• A linear equation in the standard form, ax+by+c=0, comprises a pair of solutions in form x, y; you can represent that in the coordinate plane.
• If the equation is represented graphically, it is a straight line that may or may not cut the coordinate axes.

Lines passing through the source

• There are a few linear equations; their solution is (0, 0). Such equations, when illustrated graphically, pass through the source.
• The coordinate axes, i.e., the x-axis and y-axis, can be depicted as y=0 and x=0.

Lines parallel to coordinate axes

• Linear equations of the form y=a, when illustrated graphically, are lines lateral to the x-axis, and a is the y-coordinate of the points in that line.
• Linear equations of the form x=a, when depicted graphically, are lines lateral to the y-axis, and a is the x-coordinate of the points in that line.

Linear Equations in Two Variables

When an equation can be expressed in terms of ax+by+c=0 with a, b, and c as real numbers and the value of a and b is not equivalent to 0, it is called a linear equation in two variables.

The product of a linear equation is not affected, given:

• The same number is either added or subtracted from both sides of the equation
• The same non-zero number is applied for multiplication or division involving both sides of the equation.

A linear equation in two variables offers many solutions. The graph representing every linear equation in two variables is a straight line.

Exercise 4.1

1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

Solution: Given that the cost of the notebook is = x and the pen = y 

The cost of notebook = 2x price of pen

Therefore, x=2y

x-2y =0

This is the linear equation in two variables.

2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b, and c in each  case: 

i) 2x+3y= 9.35

Solution: 2x+3y= 9.35

By rearranging the equation, you will get: 

2x+3y-9.35=0

This equation can also be written as 2x+3y+ (-9.35)=0, which makes ax+by+c=0

This means a=2, b=3 and c= -9.35

ii) x –(y/5)–10 = 0

Solution:

The equation x –(y/5)-10 = 0 can be written as,

1x+(-1/5)y +(–10) = 0

Now comparing x+(-1/5)y+(–10) = 0 with ax+by+c = 0

We get a = 1, b = -(1/5) and c = -10

iii) –2x+3y = 6

Solution: –2x+3y = 6

By rearranging the equation, we get –2x+3y–6 = 0

The equation –2x+3y–6 = 0 can be written as (–2)x+3y+(– 6) = 0

Now comparing (–2)x+3y+(–6) = 0 with ax+by+c = 0

We get a = –2, b = 3, c =-6

iv) x = 3y

Solution: x = 3y

By rearranging the equation, we get x-3y = 0

The equation x-3y=0 can be written as 1x+(-3)y+(0)c = 0

Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0

We get a = 1, b = -3, c =0

v) 2x = –5y

Solution: 2x = –5y

By rearranging the equation, we get 2x+5y = 0

The equation 2x+5y = 0 can be written as 2x+5y+0 = 0

Now, comparing 2x+5y+0= 0 with ax+by+c = 0

We get, a = 2, b = 5, c = 0

vi) 3x+2 = 0

Solution: 3x+2 = 0

The equation 3x+2 = 0 can be written as 3x+0y+2 = 0

Now comparing 3x+0+2= 0 with ax+by+c = 0

We get, a = 3, b = 0, c = 2

vii) y–2 = 0

Solution: y–2 = 0

The equation y–2 = 0 can be written as 0x+1y+(–2) = 0

Now comparing 0x+1y+(–2) = 0 with ax+by+c = 0

We get, a = 0, b = 1, c = –2

viii) 5 = 2x

Solution: 5 = 2x

By rearranging the equation, we get 2x = 5

i.e., 2x–5 = 0

The equation 2x–5 = 0 can be written as 2x+0y–5 = 0

Now comparing 2x+0y–5 = 0 with ax+by+c = 0

We get, a = 2, b = 0, c = -5

Exercise 4.2.

The values for x in the linear equation y = 3x+5

x	0	1	2	\( \cdots\cdots \)	100
y when y=3x+5	5	8	11	\( \cdots\cdots \)	305

 

Solution:

There are infinite values for x and y.
Hence, the answer is that infinitely many solutions are there for linear equation y=3x+5

1. Write four solutions for each of the following equations:

(i) 2x+y = 7
 
Solution: For finding the four solutions of 2x+y =7 you will have to substitute different values for x and y.

Let x = 0 then 2x+y = 7

(2×0)+y = 7

y = 7

So, one solution is (0,7)

Let x = 1 then 2x+y = 7

(2\( \times \)1)+y = 7 => 2+y = 7

y = 7-2; y = 5

So, one solution is (1,5)

Let y = 1 then 2x+y = 7

(2x)+1 = 7 => 2x = 7-1

2x = 6

x = 6/2

x = 3

So, another solution is (3,1)

Let x = 2, then 2x+y = 7

(2 \( \times \) 2)+y = 7

4+y = 7

y =7-4

y = 3

So, the fourth solution is (2,3)

The solutions are (0, 7), (1,5), (3,1), (2,3).
 
ii) \( \pi \) x+y = 9
 
Solution: For finding the four solutions of \( \pi \) x+y = 9, you will have to substitute different values for x and y

Let x = 0, then \( \pi \) x+y = 9

(\( \pi \) \( \times \) 0)+y = 9

y = 9

(0,9)

Let x = 1, then πx +y = 9

(\( \pi \times \)1)+y = 9

\( \pi \)+y = 9

y = 9-\( \pi \)

(1, 9-\( \pi \))

Let y = 0, then \( \pi \) x+y = 9

\( \pi \) x+0 = 9

\( \pi \) x = 9

x = 9/\( \pi \)

(9/\( \pi \),0)

Let x = -1, then \( \pi \)x + y = 9

(\( \pi \) \( \times \)-1) + y = 9

-\( \pi \)+y = 9

y = 9+\( \pi \)

(-1,9+\( \pi \))

The solutions are (0,9), (1,9-\( \pi \)), (9/\( \pi \), 0), and (-1,9+\( \pi \)).

Exercise 4.3.   

1. Before drawing the graph of the given linear equation in two variables, it is important to find out the points to plot.

For that, the values of x and y need to be found, as per the equation x+y=4

Substituting the values for x:

If x=0

x+y=4

0+y=4

y=4

On the other hand, if x=4

x+y=4

4+y=4

y=4-4

y=0

x	y
0	4
4	0

 

Points to be plotted are (0, 4) and (4, 0).

. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units.

Also check from the graph the work done when the distance travelled by the body is-

(i) 2 units

(ii) 0 unit

Solution:

Assume that the distance travelled by the body be x and the force applied on the body be y.

You already know that,

The work done by a body is directly proportional to the distance travelled by the body.

According to the question, y \( \infty \)x

y = 5x (5 is a constant of proportionality)

Solving the equation,

(i) when x = 2 units, then y = 5 \( \times \) 2 = 10 units

One solution is (2, 10)

(ii) when x = 0 units, then y = 5 \( \times \) 0 = 0 units.

Another solution is (0, 0)

The points to be plotted are (2, 10) and (0, 0).

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