**Solving Word Problems Using Graphical Method :**

In this section, we will learn how to solve word problems using graphical method.

**Example 1 :**

Half the perimeter of a rectangular garden, whose length is 4 m more its width is 36 m. Find the dimensions of the garden.

**Solution :**

Let "x" be the width of the rectangular garden

Let "y" be its length

y = x + 4 -------- (1)

Half the perimeter of the rectangular garden = 36 m

perimeter of rectangle = 2 (L + b)

L + b = 36

y + x = 36

y = 36 - x -------- (2)

Now let us find x and y intercepts to draw the graph.

**Graphing 1 ^{st} line :**

y = x + 4

x-intercept : put y = 0 x + 4 = 0 x = -4 (-4, 0) |
y - intercept put x = 0 y = 0 + 4 y = 4 (0, 4) |

**Graphing 2 ^{nd} line :**

y = 36 - x

x-intercept : put y = 0 36 - x = 0 x = 36 (36, 0) |
y - intercept put x = 0 y = 36 - 0 y = 36 (0, 36) |

Thew above lines are intersecting at the point (16, 20). So, length of the rectangular garden is 20 m and width of the rectangular garden is 16 m.

**Example 2 :**

Given the linear equation 2 x + 3 y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

**Solution :**

(i) intersecting lines

The condition for intersecting lines is a₁/a₂ ≠ b₁/b₂.

According to the above condition, we have to form an equation.

a₁ = 2 b₁ = 3 c₁ = -8

The values of a₂,b₂ and c₂ be any real values but the simplified values of a₁/a₂ and b₁/b₂ shouldn't be equal.

a₂ = 3 b₂ = -3 c₂ = -16

So, one of the required possible equation is

3 x - 3 y - 16 = 0.

(ii) Parallel lines

The condition for two parallel lines is a₁/a₂ = b₁/b₂ ≠ c₁/c₂. According to the above condition, we have to form a equation.

a₁ = 2 b₁ = 3 c₁ = -8

if the value of a₂ is 4, the value of b₂ will be 6. The value of c₂ must be any value other than -16.

a₂ = 4 b₂ = 6 c₂ = 6

So one of the required possible equation is

4x + 6y - 6 = 0.

(iii) Coincident lines

The condition for coincident lines is a₁/a₂ = b₁/b₂ = c₁/c₂.According to the above condition,we have to form a equation.

a₁ = 2 b₁ = 3 c₁ = -8

the values of a₂, b₂ and c₂ will be

a₂ = 4 b₂ = 6 c₂ = -16

So one of the required possible equation is

4x + 6y - 16 = 0.

After having gone through the stuff given above, we hope that the students would have understood, solving word problems using graphical method.

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