Class 9 Math Worksheets and Problems: Linear Equations in Two ... - Free Printable

Let's solve each question step by step and explain the reasoning.

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(1) In the graph of the linear equation $ 2x + 5y = 41 $, there is a point such that its ordinate is 3 less than its abscissa. Find coordinates of that point.

Step-by-step:

- Let the abscissa (x-coordinate) be $ x $
- Then the ordinate (y-coordinate) is $ y = x - 3 $ (since it's 3 less than abscissa)

Now substitute $ y = x - 3 $ into the equation:

$$
2x + 5(x - 3) = 41
$$

$$
2x + 5x - 15 = 41
$$

$$
7x - 15 = 41
$$

$$
7x = 56 \Rightarrow x = 8
$$

Then $ y = x - 3 = 8 - 3 = 5 $

✔ So the coordinates are: $ (8, 5) $

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(2) A line passes through points $ (-4, -3) $ and $ (1, 2) $. Find the x-intercept of the line.

Step-by-step:

We need to find the equation of the line passing through two points.

First, find the slope $ m $:

$$
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-3)}{1 - (-4)} = \frac{5}{5} = 1
$$

Now use point-slope form with point $ (1, 2) $:

$$
y - 2 = 1(x - 1)
\Rightarrow y = x + 1
$$

To find x-intercept, set $ y = 0 $:

$$
0 = x + 1 \Rightarrow x = -1
$$

✔ So, the x-intercept is $ (-1, 0) $, or simply $ x = -1 $

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(3) Equation $ 2x + 5y = 7 $ has a unique solution if x and y are:

Options:
a. Real Numbers
b. Positive Real Numbers
c. Natural Numbers
d. Rational Numbers

Explanation:

A linear equation in two variables has infinitely many solutions over the real numbers — unless restricted.

But the question asks when it has a unique solution.

This only happens if we restrict the domain so that only one pair $ (x, y) $ satisfies the equation.

- Over Real Numbers: Infinite solutions → Not unique.
- Over Positive Real Numbers: Still infinite solutions (e.g., multiple values satisfying).
- Over Natural Numbers: Only integer solutions ≥ 1. Let's check:
Try small natural numbers:
- $ x = 1 $: $ 2(1) + 5y = 7 \Rightarrow 5y = 5 \Rightarrow y = 1 $ → (1,1) valid
- $ x = 2 $: $ 4 + 5y = 7 \Rightarrow y = 0.6 $ → not natural
- $ x = 3 $: $ 6 + 5y = 7 \Rightarrow y = 0.2 $ → no
- $ x = 0 $: not natural
So only (1,1) is a natural number solution? Wait, what about $ x=6 $? $ 2(6)=12 > 7 $ → no
Actually, only (1,1) works? Let’s see:
$ 2x + 5y = 7 $
Try $ y=1 $: $ 2x = 2 \Rightarrow x=1 $
$ y=2 $: $ 5(2)=10 > 7 $ → too big
$ y=0 $: $ 2x=7 \Rightarrow x=3.5 $ → not natural
So only solution in natural numbers is (1,1)

So only one solution in natural numbers.

But wait — does the equation have a unique solution if x and y are natural numbers?

Yes, because only $ (1,1) $ satisfies it with both x and y being natural numbers.

Now consider Rational Numbers: There are infinitely many rational solutions (e.g., $ x=0, y=7/5 $; $ x=1, y=1 $; etc.)

So only in Natural Numbers is the solution unique.

✔ Answer: c. Natural Numbers

> Note: This is a bit tricky — technically, a linear equation in two variables never has a *unique* solution over an infinite field like reals or rationals. But here, "has a unique solution" implies under the restriction of the domain.

So the correct choice is: c. Natural Numbers

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(4) If both sides of an equation are divided by a non-zero number, then solution of the equation:

a. Will also be divided by same number
b. May or may not change depending on the equation
c. Changes
d. Remains the same

Explanation:

When you divide both sides of an equation by a non-zero number, you're performing a valid algebraic operation.

For example:
$$
2x + 4 = 6 \Rightarrow \text{Divide by 2: } x + 2 = 3
$$
Same solution: $ x = 1 $

The solution set does not change.

✔ Correct answer: d. Remains the same

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(5) The equation of x-axis is:

a. $ x = 0 $
b. $ x = y $
c. $ x + y = 0 $
d. $ y = 0 $

Explanation:

The x-axis consists of all points where the y-coordinate is 0.

So for any point $ (x, 0) $, $ y = 0 $

✔ Answer: d. $ y = 0 $

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(6) A telecom operator charges Rs. 0.9 for the first minute and Rs. 0.8 per minute for subsequent minutes. If duration of call is represented as $ d $, and amount charged is $ c $, find the linear equation.

Options:
a. $ c = 0.9d + 0.1 $
b. $ c = 0.8d + 0.1 $
c. $ c = 0.8d + 0.9 $
d. $ c = 0.9d + 0.8 $

Step-by-step:

- First minute: Rs. 0.9
- Each additional minute (after first): Rs. 0.8
- Total duration: $ d $ minutes

So:
- For $ d = 1 $: cost = 0.9
- For $ d = 2 $: cost = 0.9 + 0.8 = 1.7
- For $ d = 3 $: cost = 0.9 + 2×0.8 = 0.9 + 1.6 = 2.5

General formula:
$$
c = 0.9 + 0.8(d - 1)
$$

Simplify:
$$
c = 0.9 + 0.8d - 0.8 = 0.8d + 0.1
$$

Wait! That gives $ c = 0.8d + 0.1 $

Check:
- $ d = 1 $: $ 0.8(1) + 0.1 = 0.9 $ ✔
- $ d = 2 $: $ 1.6 + 0.1 = 1.7 $ ✔
- $ d = 3 $: $ 2.4 + 0.1 = 2.5 $ ✔

So the correct equation is: $ c = 0.8d + 0.1 $

✔ Answer: b. $ c = 0.8d + 0.1 $

> Note: Option b says $ c = 0.8d + 0.1 $, which matches.

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✔ Final Answers:

1. (8, 5)
2. x-intercept = -1
3. c. Natural Numbers
4. d. Remains the same
5. d. $ y = 0 $
6. b. $ c = 0.8d + 0.1 $

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