Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.10

 

Question 1. Draw the graph for the following:
(i) \( y = 2x \)
Answer:
To draw the graph for the equation \( y = 2x \), we need to find some coordinate pairs (x, y) that satisfy the equation. We can pick different values for \( x \) and calculate the corresponding values for \( y \). Let's create a table of values:
When \( x = -2 \), \( y = 2(-2) = -4 \)
When \( x = 0 \), \( y = 2(0) = 0 \)
When \( x = 2 \), \( y = 2(2) = 4 \)
When \( x = 3 \), \( y = 2(3) = 6 \)

Now, we plot these points: (-2, -4), (0, 0), (2, 4), and (3, 6) on a graph sheet. After plotting, when we draw a line connecting these points, we will get a straight line representing the equation \( y = 2x \). The graph for this equation will always pass through the origin (0,0).
In simple words: To show \( y = 2x \) on a graph, first pick some 'x' values like -2, 0, 2, 3 and find their 'y' partners by multiplying 'x' by 2. Then, mark these (x, y) pairs as dots on graph paper and draw a straight line through all the dots.

🎯 Exam Tip: When plotting linear equations, always calculate at least three points to ensure accuracy, as any two points define a line, and a third helps to check for calculation errors.

 

Question 1. (ii) \( y = 4x - 1 \)
Answer:
To draw the graph for the equation \( y = 4x - 1 \), we need to find some coordinate pairs (x, y) that satisfy the equation. We can choose different values for \( x \) and calculate the corresponding values for \( y \). Let's create a table of values:
When \( x = -1 \), \( y = 4(-1) - 1 = -4 - 1 = -5 \)
When \( x = 0 \), \( y = 4(0) - 1 = 0 - 1 = -1 \)
When \( x = 2 \), \( y = 4(2) - 1 = 8 - 1 = 7 \)

Now, we plot these points: (-1, -5), (0, -1), and (2, 7) on a graph sheet. After plotting, when we draw a line connecting these points, we will get a straight line representing the equation \( y = 4x - 1 \). This line will intercept the y-axis at -1.
In simple words: To graph \( y = 4x - 1 \), pick 'x' values like -1, 0, 2, and use the rule to find 'y' (multiply 'x' by 4, then subtract 1). Mark these (x, y) dots on the graph and draw a straight line through them.

🎯 Exam Tip: Pay close attention to negative signs in calculations, as a small error can lead to an incorrect point and a wrongly drawn graph.

 

Question 1. (iii) \( y = \frac{3}{2}x + 3 \)
Answer:
To draw the graph for the equation \( y = \frac{3}{2}x + 3 \), we need to find some coordinate pairs (x, y) that satisfy the equation. We can choose values for \( x \) that are multiples of 2 to avoid fractions in \( y \). Let's create a table of values:
When \( x = -2 \), \( y = \frac{3}{2}(-2) + 3 = -3 + 3 = 0 \)
When \( x = 0 \), \( y = \frac{3}{2}(0) + 3 = 0 + 3 = 3 \)
When \( x = 2 \), \( y = \frac{3}{2}(2) + 3 = 3 + 3 = 6 \)

Now, we plot these points: (-2, 0), (0, 3), and (2, 6) on a graph sheet. After plotting, when we draw a line connecting these points, we will get a straight line representing the equation \( y = \frac{3}{2}x + 3 \). This line has a positive slope, meaning it goes upwards from left to right.
In simple words: To graph \( y = \frac{3}{2}x + 3 \), choose 'x' values like -2, 0, 2 (multiples of the denominator help). Calculate 'y' for each. Mark these (x, y) dots on the graph and connect them with a straight line.

🎯 Exam Tip: When dealing with fractional coefficients, choose x-values that are multiples of the denominator to make y-values whole numbers, simplifying plotting.

 

Question 1. (iv) \( 3x + 2y = 14 \)
Answer:
To draw the graph for the equation \( 3x + 2y = 14 \), first, we rearrange the equation to express \( y \) in terms of \( x \):
\( 2y = 14 - 3x \)
\( y = \frac{14 - 3x}{2} \)
Now, we can find some coordinate pairs (x, y) by picking values for \( x \) and calculating the corresponding values for \( y \). Choosing even values for \( x \) often helps avoid fractions for \( y \):
When \( x = -2 \), \( y = \frac{14 - 3(-2)}{2} = \frac{14 + 6}{2} = \frac{20}{2} = 10 \)
When \( x = 0 \), \( y = \frac{14 - 3(0)}{2} = \frac{14 - 0}{2} = \frac{14}{2} = 7 \)
When \( x = 2 \), \( y = \frac{14 - 3(2)}{2} = \frac{14 - 6}{2} = \frac{8}{2} = 4 \)

Now, we plot these points: (-2, 10), (0, 7), and (2, 4) on a graph sheet. After plotting, when we draw a line connecting these points, we will get a straight line representing the equation \( 3x + 2y = 14 \). This line has a negative slope, going downwards from left to right.
In simple words: First, change the equation \( 3x + 2y = 14 \) to get 'y' by itself. Then, choose some 'x' values, like -2, 0, 2, and use the new rule to find their 'y' partners. Put these dots on a graph and draw a straight line through them.

🎯 Exam Tip: Always rearrange the equation into the \( y = mx + c \) form before calculating points, as this makes it easier to substitute x-values and find y-values correctly.

 

Question 2. Solve graphically: \( x + y = 7, x - y = 3 \)
Answer:
To solve a system of linear equations graphically, we need to draw the graph for each equation on the same coordinate plane and find their point of intersection. That point will be the solution to the system.

**For Equation 1: \( x + y = 7 \)**
First, we rearrange the equation to express \( y \) in terms of \( x \):
\( y = 7 - x \)
Now, we calculate some coordinate pairs for this line:
When \( x = -2 \), \( y = 7 - (-2) = 9 \)
When \( x = 0 \), \( y = 7 - 0 = 7 \)
When \( x = 3 \), \( y = 7 - 3 = 4 \)

**For Equation 2: \( x - y = 3 \)**
Rearrange this equation to express \( y \) in terms of \( x \):
\( -y = 3 - x \)
\( y = x - 3 \)
Now, we calculate some coordinate pairs for this second line:
When \( x = -2 \), \( y = -2 - 3 = -5 \)
When \( x = 0 \), \( y = 0 - 3 = -3 \)
When \( x = 4 \), \( y = 4 - 3 = 1 \)

We plot the points (-2, 9), (0, 7), (3, 4) for the first line and (-2, -5), (0, -3), (4, 1) for the second line on the same graph sheet. We then draw a straight line through each set of points. We observe that the two lines intersect at the point (5, 2). This intersection point is the solution to the system of equations.
Therefore, the solution set is (5, 2).
In simple words: First, make a table of 'x' and 'y' values for each equation. Plot these points on the same graph paper and draw a line for each equation. The spot where the two lines cross each other (the intersection point) is the answer. In this case, the lines meet at (5, 2).

🎯 Exam Tip: Always double-check your intersection point by substituting its coordinates back into both original equations to ensure it satisfies both simultaneously.

 

Question 2. (ii) Solve graphically: \( 3x + 2y = 4; 9x + 6y - 12 = 0 \)
Answer:
To solve this system graphically, we will find points for each equation and plot them.

**For Equation 1: \( 3x + 2y = 4 \)**
Rearrange the equation to express \( y \) in terms of \( x \):
\( 2y = 4 - 3x \)
\( y = \frac{4 - 3x}{2} \)
Now, we calculate some coordinate pairs:
When \( x = -2 \), \( y = \frac{4 - 3(-2)}{2} = \frac{4 + 6}{2} = \frac{10}{2} = 5 \)
When \( x = 0 \), \( y = \frac{4 - 3(0)}{2} = \frac{4 - 0}{2} = \frac{4}{2} = 2 \)
When \( x = 2 \), \( y = \frac{4 - 3(2)}{2} = \frac{4 - 6}{2} = \frac{-2}{2} = -1 \)

**For Equation 2: \( 9x + 6y - 12 = 0 \)**
First, simplify this equation by dividing all terms by 3:
\( \frac{9x}{3} + \frac{6y}{3} - \frac{12}{3} = 0 \)
\( 3x + 2y - 4 = 0 \)
\( 3x + 2y = 4 \)
We notice that this simplified equation is identical to Equation 1. This means both equations represent the same line. Therefore, any point on this line is a solution, resulting in infinitely many solutions.
The coordinate pairs will be the same as for Equation 1:

When we plot these points and draw the lines, we will find that both equations result in the exact same line. Since the lines are the same, they intersect at every point, meaning there are infinitely many solutions.
In simple words: We find points for the first equation. When we simplify the second equation, we find it's exactly the same as the first one. This means both equations draw the same line on the graph. When two lines are the same, they touch everywhere, so there are endless solutions.

🎯 Exam Tip: Always simplify equations before plotting. If two equations simplify to the same form, they are coincident lines with infinitely many solutions.

 

Question 2. (iii) Solve graphically: \( \frac{x}{2} + \frac{y}{4} = 1; \frac{x}{2} + \frac{y}{4} = 2 \)
Answer:
To solve this system graphically, we will find points for each equation and plot them.

**For Equation 1: \( \frac{x}{2} + \frac{y}{4} = 1 \)**
First, clear the denominators by multiplying the entire equation by 4:
\( 4(\frac{x}{2}) + 4(\frac{y}{4}) = 4(1) \)
\( 2x + y = 4 \)
Now, rearrange to express \( y \) in terms of \( x \):
\( y = 4 - 2x \)
Calculate some coordinate pairs:
When \( x = -3 \), \( y = 4 - 2(-3) = 4 + 6 = 10 \)
When \( x = -1 \), \( y = 4 - 2(-1) = 4 + 2 = 6 \)
When \( x = 0 \), \( y = 4 - 2(0) = 4 \)
When \( x = 2 \), \( y = 4 - 2(2) = 4 - 4 = 0 \)

**For Equation 2: \( \frac{x}{2} + \frac{y}{4} = 2 \)**
Again, clear the denominators by multiplying the entire equation by 4:
\( 4(\frac{x}{2}) + 4(\frac{y}{4}) = 4(2) \)
\( 2x + y = 8 \)
Now, rearrange to express \( y \) in terms of \( x \):
\( y = 8 - 2x \)
Calculate some coordinate pairs:
When \( x = -2 \), \( y = 8 - 2(-2) = 8 + 4 = 12 \)
When \( x = -1 \), \( y = 8 - 2(-1) = 8 + 2 = 10 \)
When \( x = 0 \), \( y = 8 - 2(0) = 8 \)
When \( x = 2 \), \( y = 8 - 2(2) = 8 - 4 = 4 \)

When we plot these points and draw the lines, we will find that both lines have the same slope (\( m = -2 \)) but different y-intercepts (c = 4 and c = 8). This means the two lines are parallel and will never intersect. Therefore, there is no solution to this system of equations.
In simple words: First, rewrite both equations to get rid of fractions and solve for 'y'. Make a table of points for each equation. When you plot these lines, you'll see they run next to each other, never crossing. This means there's no answer that works for both.

🎯 Exam Tip: Recognize that if two linear equations, when simplified, have the same slope but different y-intercepts, their graphs will be parallel lines, and there will be no solution.

 

Question 2. (iv) Solve graphically: \( x - y = 0; y + 3 = 0 \)
Answer:
To solve this system graphically, we will find points for each equation and plot them.

**For Equation 1: \( x - y = 0 \)**
Rearrange the equation to express \( y \) in terms of \( x \):
\( x = y \)
\( y = x \)
Now, we calculate some coordinate pairs for this line. This is a special line where the x and y values are always the same:
When \( x = -2 \), \( y = -2 \)
When \( x = 0 \), \( y = 0 \)
When \( x = 1 \), \( y = 1 \)
When \( x = 3 \), \( y = 3 \)

**For Equation 2: \( y + 3 = 0 \)**
Rearrange this equation to express \( y \):
\( y = -3 \)
This equation represents a horizontal line where the y-coordinate is always -3, regardless of the x-coordinate. It is parallel to the x-axis.
Now, we calculate some coordinate pairs:
When \( x = -2 \), \( y = -3 \)
When \( x = 0 \), \( y = -3 \)
When \( x = 1 \), \( y = -3 \)
When \( x = 2 \), \( y = -3 \)

We plot the points for both lines on the same graph sheet. For \( y=x \), we plot (-2,-2), (0,0), (1,1), (3,3). For \( y=-3 \), we plot (-2,-3), (0,-3), (1,-3), (2,-3). We draw straight lines through these points. We observe that the two lines intersect at the point (-3, -3). This intersection point is the solution to the system of equations.
Therefore, the solution set is (-3, -3).
In simple words: For the first line, 'y' is always the same as 'x'. For the second line, 'y' is always -3. Draw both lines on the graph. The spot where they cross, which is (-3, -3), is the answer.

🎯 Exam Tip: Remember that equations like \( y = \text{constant} \) represent horizontal lines, and \( x = \text{constant} \) represent vertical lines. This simplifies plotting significantly.

 

Question 2. (v) Solve graphically: \( y = 2x + 1; 3x - 6 = 0 \)
Answer:
To solve this system graphically, we will find points for each equation and plot them.

**For Equation 1: \( y = 2x + 1 \)**
Calculate some coordinate pairs for this line:
When \( x = -3 \), \( y = 2(-3) + 1 = -6 + 1 = -5 \)
When \( x = -1 \), \( y = 2(-1) + 1 = -2 + 1 = -1 \)
When \( x = 0 \), \( y = 2(0) + 1 = 0 + 1 = 1 \)
When \( x = 2 \), \( y = 2(2) + 1 = 4 + 1 = 5 \)

**For Equation 2: \( 3x - 6 = 0 \)**
Rearrange this equation to express \( x \):
\( 3x = 6 \)
\( x = 2 \)
This equation represents a vertical line where the x-coordinate is always 2, regardless of the y-coordinate. It is parallel to the y-axis.
Now, we calculate some coordinate pairs, ensuring to include the intersection y-value from the first line for \( x=2 \):
When \( x = 2 \), \( y = -1 \) (arbitrary point)
When \( x = 2 \), \( y = 0 \) (arbitrary point)
When \( x = 2 \), \( y = 1 \) (arbitrary point)
When \( x = 2 \), \( y = 5 \) (intersection point from first line)

We plot the points for both lines on the same graph sheet. For \( y=2x+1 \), we plot (-3,-5), (-1,-1), (0,1), (2,5). For \( x=2 \), we plot (2,-1), (2,0), (2,1), (2,5). We draw straight lines through these points. We observe that the two lines intersect at the point (2, 5). This intersection point is the solution to the system of equations.
Therefore, the solution set is (2, 5).
In simple words: For the first line, find 'y' by multiplying 'x' by 2 and adding 1. For the second line, 'x' is always 2. Draw both lines on a graph. The point where they cross, which is (2, 5), is the answer.

🎯 Exam Tip: Be cautious with equations like \( 3x - 6 = 0 \); they represent vertical lines \( x = 2 \), which can be easily mistaken for other forms if not simplified correctly.

 

Question 2. (vi) Solve graphically: \( x = -3; y = 3 \)
Answer:
To solve this system graphically, we will find points for each equation and plot them.

**For Equation 1: \( x = -3 \)**
This equation represents a vertical line where the x-coordinate is always -3, regardless of the y-coordinate. It is parallel to the y-axis.
Now, we calculate some coordinate pairs:
When \( x = -3 \), \( y = -3 \) (arbitrary point)
When \( x = -3 \), \( y = -2 \) (arbitrary point)
When \( x = -3 \), \( y = 2 \) (arbitrary point)
When \( x = -3 \), \( y = 3 \) (intersection point from second line)

**For Equation 2: \( y = 3 \)**
This equation represents a horizontal line where the y-coordinate is always 3, regardless of the x-coordinate. It is parallel to the x-axis.
Now, we calculate some coordinate pairs:
When \( x = -3 \), \( y = 3 \) (intersection point from first line)
When \( x = -1 \), \( y = 3 \) (arbitrary point)
When \( x = 0 \), \( y = 3 \) (arbitrary point)
When \( x = 2 \), \( y = 3 \) (arbitrary point)

We plot the points for both lines on the same graph sheet. For \( x=-3 \), we plot (-3,-3), (-3,-2), (-3,2), (-3,3). For \( y=3 \), we plot (-3,3), (-1,3), (0,3), (2,3). We draw straight lines through these points. We observe that the two lines intersect at the point (-3, 3). This intersection point is the solution to the system of equations.
Therefore, the solution set is (-3, 3).
In simple words: The first line is a vertical line where 'x' is always -3. The second line is a horizontal line where 'y' is always 3. When you draw them, they cross at the point (-3, 3), which is the answer.

🎯 Exam Tip: Visualizing vertical (x=constant) and horizontal (y=constant) lines quickly can save time and prevent errors when solving graphically.

 

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.
Answer:
Let the speed of the first car be \( x \) miles per hour (mph).
Let the speed of the second car be \( y \) miles per hour (mph).

**Condition 1: Cars driving towards each other**
When cars drive towards each other, their relative speed is the sum of their individual speeds. They meet in 1 hour, and the distance is 100 miles.
Formula: Distance = Speed \( \times \) Time
\( 100 = (x + y) \times 1 \)
So, Equation 1 is: \( x + y = 100 \)
Rearrange to express \( y \) in terms of \( x \):
\( y = 100 - x \)
Let's find some coordinate pairs for this line:
When \( x = 30 \), \( y = 100 - 30 = 70 \)
When \( x = 50 \), \( y = 100 - 50 = 50 \)
When \( x = 60 \), \( y = 100 - 60 = 40 \)
When \( x = 70 \), \( y = 100 - 70 = 30 \)

**Condition 2: Cars driving in the same direction**
When cars drive in the same direction, their relative speed is the difference between their speeds (assuming \( x > y \)). They meet in 2 hours, and the distance is 100 miles.
Formula: Distance = Speed \( \times \) Time
\( 100 = (x - y) \times 2 \)
Divide by 2:
\( 50 = x - y \)
So, Equation 2 is: \( x - y = 50 \)
Rearrange to express \( y \) in terms of \( x \):
\( -y = 50 - x \)
\( y = x - 50 \)
Let's find some coordinate pairs for this line:
When \( x = 40 \), \( y = 40 - 50 = -10 \)
When \( x = 50 \), \( y = 50 - 50 = 0 \)
When \( x = 60 \), \( y = 60 - 50 = 10 \)
When \( x = 70 \), \( y = 70 - 50 = 20 \)

Plot the points for both lines on a graph sheet. For the first equation, plot (30, 70), (50, 50), (60, 40), (70, 30). For the second equation, plot (40, -10), (50, 0), (60, 10), (70, 20). Draw straight lines through these points. The point where the two lines intersect on the graph is (75, 25). This point represents the speeds of the two cars.
Thus, the speed of the first car is 75 mph, and the speed of the second car is 25 mph.
In simple words: We set up two math rules (equations) based on how the cars move. Then, we find dots for each rule on a graph. Where the two lines of dots cross, that point tells us the speeds of the cars. The first car travels at 75 miles per hour, and the second car travels at 25 miles per hour.

🎯 Exam Tip: When solving word problems graphically, clearly define your variables (x and y), correctly formulate the equations, and interpret the intersection point in the context of the problem's units.

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