# Examples of linear equations

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## Examples of Linear Equations:

Example 1: Draw the graph of y=2x + 4. Use your graph to find (i) the slop (ii) the intercept on y-axis (iii) the area between the line and the axes.

Solution. Given y = 2x +4

Required Table is:

We plot the points (0,4), (-2,0) and (-1,2) and join them. Thus the line I represents the equation y =2x + 4.

Result: (i) Slope= -\frac { OA }{ OB } =-\frac { 4 }{ -2 } =2

(ii) Intercept on the y-axis =OA =4 units.

(iii) Area= \frac { 1 }{ 2 } \times OA\times OB

= \frac { 1 }{ 2 } \times 2\times 4=4sq. units.

Example 2: Aftab tells his daughter, “seven times as old you were there. Also three years from now, I shall be three times as old as you will be. Represent this situation algebraically and graphically.

Solution: Let the present age of the daughter= x years.

7 years ago daughter’s age = (x – 7) years

3 years from now, daughter’s age= (x + 3) years

Let the present age of father = y years

7 years ago father’s age    = (y – 7) years

3 years from now father’s age  = (y +3) years

By giving conditions, we have two algebraic equations:

Y – 7= 7(x – 7)

Y =7x – 42        …..(1)

Y + 3= 3(x +3)

y= 3x + 6         …..(2)

Required table for y= 7x – 42

Required table for y= 3x + 6

Result: (i) Algebraic representation:

y= 7 – 42 ; y = 3x + 6

(ii) After plotting the point of both the tables, we get the required graphical representation of the ages of the daughter and her father.

From the graphs the point of intersection of two lines is(12, 42) therefore, (12,42) is the solution. Hence

1.  Present age of father is 42 years on y-axis.
2.  Present age of his daughter is 12 years.

Example 3: The coach of cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of some kind for Rs.1500. Represent this situation algebraically and geometrically.

Solution: Let the cost of one bat be Rs. x

and                 the cost of one ball be Rs. y

Then according to given condition, we have

3x + 6y= 3900  \Rightarrow   x + 2y= 1300  ….(1)

x + 3y= 1500   \Rightarrow   x + 3y= 1500  ….(2)

(i) Algebraically representation is: x + 2y= 1300 ;  x + 3y= 1500

(ii) For graphical representation, we have required tables:

x= 130 – 2y

x=1500 – 3y

Example 4:  Gloria is walking along the path joining(-2,3) and (2, -2), while suresh is walking along the path joining (0,5) and (4,0). Represent this situation graphically.

Solution:  Plot the point A (-2, 3) and B (2, -2) on the graph paper. Join these points. Obtain the path AB.

Also plot the points C (0, 5) and D (4, 0). Join these points. Obtain the path CD.

Since paths AB and CD have no common point so, they do not meet at all.

\therefore   Path AB \parallel CD.

Example 5:  Romila went to a stationary stall and purchased 2 pencils and 3 erasers for Rs. 9. Her friend Sonali saw the new variety of pencils and erases with Romila and she also bought 4 pencils and 6 erasers of the same kind for Rs. 18. Represent this situation algebraically and solve it graphically.

Solution:  Let the cost of pencil

= Rs. x

And the cost of an eraser= Rs. y

(i) Then according to given condition, we have

2x + 3y= 9   …..(i)

4x + 6y= 8  ……(ii)

1.  and (2) provide us algebraical representation  of given situation.

(ii) For graphical representation, we have required tables:

y=\frac { 9-2x }{ 3 } y=\frac { 18-4x }{ 6 }

Result:  After plotting the points of both the tables on the graph paper, we see that both the lines coincide.

From the above graph, we have:

1. If the graph of two lines intersect at a point, the system is known as consistent. Only one solution.
2. If graph of two lines is parallel to lines, the system is known as inconsistent  (no common point). No solution.
3. If the graph of two lines is the same i.e., coincides, the system is known as dependent (many common points). Many solutions.