 = 1 + 2

⇒ 3 = 3

Also, for (2, 0)

⇒ 0 = -2 + 2

⇒ 0 = 0

Thus, y = -x + 2 is the equation

(iv) x + 2y = 6

⇒ -1 + 6 = 6

⇒ 5 = 6 which is not true

Given, the point (2,-2) lies on the graph of the linear equation 5x+ky=4

⇒ 5 × 2 – 2k = 4

⇒ 2k = 6

⇒ k = 3

2x+3y=12

It passes through (6, 0) and (0, 4)

From the graph, at y = 3, x = 3/2

And at x = -3, y = 6

(i) (3/2, 3)

(ii) (-3, 6)

(i) 6x – 3y = 12

It cuts the coordinate axis at (2, 0) and (0, -4)

(ii) –x + 4y = 8

It cuts the coordinate axis at (-8, 0) and (0, 2)

(iii) 2x + y = 6

It cuts the coordinate axis at (0, 6) and (3, 0)

(iv) 3x + 2y + 6 = 0

It cuts the coordinate axis at (-2, 0) and (0, -3)

2x + y = 6

It cuts the coordinate axis at (3, 0) and (0, 6)

Area of the region = (1/2) × 3 × 6 = 9 sq. unit

x/3 + y/4 = 1

The line cuts the axes at (3, 0) and (4, 0)

Area of the triangle fomed

Y = |x|

For every x, y is positive

Y = x for x > 0 and y = -x for x < 0

Y = |x| + 2

y = x + 2 for x > 0

And y = -x + 2 for x < 0

X-Y=1

2x + 3y =12

2x + 3y = 12 passes through (6, 0) and (0, 4)

x – y = 1 passes through (1, 0) and (0, -1)

Coordinates of the vertices of the triangle formed with the y axis are (0, 4), (0, -1) and (3, 2)

Base of the triangle = 4 + 1 =

Height of the triangle = 3

Area of the triangle

⇒ Area of the triangle

4x-3y+4=0 passes through (-1, 0) and (0, 4/3)

4x + 3y – 20 = 0 passes through (5, 0) and (0, 20/3)

Coordinates of the vertices of triangle with x-axis, (-1, 0), (5, 0) and (2, 4)

Height of the triangle = 4

Base of the triangle = 5 + 1 = 6

Area of the triangle = 1/2 × 4 × 6 = 12 sq. unit

Given, path of a train A is given by the equation 3x+4y-12=0 and the path of another train B is given by the equation 6x+8y-48=0.

3x + 4y – 12 = 0 passes through (4, 0) and (0, 3)

6x + 8y – 48 = 0 passes through (8, 0) and (0, 6)

Given, present ages of Aarushi and Ravish are x and y years respectively.

Eq1: (y – 7) = 7(x – 7 )

⇒ y = 7x – 42 which passes through (0, -42) and (6, 0)

Eq2: (y + 3) = 3(x + 3)

⇒ y = 3x + 6 which passes through (0, 6) and (-2, 0)

Uniform speed is 60 km/hr

Speed = distance/time

⇒ distance = 60 × time

Slope of diatance time graph is 60.

(i) Distance travelled in 2.5 hours = 150 km

(ii) Distance travelled in 0.5 hours = 30 km

(i) x = 2

(ii) y + 3 = 0

(iii) y = 3

(iv) 2x + 9 = 0

(v) 3x – 5 =0

2x + 13 = 0

(i) One variable

(ii) two variables

3x + 2 = x – 8

⇒ 2x = -10

⇒ x = -5

(i) on number line

(ii) on the Cartesian plane

Slope of the line parallel to x – axis is 0

Eq of line parallel to x-axis and passing through (a, b) is y – b = 0

(i) (0, 3)

⇒ y – 3 = 0

(ii) (0, 4)

⇒ y – 4 = 0

(iii) (2, - 5)

⇒ y + 5 = 0

(iv) (-4, -3)

⇒ y + 3 = 0

Slope of the line that is parallel to y-axis is infinity.

Eq of the line parallel to y-axis passing through (a, b) is x – a = 0

(i) (4, 0)

⇒ x – 4 = 0

(ii) (-2, 0)

⇒ x + 2 = 0

(iii) (3, 5)

⇒ x – 3 = 0

(iv) (-4, - 3)

⇒ x + 4 = 0

Equation of the line representing x-axis is y = 0

Given, (4, 19) is a solution of the equation y=ax+3

⇒ 19 = 4a + 3

⇒ a = 4

Given, (a,4) lies on the graph of 3x+y=10

Thus it is a solution

⇒ 3a + 4 = 10

⇒ a = 2

Equation of the line representing y-axis is x = 0

2x – y = 4

At y = 0, x = 2

Thus the line cuts the x-axis at (2, 0)

Equation of a line parallel to x-axis passing through (a, b) is y = b

Thus, equation of a line passing through the point (0,4) and parallel to x-axis is y = 4

Infinitely many equations satisfy x = 2 and y = 3 as infinitely many lines pass through a single point.

Equation of a line parallel to x-axis passing through (a, b) is y = b

Thus, equation of a line passing through the point (3, 5) and parallel to x-axis is y = 5

X – 2 = 0

X = 2 is a point on the number line

Equation of a line parallel to y-axis passing through (a, b) is x = a

Thus, equation of a line passing through the point (-3, -7) and parallel to y-axis is x = -3

X = 2 and y = -1

We will check by substituting the values in the given equations

(a) 2 - 2 = 0 which is true

Thus x + 2y = 0 is the equation

(b) 2 - 2 = 4 which is not true

(c) 4 - 1 = 0 which is not true

(d) 2 × 2 - 1 = 5 which is not true

Equation of a line parallel to y-axis passing through (a, b) is x = a

Thus, equation of a line passing through the point (-4, 6) and parallel to y-axis is x = -4

Given, (2k-1,k) is a solution of the equation 10x-9y=12

⇒ 20x – 10 – 9k = 12

⇒ 11k = 22

⇒ k = 2

3x – 2 = 2x + 3

⇒ x = 5

2y – 1 = y + 1

⇒ y = 2

Distance between the graph of the equations x = -3 and x=2 is = 2 – (-3) = 5 units

Given, point (a,2) lies on the graph of the linear equation 2x-3y+8=0

Thus, (a, 2) satisfies the equation

⇒ 2a – 6 + 8 = 0

⇒ 2a = -2

⇒ a = -1

Distance between the graphs of the equations y = -1 and y = 3 is = 3 – (-1) = 4 units

Given, (1,-2) lies on the graph of the linear equation x-2y+k=0

Thus, (1, -2) satisfies the equation

⇒ 1 + 4 + k = 0

⇒ k = - 5

4x + 3y = 12

A is (3, 0)

B is (0, 4)

Base of triangle AOB = OA = 3 untis

Perpendicualr of triangle AOB = OB = 4 units

Hypotenuse² = perepndicular² + base²

⇒ Hypotenuse² = 16 + 9 = 25 sq units

⇒ Hypotenuse = 5 units