Linear Equations In Two Variables

Let the cost of pen be y and the cost of notebook be x

As per the question,

Cost of a notebook = Twice the cost of pen = 2y

So,

2y = x

x – 2y = 0

This is a linear equation in two variables to represent this statement.

(i) 2x + 3y = 9.35̅

2x + 3y - 9.35̅ = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = 2,

b = 3

and c = - 9.35̅

(ii) x – y/5 - 10 = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = 1,

b =

and c = - 10

(iii) - 2x + 3y = 6

- 2x + 3y – 6 = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = - 2,

b = 3

and c = -6

(iv) x = 3 y

x – 3 y = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = 1,

b = -3

and c = 0

(v) 2x = -5y

2x + 5y = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = 2,

b = 5

and c = 0

(vi) 3x + 2 = 0

3x + 0y + 2 = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = 3,

b = 0

And c = 2

(vii) y – 2 = 0

0x + y – 2 = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = 0,

b = 1

And c = -2

(viii) 5 = 2x

-2x + 0y + 5 = 0

Now,

On comparing this equation with ax + by + c = 0

We get,

a = -2,

b = 0

And c = 5

Equation y = 3x + 5 is a linear equation in two variables.

This means that we can put any value of x and in return, we will get some other value of y.

For example, at x = 0, y = 3 x 0 + 5, y = 5

at x = 1000, y = 3 x 1000 + 5, y = 3005

we can put infinitely many values of x and we will get infinitely many solutions for y.

Thus there are infinite solutions for the equation.

Graphically:

When plotted graphically, it will give a straight line extending to infinity and thus will give infinite solutions.

For finding out solutions for any linear equation in 2 variables, just put one value in any of the variable in the equation and obtain the other value.

For example : For equation x + y = 5

Let x = 1, then 1 + y = 5, y = 5 - 1, y = 4

Thus x =1 and y = 4 will be a solution for the equation x + y = 5.


(i)2x + y = 7

Given Equation:

2x + y = 7

rearranging the equation we get,

y = 7 – 2x

Put x = 0,

y = 7 – 2 × 0

y = 7

(0, 7) is the solution of the equation

Now,

Put x = 1

y = 7 – 2 × 1

y = 5

(1, 5) is the solution of the equation

Now,

Put x = 2

y = 7 – 2 × 2

y = 3

(2, 3) is the solution of the equation

Now,

Put x = -1

y = 7 – 2 × - 1

y = 9

(-1, 9) is the solution of the equation

The four solutions of the equation 2x + y = 7 are:

(0, 7),

(1, 5),

(2, 3)

and (-1, 9)

(ii) πx + y = 9

πx + y = 9

y = 9 – πx

Now,

Put x = 0

y = 9 – π × 0

y = 9

(0, 9) is the solution of the equation

Now,

Put x = 1

y = 9 – π × 1

y = 9 - π

(1, 9 - π) is the solution of the equation

Now,

Put x = 2

y = 9 – π × 2

y = 9 – 2 π

(2, 9 - 2 π) is the solution of the equation

Now,

Put x = -1

y = 9 – π × (-1)

y = 9 + π

(-1, 9 + π) is the solution of the equation

The four solutions of the equation πx + y = 9 are:

(0, 9),

(1, 9 – π),

(2, 9 - 2 π)

and (-1, 9 + π)

(iii) x = 4y

Now,

Put x = 0

0 = 4y

y = 0

(0, 0) is the solution of the equation

Now,

put x = 1

1 = 4y

y =

(1, ) is the solution of the equation

Now,

Put x = 4

4 = 4y

y = 1

(4, 1) is the solution of the equation

Now,

Put x = 8

8 = 4y

y = 2

(8, 2) is the solution of the equation

The four solutions of the equation πx + y = 9 are:

(0, 0),

(1, ),

(4, 1),

and (8, 2)

For a linear equation a point will be its solution, if it satisfies the equation.

So put the points directly in the equation given, if the equation is satisfied, it is a solution or else it is not.

Given Equation: x - 2 y = 4

(i)
Put x = 0

And

y = 2

In the equation x – 2y = 4

0 – 2 * 2 = 4

-4 ≠ 4

Therefore,

(0, 2) isn’t a solution of the given equation

(ii) Put x = 2

And

y = 0

In the equation x – 2y = 4

2 – 2 * 0 = 4

2 ≠ 4

Therefore,

(2, 0) isn’t a solution of the given equation

(iii) Put x = 4

And

y = 0

In the equation x – 2y = 4

4 – 2 * 0 = 4

4 = 4

Therefore,

(4, 0) is a solution of the equation

(iv) Put x =

And

In the equation x – 2y = 4

(1 – 8) = 4

≠ 4

Therefore,

is n’t a solution of the given equation

(v) Put x = 1

And

y = 1

In the equation x – 2y = 4

1 – 2 * 1 = 4

-1 ≠ 4

Therefore, (1, 1) isn’t a solution of the given equation

The Given equation is 2 x + 3 y = k

The solution of the given equation are:

x = 2 and y = 1, this means that the equation will satisfy these points.

Now,

After putting the value of x and y in the equation, we get

2 × 2 + 3 × 1 = k

k = 4 + 3

k = 7

(i) Given equation: x + y = 4

Put x = 0 hence, y = 4
Put x = 4 hence, y = 0

(ii)Give Equation:
x – y = 2

Put x = 0 hence, y = -2

Put x = 2 hence, y = 0

Now we have two points A(0, -2) and B(2, 0) to plot on the graph. Plot these points and join them with a line to obtain the graph of equation.

(iii) Given Equation:
y = 3x

Put x = 0 hence, y = 0

Put x = 1 hence, y = 3

Now we have two points A(0, 0) and B(1, 3) to plot on graph. Join these points together to obtain graph of the equation.

(iv) Given equation:
3 = 2x + y

Put x = 0 hence, y = 3

Put x = 1 hence, y = 1

Now we have two points A(0, 3) and B(1, 1) to plot on the graph. Join the points and the line obtained will be the graph of the equation.

Note: You have to take at least two points to draw a graph.

Here,

x = 2 and y = 14

Hence,

x + y = 16

Also,

y = 7x

y – 7x = 0

Therefore,

The equations of two lines passing through (2, 14) are:

x + y = 16

And,

y – 7x = 0

There would be infinite such lines because infinite number of lines can pass through a given point.

We can also see this graphically:


From a single point there can be infinite number of lines that can pass.

The point (3, 4) lies on the graph of the equation 3y = ax + 7

Therefore,

Put x = 3 and y = 4 in the equation,

3y = ax + 7,

Hence,

3 x 4 = a x 3 + 7

12 = 3a + 7

3a = 12 – 7

3a = 5

a = 5/3

Total fare of the taxi = y

Total distance covered = x

Fare for the subsequent distance after 1^st kilometer = Rs 5

Fare for 1^st kilometer = Rs 8

As per the question,

Total fare = fare of 1st km + Fare of subsequent km x Distance covered
Now for first km, Fare = Rs. 8
For 2nd km, Fare = 8 + 5
For 3rd km, Fare = 8 + 2 × 5
And so on,
Let total fare by y and distance travel be x.
We can see that x is one less than the number of km travelled in total. So,
y = 8 + 5 (x - 1)
= 8 + 5x - 5
= 5x + 3

Now for plotting graph of the given linear equation, we would need different points,

So the two points two plot are A(0, 3) and B(1,8). By joining these points we will get the graph of the linear equation.

In figure 4.6,

Points plotted are (0, 0), (-1, 1) and (1, -1)

Now let us take equations one by one.
(i) y = x.
at x = 0, y = 0
at x = - 1, y = -1 [(-1, 1) point is not satisfied]
Therefore, the equation does not follow the graph.

(ii) x + y = 0
at x = 0, y = 0
at x = -1, y = 1
at x = 1, y = - 1
All the points are satisfied and Hence, the equation follows the graph.

In figure 4.7, points are (-1, 3), (0, 2) and (2, 0)

Hence,

(iii) y = - x + 2
at x = -1, y = 3
at x = 0, y = 2
at x = 2, y = 0
Therefore, all the points are satisfied and the equation follows the graph.
(iv) x + 2 y = 6

at x = - 1, y = 7/2 [(-1, 3) point is not satisfied]
Hence, the equation does not follow the graph.

Let the distance travelled be x by the body and y be the work done by the force

Given: y ∝ x

To equate the proportional, we need a constant.
Hence, y = k x
where k is a constant

Here, it is given 5

Hence,

y = 5 x

According to question,

(i) When x = 2 units then y = 5 x 2 = 10 units

(ii) When x = 0 unit then y = 5 x 0 = 0 unit

Let the amount contributed by Yamini be x and amount contributed by Fatima be y

Now, According to question,

x + y = 100

When x = 0 then y = 100

When x = 50 then y = 50

When x = 100 then y = 0

(i) Given:

When C = 0 then F = 32,

And

When C = -10 then
F = 9(-2) + 32
= - 18 + 32
= 14

(ii) Putting the value of C = 30 in F = () C + 32, we get

F = × 30 + 32

F = 54 + 32

F = 86

(iii) Putting the value of F = 95 in F = ()C + 32, we get

95 = ()C + 32

()C = 95 - 32

C = 63 ×

C = 35

(iv) After putting the value of F = 0 in

F = () C + 32,

We get:

0 = () C + 32

() = -32

C = -32 × 5/9

C =

Putting the value of C = 0 in

F = () C + 32,

We get:

F = () × 0 + 32

F = 32

(v) We have to find when F = C

Hence,

After putting F = C in

F = ()C + 32,

We get:

F = ()F + 32

F – ×F = 32

F = 32

F = – 40

Therefore at value -40,

Fahrenheit and Celsius both are numerically the same.

(i) It is represented as y = 3 in one variable. So for that, we need only one direction, take it as shown below and mark different points on the line.
For that, we can represent a point marked on the number line. And for y = 3, mark y at y = 3.

(ii) In two variables, it is represented as a line that is parallel to X-axis

0 x + y = 3, which shows that the y component of the graph will always remain constant and will be equal to 3.

(i)
First, we solve the given equation

2 x + 9=0

or, 2 x = -9

or, x = -9/2

In one variable, it can be represented on a straight line with one dimension or coordinate only, so it will be represented as x = -9/2

(ii) In two variables, it will be represented as a line parallel to Y-axis

2 x + 0 y + 9 = 0
Now if we put y = 0
2x + 9 = 0
x = -9/2
If we put y = 1
2x + 0(1) + 9 = 0
2x + 9 =0
x = -9/2