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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.10

**9th Maths Exercise 3.10 Solutions Question 1.**

Draw the graph for the following

(i) y = 2x

(ii) y = 4x – 1

(iii) y = \(\left(\frac{3}{2}\right)\) x + 3

(iv) 3x + 2y = 14

Solution:

(i) Put x = -1, y = 2 × -1 = -2

When x = 0, y = 2 × 0 = 0

When x = 1, y = 2 × 1 = 2

The points (x, y) to be plotted: (-1, -2), (0, 0), (1, 2)

(ii) When x = -1 ⇒ y = 4 (-1) -1

y = – 4 – 1 = – 5

x = 0 ⇒ y = 4 × 0 – 1 = -1

x = 1 ⇒ y = 4 × 1 – 1 = 3

The points (x, y) to be plotted: (-1, -5), (0, -1), (1, 3)

The points to be plotted: (-2, 0), (0, 3), (2, 6)

The points to be plotted: (-2, 10), (0, 7), (2, 4)

**9th Maths Graph Question 2.**

Solve graphically

(i) x + y = 7; x – y = 3

(ii) 3x + 2y = 4; 9x + 6y – 12 =0

(iii) \(\frac{x}{2}+\frac{y}{4}=1 ; \frac{x}{2}+\frac{y}{4}=2\)

(iv) x – y = 0; y + 3 = 0

(v) y = 2x + 1; y + 3x – 6 = 0

(vi) x = -3, y = 3

Solution:

(i) We can find x and y intericepts and thus of the two points on the lines (1), (2)

x + y = 7 ……… (1), x – y = 3 …………. (2)

To draw the graph of (1)

Put x = 0 in (1)

0 + y = 7 ⇒ y = 7

Thus A (0, 7) is a point on the line

Put y = 0 in (1)

x + 0 = 7 ⇒ x = 7

Thus B (7, 0) is another point on the line

Plot A and B. Join them to produce the line (1).

To draw the graph of (2), we can adopt the same procedure.

When x = 0,(2) ⇒ x – y = 3

0 – y = 3 ⇒ y = -3

P (0, -3) is a point on the line.

Put y = 0 in (2); x – 0 = 3

x = 3

∴ Q (3, 0) is another point on the line (2)

Plot P, Q

1 The point of intersection (5, 2) of lines (1), (2) is a solution

(ii) 3x + 2y = 4 ……. (1)

9x + 6y= 12 ………. (2)

To draw the graph of (1)

Put x = 0 in (1) ⇒ 3 (0) + 2y = 4

2y = 4

y =2

∴ A (0, 2) is a point on the line (1)

Put y = 0 in (1) ⇒ 3x + 2(0) = 4

3x = 4

x = \(\frac{4}{3}\) = 1.3

∴ B (1.3, 0) is another point on the line (1)

Plot the points A, B. Join them to produce the line (1)

To draw the graph of (2)

Put x = 0 in (2) ⇒ 9 (0) + 6y = 12

6y = 12

y = 2

∴ P (0, 2) is a point on the line (2)

Put y = 0 in (2) ⇒ 9x + 6(0) = 12

9x = 12

x = \(\frac{12}{9}=\frac{4}{3}\)

x = 1.3

Q (1.3, 0) is another point on the line (2)

Plot P, Q. Join them to produce the line (2).

The point of intersection of the lines (1), (2) is a solution.

[A, B],[P, Q] represent the same line.

∴ All the points on one line are also on the other.

This means we have an infinite number of solutions.

∴ Comparing (1), (2) we can conclude their slopes are equal

∴ The lines are parallel and will not meet at any point and hence no solution exists.

Let us draw the graphs of (1) & (2)

(1) ⇒ 2x + y = 4, Put x = 0 in (1) ⇒ y = 4

∴ A (0, 4) is a point on (1)

Put y = 0 in (1) ⇒ 2x = 4

x = 2

∴ B (2, 0) is another point on (1)

Plot A, B ; Join them to produce the line (1)

(2) ⇒ 2x + y = 8, Put x = 0 in (2)

2(0) + y = 8

y = 8, P (0,8) is a point

Put y = 0 in (2) ⇒ 2x + 0 = 8

2x = 8

x = 4, Q (4, 0) is another point on the line (2)

Plot P, Q: Join them to produce the line (2)

(iv) x – y = 0 ………….. (1) ⇒ -y = -x ⇒ y = x

Put x = 0 in (1), 0 – y = 0

-y = 0 ⇒ y = 0

A (0, 0) is a point on the line (1)

Put y = 0 in (1) ⇒ x – 0 = 0 ⇒ x = 0

B (-3, -3) is also the same point as A

(2) ⇒ y + 3 = 0

y = -3

∴ from (1) y = -3 = x

∴ B (-3, -3) is the solution

(v) y = 2x + 1 …………… (1)

The points of intersection (1, 3) of the lines (1) and (2) is a solution. The solution is the point that is common to both the lines.

∴ The solution is as x = 1, y = 3.

(vi) The point of intersection (-3, 3) is a solution.

x + y = 7 ……… (1), x – y = 3 …………. (2)

To draw the graph of (1)

Put x= 0 in (1)

0 + y = 7 ⇒ y =7

Thus A (0, 7) is a point on the line

Put y = 0 in (1)

x + 0 = 7 ⇒ x =7

Thus B (7, 0) is another point on the line

Plot A and B. Join them to produce the line (1).

To draw the graph of (2), we can adopt the same procedure.

When x = 0, …….. (2) ⇒ x – y = 3

0 – y = 3 ⇒ y = -3

P (0, -3) is a point on the line.

Put y = 0 in (2) ; x – 0 = 3

x = 3

∴ Q (3, 0) is another point on the line (2) Plot P, Q

**9th Graph Exercise 3.10 Question 3.**

Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.

Solution:

Let x, y be the speed of the two cars. If the two cars travel towards each other they will meet in 1 hr. The distance between them d = 100; \(\frac{d}{s}\) = t

i.e., \(\frac{100}{x+y}\) = 1 ⇒ x + y = 100 ………. (1)

If the two cars travel in the same direction they will meet in 2 hrs.

x + y = \(\frac{100}{2}\) ⇒ x – y = 50 ………….. (2)

x + y = 10 ………….. (1)

Put x = 0 in (1), then 0 + y = 100 ⇒ y =100

A (0, 100) is a point on (1)

Put y = 0 in (1), then x + 0 = 100 ⇒ x =100

B (100, 0) is another point on (1)

Plot A & B. Join them to produce the line (1)

Similarly by x – y = 50

Put x = 0 in (2), then 0 – y = 50 ⇒ y = -50

P (0, -50) is a point on (2)

Put y = 0 in (2), then x – 0 = 50 ⇒ x = 50

Q (50, 0) is another point on (2)

Plot P & Q. Join them to produce the line (2)

The point of intersection (75, 25) of the two lines (1) & (2) is the solution.

∴ The solution i.e., the speed of the two cars x and y is given by x = 75 km and y = 25 km