CBSE Class 9 Mathematics Linear Equations in two variables ... - Free Printable

Let's solve each question from the given worksheet step by step, with clear explanations.

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SECTION A: (1 MARK)

#### 1. Find $ m $, if point $ (7, -3) $ lies on the equation $ y - \frac{3}{2} = m(x - \frac{1}{2}) $.

We are given a point $ (7, -3) $ that lies on the line:
$$
y - \frac{3}{2} = m\left(x - \frac{1}{2}\right)
$$

Substitute $ x = 7 $, $ y = -3 $ into the equation:

$$
-3 - \frac{3}{2} = m\left(7 - \frac{1}{2}\right)
$$

$$
-\frac{6}{2} - \frac{3}{2} = m\left(\frac{14}{2} - \frac{1}{2}\right)
$$

$$
-\frac{9}{2} = m \cdot \frac{13}{2}
$$

Now solve for $ m $:

$$
m = \frac{-\frac{9}{2}}{\frac{13}{2}} = -\frac{9}{13}
$$

But the answer provided is $ m = -\frac{24}{47} $. This suggests a typo in the problem or the expected answer. Let's double-check.

Wait — let's recheck the original equation. Is it possible the equation was meant to be:

$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$

Yes, and we substituted correctly.

But our result is $ m = -\frac{9}{13} $, not $ -\frac{24}{47} $.

Let’s verify if the point satisfies the equation with $ m = -\frac{24}{47} $:

Left side: $ y - \frac{3}{2} = -3 - 1.5 = -4.5 = -\frac{9}{2} $

Right side: $ m(x - \frac{1}{2}) = -\frac{24}{47}(7 - 0.5) = -\frac{24}{47} \cdot 6.5 = -\frac{24}{47} \cdot \frac{13}{2} = -\frac{312}{94} = -\frac{156}{47} \approx -3.32 $

But $ -\frac{9}{2} = -4.5 $, which is not equal. So either the answer key is wrong, or the equation is different.

Wait — perhaps the equation is:

$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$

But maybe the point is $ (7, -3) $? Yes.

So unless there's a mistake in the question, our calculation seems correct.

But let’s suppose the equation is actually:

$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$

Then as above:

$$
-3 - \frac{3}{2} = m(7 - \frac{1}{2}) \Rightarrow -\frac{9}{2} = m \cdot \frac{13}{2} \Rightarrow m = -\frac{9}{13}
$$

So the correct value is $ m = -\frac{9}{13} $.

But the given answer is $ m = -\frac{24}{47} $. That does not match.

Possibility: Maybe the equation is $ y - 3 = m(x - 7) $? But no, it says $ y - \frac{3}{2} = m(x - \frac{1}{2}) $.

Alternatively, could the point be $ (7, -3) $, but the equation has different constants?

Wait — perhaps the equation is:

$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$

and the point is $ (7, -3) $, then:

$$
-3 - 1.5 = m(7 - 0.5) \Rightarrow -4.5 = m(6.5) \Rightarrow m = -4.5 / 6.5 = -9/13
$$

So again, $ m = -\frac{9}{13} $

Thus, the provided answer $ m = -\frac{24}{47} $ appears incorrect.

But since the problem says "NCERT Exemplar", let's assume there's a typo in the question or answer. We'll proceed with correct math.

✔ Answer: $ m = -\frac{9}{13} $

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#### 2. Find the value of $ a $ in the equation $ ax + y = 5 $ if $ x = 2 $ and $ y = 3 $.

Given: $ ax + y = 5 $, $ x = 2 $, $ y = 3 $

Substitute:

$$
a(2) + 3 = 5 \Rightarrow 2a = 2 \Rightarrow a = 1
$$

✔ Answer: $ a = 1 $ ✔️

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#### 3. If $ x - 4 = \sqrt{3} y $ is written in standard form $ ax + by + c = 0 $, then find values of $ a, b, c $.

Start with:

$$
x - 4 = \sqrt{3} y
$$

Bring all terms to one side:

$$
x - \sqrt{3} y - 4 = 0
$$

Compare with $ ax + by + c = 0 $:

- $ a = 1 $
- $ b = -\sqrt{3} $
- $ c = -4 $

✔ Answer:
- $ a = 1 $
- $ b = -\sqrt{3} $
- $ c = -4 $ ✔️

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SECTION B: (2 MARKS)

#### 4. Represent an equation of a straight line which is parallel to x-axis and at a distance of 2.5 units below it.

A line parallel to the x-axis is horizontal. Its equation is $ y = k $.

Since it's 2.5 units below the x-axis, $ y = -2.5 $

So the equation is:

$$
y = -2.5 \quad \text{or} \quad y = -\frac{5}{2}
$$

In standard form: $ 0x + 1y + \frac{5}{2} = 0 $, or simply $ y + \frac{5}{2} = 0 $

✔ Answer: $ y = -2.5 $ or $ y + \frac{5}{2} = 0 $

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#### 5. For the first km, fare is Rs 15 and for successive distance, it is Rs 8 per km. Taking distance covered as $ x $ km and total fare as $ y $ (Rs), represent a linear equation in two variables.

Fare = Fixed charge + Variable charge

- First km: Rs 15
- Each additional km: Rs 8

So for $ x $ km:

- If $ x = 1 $, fare = 15
- If $ x > 1 $, fare = 15 + 8(x - 1)

Simplify:

$$
y = 15 + 8(x - 1) = 15 + 8x - 8 = 8x + 7
$$

So the equation is:

$$
y = 8x + 7
$$

Or in standard form: $ 8x - y + 7 = 0 $

✔ Answer: $ y = 8x + 7 $
Or as given: $ y = 15 + 8(x - 1) $ → $ y = 8x + 7 $

✔️ Matches given answer: $ y = 15 + 8(x - 1) $

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#### 6. If $ (2, 3) $ and $ (4, 0) $ lie on the graph of $ ax + by + 1 = 0 $, then find $ a $ and $ b $.

Plug both points into the equation:

Point (2, 3):

$$
a(2) + b(3) + 1 = 0 \Rightarrow 2a + 3b = -1 \quad \text{(1)}
$$

Point (4, 0):

$$
a(4) + b(0) + 1 = 0 \Rightarrow 4a + 1 = 0 \Rightarrow a = -\frac{1}{4}
$$

Now substitute $ a = -\frac{1}{4} $ into (1):

$$
2(-\frac{1}{4}) + 3b = -1 \Rightarrow -\frac{1}{2} + 3b = -1 \Rightarrow 3b = -\frac{1}{2} \Rightarrow b = -\frac{1}{6}
$$

But the given answer is $ a = \frac{1}{4}, b = \frac{1}{6} $. Let's check.

Wait — our result is $ a = -\frac{1}{4}, b = -\frac{1}{6} $

But let's test with the equation:

$ ax + by + 1 = 0 $

With $ a = -\frac{1}{4}, b = -\frac{1}{6} $:

For (2, 3): $ -\frac{1}{4}(2) - \frac{1}{6}(3) + 1 = -0.5 - 0.5 + 1 = 0 $ ✔

For (4, 0): $ -\frac{1}{4}(4) + 0 + 1 = -1 + 1 = 0 $ ✔

So our solution is correct.

But the given answer says $ a = \frac{1}{4}, b = \frac{1}{6} $

Try that: $ \frac{1}{4}(2) + \frac{1}{6}(3) + 1 = 0.5 + 0.5 + 1 = 2 \neq 0 $

✘ Incorrect.

So the correct answer is $ a = -\frac{1}{4}, b = -\frac{1}{6} $

But the provided answer is $ a = \frac{1}{4}, b = \frac{1}{6} $ — likely a sign error.

Wait — maybe the equation is $ ax + by = 1 $? But it says $ ax + by + 1 = 0 $

So $ ax + by = -1 $

Our values satisfy it.

✔ Correct Answer: $ a = -\frac{1}{4}, b = -\frac{1}{6} $

But the given answer is $ a = \frac{1}{4}, b = \frac{1}{6} $ — possibly typo.

Let’s see if they meant $ ax + by - 1 = 0 $? Then $ ax + by = 1 $

Then for (4,0): $ 4a = 1 \Rightarrow a = 1/4 $

For (2,3): $ 2a + 3b = 1 \Rightarrow 2(1/4) + 3b = 1 \Rightarrow 0.5 + 3b = 1 \Rightarrow 3b = 0.5 \Rightarrow b = 1/6 $

Ah! So if the equation were $ ax + by - 1 = 0 $, then $ a = 1/4, b = 1/6 $

But the question says $ ax + by + 1 = 0 $

So unless there's a typo in the question, the correct answer should be negative.

But since the given answer is $ a = 1/4, b = 1/6 $, and it matches if the constant is $ -1 $, probably the equation is $ ax + by = 1 $, i.e., $ ax + by - 1 = 0 $

So likely a typo in the question.

Assuming the intended equation is $ ax + by = 1 $, then:

From (4,0): $ 4a = 1 \Rightarrow a = 1/4 $

From (2,3): $ 2(1/4) + 3b = 1 \Rightarrow 0.5 + 3b = 1 \Rightarrow b = 1/6 $

✔ So if the equation is $ ax + by = 1 $, then $ a = \frac{1}{4}, b = \frac{1}{6} $

But as written: $ ax + by + 1 = 0 $ → $ ax + by = -1 $ → $ a = -1/4, b = -1/6 $

So either the question has a typo, or the answer key does.

We’ll go with what matches: if the answer is $ a = 1/4, b = 1/6 $, then the equation must be $ ax + by = 1 $

So assuming the equation is $ ax + by = 1 $, then:

✔ $ a = \frac{1}{4}, b = \frac{1}{6} $ ✔️

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#### 7. Find the co-ordinates of the points where the graph of $ 7x - 3y = 4 $ cuts the x-axis and y-axis.

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

$$
7x - 3(0) = 4 \Rightarrow 7x = 4 \Rightarrow x = \frac{4}{7}
$$

So x-intercept: $ \left( \frac{4}{7}, 0 \right) $

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

$$
7(0) - 3y = 4 \Rightarrow -3y = 4 \Rightarrow y = -\frac{4}{3}
$$

So y-intercept: $ \left( 0, -\frac{4}{3} \right) $

But the given answer says:

- X axis: $ \left( \frac{4}{7}, 0 \right) $ ✔
- Y axis: $ \left( 0, -\frac{4}{3} \right) $ — but the given says $ \left( 0, -\frac{1}{3} \right) $ ✘

Wait — the answer says $ \left( 0, -\frac{1}{3} \right) $ — this is wrong.

Check: $ -3y = 4 \Rightarrow y = -4/3 $

So correct y-intercept is $ (0, -4/3) $

✔ Correct Answer:
- X-axis: $ \left( \frac{4}{7}, 0 \right) $
- Y-axis: $ \left( 0, -\frac{4}{3} \right) $

Given answer has $ (0, -1/3) $ — incorrect.

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SECTION C: (3 MARKS)

#### 8. Solve: $ \frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3} $

First, find LCM of denominators: 7, 5, 3 → LCM = 105

Multiply both sides by 105:

$$
105 \left( \frac{3x+2}{7} + \frac{4x+1}{5} \right) = 105 \cdot \frac{2(2x+1)}{3}
$$

$$
15(3x+2) + 21(4x+1) = 35 \cdot 2(2x+1)
$$

Compute each term:

- $ 15(3x+2) = 45x + 30 $
- $ 21(4x+1) = 84x + 21 $
- RHS: $ 70(2x+1) = 140x + 70 $

Now:

$$
45x + 30 + 84x + 21 = 140x + 70
\Rightarrow 129x + 51 = 140x + 70
$$

Bring like terms together:

$$
51 - 70 = 140x - 129x \Rightarrow -19 = 11x \Rightarrow x = -\frac{19}{11}
$$

But the given answer is $ x = 4 $

Let’s check if $ x = 4 $ satisfies the original equation:

LHS:
- $ \frac{3(4)+2}{7} = \frac{14}{7} = 2 $
- $ \frac{4(4)+1}{5} = \frac{17}{5} = 3.4 $
- Sum: $ 2 + 3.4 = 5.4 $

RHS:
- $ \frac{2(8+1)}{3} = \frac{2(9)}{3} = 6 $

5.4 ≠ 6 → Not valid.

So $ x = 4 $ is not a solution.

Let’s solve again carefully.

Original:

$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

LCM = 105

Multiply both sides:

$$
15(3x+2) + 21(4x+1) = 35 \cdot 2(2x+1)
$$

Wait: $ 105 / 7 = 15 $, $ 105 / 5 = 21 $, $ 105 / 3 = 35 $

So RHS: $ 35 \cdot 2(2x+1) = 70(2x+1) = 140x + 70 $

LHS:
- $ 15(3x+2) = 45x + 30 $
- $ 21(4x+1) = 84x + 21 $
- Total: $ 129x + 51 $

So:

$$
129x + 51 = 140x + 70 \Rightarrow -11x = 19 \Rightarrow x = -\frac{19}{11}
$$

So correct answer is $ x = -\frac{19}{11} $

But given answer is $ x = 4 $ — clearly incorrect.

Possibility: Typo in equation?

Suppose instead:

$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

Let’s try $ x = 4 $:

LHS: $ \frac{14}{7} + \frac{17}{5} = 2 + 3.4 = 5.4 $

RHS: $ \frac{2(9)}{3} = 6 $ → not equal

Try $ x = 1 $:

LHS: $ \frac{5}{7} + \frac{5}{5} = 0.714 + 1 = 1.714 $

RHS: $ \frac{2(3)}{3} = 2 $ → no

Try $ x = 0 $:

LHS: $ \frac{2}{7} + \frac{1}{5} \approx 0.286 + 0.2 = 0.486 $

RHS: $ \frac{2(1)}{3} = 0.666 $ → no

So no, $ x = 4 $ is not a solution.

✔ Correct answer: $ x = -\frac{19}{11} $

But the given answer is $ x = 4 $ — likely a typo.

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#### 9. Draw the graph of $ y = x $ and $ y = -x $ on the same Cartesian plane. What do you observe?

- $ y = x $: passes through origin, slope 1, goes up-right
- $ y = -x $: passes through origin, slope -1, goes down-right

They intersect at origin $ (0,0) $

They are symmetric about the y-axis and x-axis.

They form a 'X' shape, perpendicular to each other.

Observation: The two lines intersect at the origin and are perpendicular to each other.

✔ Answer: The graphs intersect at the origin and are perpendicular to each other.

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#### 10. Draw the graph of $ 2x + 3y = 12 $ and check whether the points $ (4.5, 1) $ and $ (1.5, 3) $ lie on the graph.

Find intercepts:

- x-intercept: $ y = 0 $ → $ 2x = 12 $ → $ x = 6 $
- y-intercept: $ x = 0 $ → $ 3y = 12 $ → $ y = 4 $

Plot points $ (6,0) $ and $ (0,4) $, draw line.

Now check points:

Point (4.5, 1):

$ 2(4.5) + 3(1) = 9 + 3 = 12 $ → yes ✔

Point (1.5, 3):

$ 2(1.5) + 3(3) = 3 + 9 = 12 $ → yes ✔

✔ Both points lie on the graph.

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#### 11. Give the geometrical interpretation of $ 5x + 3 = 3x - 7 $ as an equation:

(i) In one variable

Rewrite: $ 5x + 3 = 3x - 7 $

Bring terms together:

$ 5x - 3x = -7 - 3 \Rightarrow 2x = -10 \Rightarrow x = -5 $

This represents a point on the number line at $ x = -5 $

(ii) In two variables

Write as: $ 5x + 3 = 3x - 7 $

Bring all to one side:

$ 5x - 3x + 3 + 7 = 0 \Rightarrow 2x + 10 = 0 $

But to write as two-variable equation, introduce $ y $:

Set $ y = 5x + 3 $ and $ y = 3x - 7 $

So we have two lines:

- $ y = 5x + 3 $
- $ y = 3x - 7 $

Their intersection point is the solution.

Solve: $ 5x + 3 = 3x - 7 \Rightarrow 2x = -10 \Rightarrow x = -5 $, $ y = 5(-5) + 3 = -25 + 3 = -22 $

So the geometrical interpretation is the point of intersection of the two lines $ y = 5x + 3 $ and $ y = 3x - 7 $

✔ Answer:
(i) A point on the number line at $ x = -5 $
(ii) The point of intersection of two lines representing the equations.

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✔ Summary of Correct Answers:

| Q | Correct Answer |
|---|----------------|
| 1 | $ m = -\frac{9}{13} $ (given $ -\frac{24}{47} $ is incorrect) |
| 2 | $ a = 1 $ ✔️ |
| 3 | $ a=1, b=-\sqrt{3}, c=-4 $ ✔️ |
| 4 | $ y = -2.5 $ or $ y = -\frac{5}{2} $ |
| 5 | $ y = 8x + 7 $ or $ y = 15 + 8(x-1) $ ✔️ |
| 6 | $ a = -\frac{1}{4}, b = -\frac{1}{6} $ (given $ a=1/4, b=1/6 $ likely assumes equation is $ ax+by=1 $) |
| 7 | X-axis: $ (\frac{4}{7}, 0) $, Y-axis: $ (0, -\frac{4}{3}) $ (given $ (0, -1/3) $ is wrong) |
| 8 | $ x = -\frac{19}{11} $ (given $ x=4 $ is incorrect) |
| 9 | Lines intersect at origin and are perpendicular |
| 10 | Both points lie on the graph |
| 11 | (i) Point on number line; (ii) Intersection of two lines |

Many answers in the key appear to have sign errors or typos. Always verify with substitution.