CBSE Class 9 Mathematics Linear Equations in two variables Worksheet Set 01

Question. Represent the linear equation in two variables in its standard form. 2y – 3 + 2x =5
Answer: The standard form for a linear equation in two variables is represented as \( ax + by + c = 0 \).
Let's rearrange the given equation, \( 2y - 3 + 2x = 5 \):
First, group the \( x \) and \( y \) terms together on the left side:
\( 2x + 2y - 3 = 5 \)

Next, shift the constant \( 5 \) from the right side over to the left side by subtracting it:
\( 2x + 2y - 3 - 5 = 0 \)
\( 2x + 2y - 8 = 0 \)

We can also divide the entire equation by 2 to write it in its simplest form:
\( x + y - 4 = 0 \)
In simple words: We rearrange the terms so x is first, y is second, the plain number is third, and 0 is on the right side.

Exam Tip: Ensure that the right side of your equation is always simplified to 0 to write it correctly in standard form.

 

Question. What is the value of a, b, c in the given equation 3x – 5= 2y
Answer: Let us first write the equation \( 3x - 5 = 2y \) in standard form \( ax + by + c = 0 \) by moving the \( y \) term to the left-hand side:
\( 3x - 2y - 5 = 0 \)

By comparing this rearranged equation to the general form \( ax + by + c = 0 \), we can identify each of the constants:
- \( a = 3 \)
- \( b = -2 \)
- \( c = -5 \)
In simple words: Move everything to one side first. The number in front of x is a, the number in front of y is b, and the plain number is c.

Exam Tip: Don't forget to keep the negative sign with the values of b and c when writing down your final answer.

 

Question. Find the value of b and c in the equation 3x= 15
Answer: To find the constants, let's rearrange the equation \( 3x = 15 \) into standard two-variable form \( ax + by + c = 0 \). Since there is no \( y \) term present, we write its coefficient as 0:
\( 3x + 0y - 15 = 0 \)

Comparing this directly with the standard format \( ax + by + c = 0 \) yields the constant values:
- \( a = 3 \)
- \( b = 0 \)
- \( c = -15 \)

Therefore, \( b = 0 \) and \( c = -15 \).
In simple words: Since there is no y in the equation, its coefficient b is 0, and moving 15 to the other side makes c equal to -15.

Exam Tip: When a variable is missing, always set its coefficient to zero in standard form instead of omitting it.

 

Question. Check whether (4,0) is a solution of the equation2x+3y =8
Answer: Let us test the point \( (4,0) \) by substituting \( x = 4 \) and \( y = 0 \) into the left-hand side (LHS) of the given equation \( 2x + 3y = 8 \):
\( \text{LHS} = 2(4) + 3(0) \)
\( \text{LHS} = 8 + 0 \)
\( \text{LHS} = 8 \)

Now, let us look at the right-hand side (RHS) of the equation:
\( \text{RHS} = 8 \)

Since \( \text{LHS} = \text{RHS} \), the coordinates satisfy the equation.
Therefore, \( (4,0) \) is a valid solution.
In simple words: Put 4 in place of x and 0 in place of y. Since the calculation equals 8, the point is indeed a solution.

Exam Tip: For verify-type questions, clearly state whether LHS equals RHS at the end to get full marks.

 

Question. Find if \( \sqrt{2}, -1 \) is a solution of the equation x + 3y =1
Answer: Let us check if the coordinates \( (\sqrt{2}, -1) \) satisfy the equation by substituting \( x = \sqrt{2} \) and \( y = -1 \) into the left-hand side (LHS) of \( x + 3y = 1 \):
\( \text{LHS} = \sqrt{2} + 3(-1) \)
\( \text{LHS} = \sqrt{2} - 3 \)

Since \( \sqrt{2} \approx 1.414 \), we have:
\( \text{LHS} \approx 1.414 - 3 = -1.586 \)

The right-hand side is:
\( \text{RHS} = 1 \)

Since \( \text{LHS} \neq \text{RHS} \), the coordinates do not satisfy the equation.
Therefore, \( (\sqrt{2}, -1) \) is not a solution.
*(Note: If the point was meant to be \( (4, -1) \), substituting \( x=4, y=-1 \) would give \( 4 + 3(-1) = 1 \), making it a solution.)*
In simple words: When we plug the numbers into the left side, we get a value that does not equal 1, so this point does not fit.

Exam Tip: Keep radical expressions like \( \sqrt{2} \) in their exact form during algebraic checks unless a decimal approximation is specifically requested.

 

Question. Form an equation in two variables with the given information: the number of ducks is three more than three times number of hens, and the total of all hens and ducks is 156.
Answer: Let the total number of hens be \( x \) and the total number of ducks be \( y \).
According to the first statement, "the number of ducks is three more than three times the number of hens":
\( y = 3x + 3 \)
\( \implies 3x - y + 3 = 0 \)

According to the second statement, "the total number of hens and ducks is 156":
\( x + y = 156 \)
\( \implies x + y - 156 = 0 \)

Thus, the situation is represented by the following system of linear equations in two variables:
\( y = 3x + 3 \) and \( x + y = 156 \).
In simple words: Let x be hens and y be ducks. "Ducks are three more than three times hens" means y = 3x + 3, and "total is 156" means x + y = 156.

Exam Tip: Clearly define your variables (what x and y represent) at the beginning of your response.

 

Question. Find four solutions of the given equations 3x – y=4
Answer: Let us rewrite the equation to express \( y \) explicitly in terms of \( x \):
\( y = 3x - 4 \)

Now, we can find four different solutions by choosing four different values for \( x \):
1. If \( x = 0 \):
\( y = 3(0) - 4 = -4 \)
So, \( (0, -4) \) is the first solution.

2. If \( x = 1 \):
\( y = 3(1) - 4 = -1 \)
So, \( (1, -1) \) is the second solution.

3. If \( x = 2 \):
\( y = 3(2) - 4 = 2 \)
So, \( (2, 2) \) is the third solution.

4. If \( x = 3 \):
\( y = 3(3) - 4 = 5 \)
So, \( (3, 5) \) is the fourth solution.
In simple words: Change the formula to y = 3x - 4. Pick simple numbers for x like 0, 1, 2, and 3 to calculate the matching y values.

Exam Tip: Presenting your final coordinates clearly as ordered pairs like (x, y) makes it very easy for the examiner to award full marks.

 

Question. Find the value of k, when x = -1 and y =2 in the equation 3x –7y = 3k .
Answer: Let us substitute the given values \( x = -1 \) and \( y = 2 \) directly into the equation \( 3x - 7y = 3k \):
\( 3(-1) - 7(2) = 3k \)
\( -3 - 14 = 3k \)
\( -17 = 3k \)
\( k = -\frac{17}{3} \)

Thus, the value of \( k \) is \( -\frac{17}{3} \).
In simple words: Replace x with -1 and y with 2. Solve the simple equation to find that k equals -17/3.

Exam Tip: Be careful with signs during multiplication, especially when dealing with negative values like \( 3(-1) \).

 

Question. Find two solutions for each of the following equations:
(i) 4x + 3y =12
(ii) 2x + 5y = 0
Answer: Let us calculate two unique solution sets for each equation:

**(i) For the equation \( 4x + 3y = 12 \):**
- Let us set \( x = 3 \):
\( 4(3) + 3y = 12 \implies 12 + 3y = 12 \implies 3y = 0 \implies y = 0 \)
Thus, \( (3, 0) \) is a solution.
- Let us set \( y = 4 \):
\( 4x + 3(4) = 12 \implies 4x + 12 = 12 \implies 4x = 0 \implies x = 0 \)
Thus, \( (0, 4) \) is a solution.

**(ii) For the equation \( 2x + 5y = 0 \):**
- Let us set \( x = 0 \):
\( 2(0) + 5y = 0 \implies 5y = 0 \implies y = 0 \)
Thus, \( (0, 0) \) is a solution.
- Let us set \( x = 5 \):
\( 2(5) + 5y = 0 \implies 10 + 5y = 0 \implies 5y = -10 \implies y = -2 \)
Thus, \( (5, -2) \) is a solution.
In simple words: Choose simple whole numbers for one letter and find what the other letter must be.

Exam Tip: If an equation has no constant term (like 2x + 5y = 0), the origin point (0, 0) is always one of its solutions.

 

Question. Express the following linear equations in the form ax + by + c = 0 and find the values of a , b, c.In the equation \( x - \frac{1}{5}y = -9.\overline{35} \)
Answer: To express \( x - \frac{1}{5}y = -9.\overline{35} \) in the standard format \( ax + by + c = 0 \), let's bring the constant term over to the left-hand side:
\( 1x - \frac{1}{5}y + 9.\overline{35} = 0 \)

Now we can match this directly with the general form \( ax + by + c = 0 \):
- \( a = 1 \)
- \( b = -\frac{1}{5} \) (or \( -0.2 \))
- \( c = 9.\overline{35} \)
In simple words: Move the number from the right side to the left side so the equation equals zero. Then identify the values of a, b, and c.

Exam Tip: Keep repeating decimals exactly as written with their bar symbols; do not round them off when stating the value of c.

 

Question. Write the following as linear equation in two variables.2x = 15 and –3y –4 =0
Answer: To write a single-variable equation in the standard two-variable form \( ax + by + c = 0 \), we add the missing variable with a coefficient of \( 0 \):

**First, for \( 2x = 15 \):**
Rearrange by moving the constant to the left side:
\( 2x - 15 = 0 \)
Now, insert the \( y \) term with a zero coefficient:
\( 2x + 0y - 15 = 0 \)

**Second, for \( -3y - 4 = 0 \):**
Insert the \( x \) term with a zero coefficient:
\( 0x - 3y - 4 = 0 \)
In simple words: Since there is a missing letter, we write it in the equation with a 0 in front of it.

Exam Tip: Ensure you include the signs (+ or -) correctly when presenting your final linear expressions.

 

Question. Write five solutions of the given equation: \( \pi x + y = 7 \).
Answer: Let us write the given equation as:
\( y = 7 - \pi x \)

We can find five solutions by substituting different values for \( x \):
1. If \( x = 0 \):
\( y = 7 - \pi(0) = 7 \). Solution: \( (0, 7) \)

2. If \( x = 1 \):
\( y = 7 - \pi(1) = 7 - \pi \). Solution: \( (1, 7 - \pi) \)

3. If \( x = 2 \):
\( y = 7 - \pi(2) = 7 - 2\pi \). Solution: \( (2, 7 - 2\pi) \)

4. If \( x = -1 \):
\( y = 7 - \pi(-1) = 7 + \pi \). Solution: \( (-1, 7 + \pi) \)

5. If \( x = \frac{7}{\pi} \):
\( y = 7 - \pi\left(\frac{7}{\pi}\right) = 7 - 7 = 0 \). Solution: \( \left(\frac{7}{\pi}, 0\right) \)
In simple words: Put simple numbers like 0, 1, or 2 in place of x to find the corresponding values of y.

Exam Tip: Leave \( \pi \) as a symbol in your coordinates instead of converting it to 3.14 unless the question specifically asks for an approximate value.

 

Question. Write the four solutions of the equation (2x - 1)/(3y - 5) = 1/3.
Answer: Let's simplify the fractional equation first by cross-multiplying:
\( 3(2x - 1) = 1(3y - 5) \)
\( 6x - 3 = 3y - 5 \)
\( 6x - 3y - 3 + 5 = 0 \)
\( 6x - 3y + 2 = 0 \)
\( \implies 3y = 6x + 2 \implies y = 2x + \frac{2}{3} \)

Now, let's find four different solutions by choosing values for \( x \):
1. If \( x = 0 \):
\( y = 2(0) + \frac{2}{3} = \frac{2}{3} \). Solution: \( (0, \frac{2}{3}) \)

2. If \( x = 1 \):
\( y = 2(1) + \frac{2}{3} = \frac{8}{3} \). Solution: \( (1, \frac{8}{3}) \)

3. If \( x = -1 \):
\( y = 2(-1) + \frac{2}{3} = -\frac{4}{3} \). Solution: \( (-1, -\frac{4}{3}) \)

4. If \( x = 2 \):
\( y = 2(2) + \frac{2}{3} = \frac{14}{3} \). Solution: \( (2, \frac{14}{3}) \)
In simple words: First cross-multiply to make it a simple equation, then pick four different values for x to calculate y.

Exam Tip: Simplifying the equation first before choosing coordinates prevents calculation mistakes with fractions.

 

Question. Write the solution of the equation x +y =4
Answer: A linear equation in two variables has an infinite number of solutions. We can express the general solution as:
\( y = 4 - x \)
For any real number \( x \), the ordered pair is \( (x, 4 - x) \).

Let us list some of these points as examples:
- If \( x = 0 \), then \( y = 4 \). Point: \( (0, 4) \)
- If \( x = 4 \), then \( y = 0 \). Point: \( (4, 0) \)
- If \( x = 2 \), then \( y = 2 \). Point: \( (2, 2) \)
- If \( x = 1 \), then \( y = 3 \). Point: \( (1, 3) \)
In simple words: There are endless pairs of numbers that add up to 4, like (0,4), (4,0), and (2,2).

Exam Tip: Unless a specific number of coordinates is requested, writing 2 or 3 distinct solutions is standard practice.

 

Question. Lata and Gautami together contributed Rs. 100 for a donation camp. Represent this situation graphically.
Answer: Let Lata's donation be represented by Rs. \( x \) and Gautami's donation be represented by Rs. \( y \). Since their total contribution is Rs. 100, we write:
\( x + y = 100 \)

To plot this relationship, let us find a few points:
- If \( x = 0 \), then \( y = 100 \). Point: \( (0, 100) \)
- If \( x = 100 \), then \( y = 0 \). Point: \( (100, 0) \)
- If \( x = 50 \), then \( y = 50 \). Point: \( (50, 50) \)

Since money contributions cannot be negative, we plot this line in the first quadrant:

In simple words: The equation is x + y = 100. We draw a straight line that connects 100 on the vertical axis to 100 on the horizontal axis.
Exam Tip: Since donations cannot be negative, do not extend your line into the second, third, or fourth quadrants.

 

Question. The taxi fare in a city is as follows: For the first km, the fare is Rs.8 and for every subsequent Km it is Rs. 5. Taking the distance traveled as x and the total fare as y, represent the equation graphically.
Answer: Let the total distance covered be \( x \) km and the total fare be Rs. \( y \).
For the first kilometer, the charge is Rs. 8.
For the rest of the trip, which is \( (x - 1) \) kilometers, the fare is Rs. 5 per kilometer.
This gives us the total charge equation:
\( y = 8 + 5(x - 1) \)
\( y = 8 + 5x - 5 \)
\( y = 5x + 3 \) (for \( x \ge 1 \))

Let's find some coordinate points to plot this line:
- If \( x = 1 \), then \( y = 5(1) + 3 = 8 \). Point: \( (1, 8) \)
- If \( x = 2 \), then \( y = 5(2) + 3 = 13 \). Point: \( (2, 13) \)
- If \( x = 3 \), then \( y = 5(3) + 3 = 18 \). Point: \( (3, 18) \)

We plot these coordinates to draw the graph:

In simple words: The first kilometer costs Rs. 8, and each extra kilometer adds Rs. 5. This makes the total cost equation y = 5x + 3.
Exam Tip: Since a taxi ride must start at a minimum distance, begin your solid graph line from the coordinate (1, 8) rather than the origin.

 

Question. Represent the equation 2 + 3y = 7x graphically.
Answer: Let us rearrange the equation \( 2 + 3y = 7x \) to express \( y \) in terms of \( x \):
\( 3y = 7x - 2 \)
\( y = \frac{7x - 2}{3} \)

Let's find some integer coordinate pairs to plot:
- If \( x = 2 \), then \( y = \frac{7(2) - 2}{3} = \frac{12}{3} = 4 \). Point: \( (2, 4) \)
- If \( x = -1 \), then \( y = \frac{7(-1) - 2}{3} = \frac{-9}{3} = -3 \). Point: \( (-1, -3) \)
- If \( x = 5 \), then \( y = \frac{7(5) - 2}{3} = \frac{33}{3} = 11 \). Point: \( (5, 11) \)

Here is the graphical display of this line:

In simple words: We rearrange the formula to calculate y easily. Plotting points like (2, 4) and (-1, -3) allows us to draw our line.
Exam Tip: Try to choose inputs for x that result in clean integer values for y, which makes drawing on graph paper much more accurate.

 

Question. Form an equation for the statement: The sum of cost of pens and twice the cost of pencils is Rs. 6 and represent the situation graphically.
Answer: Let the price of a pen be represented by Rs. \( x \), and the price of a pencil be represented by Rs. \( y \).
Based on the statement, "the sum of cost of pens and twice the cost of pencils is Rs. 6", we form the equation:
\( x + 2y = 6 \)

To draw this on a coordinate plane, let's determine some coordinate points:
- If \( x = 0 \), then \( 2y = 6 \implies y = 3 \). Point: \( (0, 3) \)
- If \( y = 0 \), then \( x = 6 \). Point: \( (6, 0) \)
- If \( x = 2 \), then \( 2 + 2y = 6 \implies 2y = 4 \implies y = 2 \). Point: \( (2, 2) \)

Now, let's display the graph:

In simple words: Let x be the pen cost and y be the pencil cost. The equation is x + 2y = 6. This forms a straight downward line on the graph.
Exam Tip: Since costs cannot be negative values, terminate the graph line at the axes where \( x = 0 \) and \( y = 0 \).

 

Question. Form the graph of the equation y = 2x
Answer: Let us find some points to plot for the linear equation \( y = 2x \):
- If \( x = 0 \), then \( y = 2(0) = 0 \). Point: \( (0, 0) \)
- If \( x = 1 \), then \( y = 2(1) = 2 \). Point: \( (1, 2) \)
- If \( x = -1 \), then \( y = 2(-1) = -2 \). Point: \( (-1, -2) \)
- If \( x = 2 \), then \( y = 2(2) = 4 \). Point: \( (2, 4) \)

Using these points, we construct the graph of the equation:

In simple words: The equation y = 2x means y is always twice as big as x. The line passes directly through the origin (0, 0).
Exam Tip: Any linear equation in the form y = mx has no constant term, meaning its graph will always pass directly through the origin (0,0).

 

Question. Plot the graph of the equation y – 2x = 4
Answer: Let us rearrange the equation \( y - 2x = 4 \) as:
\( y = 2x + 4 \)

Now, let us find a few coordinate points to define the line:
- If \( x = 0 \), then \( y = 2(0) + 4 = 4 \). Point: \( (0, 4) \)
- If \( y = 0 \), then \( 0 = 2x + 4 \implies 2x = -4 \implies x = -2 \). Point: \( (-2, 0) \)
- If \( x = 1 \), then \( y = 2(1) + 4 = 6 \). Point: \( (1, 6) \)

Plotting these values allows us to construct the line:

In simple words: The equation can be written as y = 2x + 4. It crosses the vertical axis at 4 and the horizontal axis at -2.
Exam Tip: Finding where the line intersects both axes (the x-intercept and y-intercept) is the most efficient way to sketch a linear graph.

 

Question. Express y =4 as linear equation in two variables.
Answer: To represent the single-variable equation \( y = 4 \) as a linear equation in two variables, we insert the missing \( x \) variable with a coefficient of \( 0 \):
\( 0 \cdot x + 1 \cdot y = 4 \)

Written in the standard standard form \( ax + by + c = 0 \), it becomes:
\( 0x + y - 4 = 0 \)
In simple words: Since there is no x in the equation, we write it with a 0 in front: 0x + y - 4 = 0.

Exam Tip: Be sure to keep the coefficient of y as 1 and write the zero coefficient for x explicitly when converting.

 

Question. Give the representation of 2x +9 =0 as an equation in
a) One variable
b) Two variable
Answer: Let us represent the given equation \( 2x + 9 = 0 \) in both formats:

**a) Representation in one variable:**
Solving the equation for \( x \):
\( 2x = -9 \)
\( x = -\frac{9}{2} = -4.5 \)
In one variable, this represents a unique point on the number line at \( -4.5 \).

**b) Representation in two variables:**
We express the equation by adding a \( y \) term with a zero coefficient:
\( 2x + 0y + 9 = 0 \)
On the 2D Cartesian plane, this equation represents a vertical line parallel to the y-axis, crossing through \( (-4.5, 0) \).

In simple words: In one variable, x = -4.5 is just a point on a line. In two variables, 2x + 0y + 9 = 0 is a vertical line on a grid.
Exam Tip: Be ready to explain the geometric difference: a one-variable solution is a point, whereas a two-variable solution represents an entire line.

 

Question. The temperature in degree Celsius is given by the following formula F = (9/5) C + 32
Answer the following questions
a. What will be the temperature in degree Celsius if the temperature is 45 °F?
b. If the temperature is 0 °C, what is the temperature in Fahrenheit?
c. Is there a temperature, which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Answer: Let us solve each part of the temperature conversion question using the given formula \( F = \frac{9}{5}C + 32 \):

**a. Find the temperature in Celsius if the temperature is 45 °F:**
Substitute \( F = 45 \) into the conversion formula:
\( 45 = \frac{9}{5}C + 32 \)
\( 45 - 32 = \frac{9}{5}C \)
\( 13 = \frac{9}{5}C \)
\( C = 13 \times \frac{5}{9} \)
\( C = \frac{65}{9} \approx 7.22 \text{ }^{\circ}\text{C} \)

**b. Find the temperature in Fahrenheit if the temperature is 0 °C:**
Substitute \( C = 0 \) into the formula:
\( F = \frac{9}{5}(0) + 32 \)
\( F = 0 + 32 \)
\( F = 32 \text{ }^{\circ}\text{F} \)

**c. Find if there is a temperature that is numerically identical in both scales:**
Let \( x \) be the temperature value where both scales are equal, so \( F = C = x \). Substitute \( x \) for both variables in the formula:
\( x = \frac{9}{5}x + 32 \)
Multiply the entire equation by 5 to clear the fraction:
\( 5x = 9x + 160 \)
\( 5x - 9x = 160 \)
\( -4x = 160 \)
\( x = \frac{160}{-4} \)
\( x = -40 \)

Yes, \( -40^{\circ} \) is numerically the same value on both the Celsius and Fahrenheit temperature scales.
In simple words: We plug the given numbers into the formula to solve each part. We find that -40 is the special temperature where both scales show the exact same number.

Exam Tip: Showing the complete algebraic derivation in part (c) is critical to receiving full marks instead of just writing down the final value of -40.

 

 

Please click on below link to download CBSE Class 9 Mathematics Linear Equations in two variables Worksheet Set A