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

Let's solve each question step-by-step from the given worksheet, explaining the reasoning and solutions clearly.

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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}) $.

Given:
Point $ (7, -3) $ lies on the line:
$$
y - \frac{3}{2} = m\left(x - \frac{1}{2}\right)
$$

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

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

$$
-\frac{6}{2} - \frac{3}{2} = m\left(\frac{14}{2} - \frac{1}{2}\right)
\Rightarrow -\frac{9}{2} = m \cdot \frac{13}{2}
$$

Now solve for $ m $:

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

But wait! The expected answer is $ m = -\frac{24}{47} $, which suggests a possible typo in the problem or the provided answer.

Let’s double-check the equation:
Is it possible the equation was meant to be:
$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$
Yes, that’s what’s written.

But with $ (7, -3) $, we get $ m = -\frac{9}{13} $, not $ -\frac{24}{47} $. Let’s verify if $ -\frac{24}{47} $ could be correct.

Try plugging $ m = -\frac{24}{47} $ into the equation at $ x = 7 $:

$$
y = -\frac{24}{47}(7 - 0.5) + 1.5 = -\frac{24}{47}(6.5) + 1.5
= -\frac{24 \times 13}{94} + 1.5 = -\frac{312}{94} + 1.5
= -3.319 + 1.5 = -1.819 \neq -3
$$

So $ m = -\frac{24}{47} $ does not satisfy the point.

Thus, likely the correct answer is $ m = -\frac{9}{13} $, but the provided answer may be incorrect.

> ✔ Correct Answer: $ m = -\frac{9}{13} $
> ✘ Given answer $ m = -\frac{24}{47} $ seems incorrect.

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

Plug in $ x = 2 $, $ y = 3 $:

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

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

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 has constant $ y $-value.

"Below" means negative direction.

So, $ y = -2.5 $

Or in fraction: $ y = -\frac{5}{2} $

Equation: $ y + \frac{5}{2} = 0 $ or simply $ y = -\frac{5}{2} $

✔ 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. Take distance covered as $ x $ km and total fare as $ y $ (Rs). Represent linear equation in two variables.

Fare = Base fare + Extra fare

- First km: Rs 15
- Remaining $ (x - 1) $ km at Rs 8 per km

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

✔ Answer: $ y = 8x + 7 $
Or $ y = 8x + 7 $ → $ 8x - y + 7 = 0 $

But the given answer says: $ y = 15 + 8(x - 1) $ — which is correct, but simplified form is better.

✔ Final: $ y = 8x + 7 $

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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 $ ax + by + 1 = 0 $

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

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

Now plug 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 expected answer is $ a = \frac{1}{4}, b = \frac{1}{6} $? Wait — let’s check sign.

We have $ a = -\frac{1}{4}, b = -\frac{1}{6} $

But let’s test in original equation:

$ ax + by + 1 = 0 $

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

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

So correct values are:
- $ a = -\frac{1}{4} $
- $ b = -\frac{1}{6} $

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

Wait — maybe they wrote the equation as $ ax + by = -1 $? But it says $ ax + by + 1 = 0 $

So our calculation is correct.

But perhaps they meant $ ax + by = 1 $? Let's see.

Suppose equation is $ ax + by = 1 $, then:

At $ (2,3) $: $ 2a + 3b = 1 $

At $ (4,0) $: $ 4a = 1 \Rightarrow a = \frac{1}{4} $

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

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

But the problem says: $ ax + by + 1 = 0 $ → so $ ax + by = -1 $

Therefore, signs must be negative.

Hence, either the problem statement is wrong, or the answer key assumes different form.

But based on given equation: $ ax + by + 1 = 0 $, the correct values are:

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

But since the expected answer is $ a = \frac{1}{4}, b = \frac{1}{6} $, likely there’s a typo in the problem or answer.

Alternatively, maybe they want the equation $ ax + by = 1 $, not $ +1 = 0 $

But as written, our solution is correct.

> ⚠️ Note: The given answer $ a = \frac{1}{4}, b = \frac{1}{6} $ corresponds to equation $ ax + by = 1 $, not $ ax + by + 1 = 0 $

So unless the equation was mistyped, our answer is correct.

But for consistency with expected answer, perhaps the equation was meant to be $ ax + by = 1 $?

Assuming that, then:

✔ $ a = \frac{1}{4}, b = \frac{1}{6} $ (if equation is $ ax + by = 1 $)

But as per given: $ ax + by + 1 = 0 $, we get $ a = -\frac{1}{4}, b = -\frac{1}{6} $

So discrepancy exists.

Let’s proceed with correct math.

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

- X-axis intercept: Set $ y = 0 $

$$
7x = 4 \Rightarrow x = \frac{4}{7}
\Rightarrow \left(\frac{4}{7}, 0\right)
$$

- Y-axis intercept: Set $ x = 0 $

$$
-3y = 4 \Rightarrow y = -\frac{4}{3}
\Rightarrow \left(0, -\frac{4}{3}\right)
$$

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

But the given answer says:
X axis $ (\frac{4}{7}, 0) $ — correct
Y axis $ (0, -\frac{4}{3}) $ — correct

Wait, in your image, it says $ (0, -\frac{4}{3}) $, but you wrote $ (0, -\frac{4}{3}) $ — yes, matches.

But in the image, it shows $ (0, -\frac{4}{3}) $ — correct.

✔ So answer is correct.

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SECTION C: (3 MARKS)

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

Find LCM of denominators: 7, 5, 3 → LCM = 105

Multiply both sides by 105:

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

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

Compute:

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

Right:
$$
35(2x+1) = 70x + 35
$$

Now:
$$
129x + 51 = 70x + 35 \\
129x - 70x = 35 - 51 \\
59x = -16 \\
x = -\frac{16}{59}
$$

But the expected answer is $ x = 4 $? That can't be.

Let’s test $ x = 4 $:

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:
$$
\frac{2(4)+1}{3} = \frac{9}{3} = 3
$$

Not equal.

So $ x = 4 $ is not a solution.

Our calculation gives $ x = -\frac{16}{59} $

But perhaps there’s a typo in the equation?

Let me recheck:

Original: $ \frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2x+1}{3} $

Maybe the right side is $ \frac{2x+1}{3} $, but let's try solving again.

Wait — maybe the equation is:

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

Yes.

We did:
- 15(3x+2) = 45x + 30
- 21(4x+1) = 84x + 21 → total 129x + 51
- 35(2x+1) = 70x + 35

129x + 51 = 70x + 35 → 59x = -16 → x = -16/59

No way this equals 4.

So either the problem is wrong or the answer is wrong.

But the expected answer is $ x = 4 $. Try to see what equation would give $ x = 4 $.

Suppose:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2x+1}{3}
$$

Try $ x = 4 $: LHS = $ \frac{14}{7} + \frac{17}{5} = 2 + 3.4 = 5.4 $

RHS = $ \frac{9}{3} = 3 $ → not equal.

So $ x = 4 $ is not a solution.

Hence, expected answer is incorrect.

✔ Correct answer: $ x = -\frac{16}{59} $

But since the answer key says $ x = 4 $, possibly the equation is different.

Wait — maybe it's:

$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2x+1}{3} + something?
$$

Or maybe the equation is:

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

Still no.

Perhaps the equation is:

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

But maybe it's a typo, and should be:

Try $ x = 4 $: suppose LHS = RHS

But it's not.

Alternatively, maybe the equation is:

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

No.

I think the answer key is wrong.

✔ Correct solution: $ x = -\frac{16}{59} $

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#### 9. Draw the graph of $ y = x $ and $ y = -x $ on 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 x-axis and y-axis.

They form a "X" shape.

✔ Observation: Both lines pass through the origin and are perpendicular to each other. Their point of intersection is the origin.

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

First, find intercepts:

- X-intercept: $ y = 0 $ → $ 2x = 12 $ → $ x = 6 $ → $ (6, 0) $
- Y-intercept: $ x = 0 $ → $ 3y = 12 $ → $ y = 4 $ → $ (0, 4) $

Plot these and draw line.

Now check points:

Point (4.5, 1):

$ 2(4.5) + 3(1) = 9 + 3 = 12 $ → ✔ Yes, lies on graph.

Point (1.5, 3):

$ 2(1.5) + 3(3) = 3 + 9 = 12 $ → ✔ Yes, lies on graph.

✔ 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 $
→ $ 5x - 3x = -7 - 3 $ → $ 2x = -10 $ → $ x = -5 $

Geometrical interpretation: It represents a point on the number line at $ x = -5 $

(ii) In two variables
Write as: $ 5x + 3 = 3x - 7 $
Bring all terms to one side:
$ 5x - 3x + 3 + 7 = 0 $ → $ 2x + 10 = 0 $

But to make it two variables, write as:
$ 5x + 3 = 3x - 7 $ → $ 2x + 10 = 0 $

This is equivalent to $ 2x + 0y + 10 = 0 $

So it represents a straight line parallel to y-axis (vertical line), at $ x = -5 $

But actually, since $ y $ doesn’t appear, it’s a vertical line.

So geometrical interpretation:
- In two variables, the equation represents a vertical line passing through $ x = -5 $, i.e., all points $ (-5, y) $ for any $ y $

✔ Answer:
(i) A point on the number line at $ x = -5 $
(ii) A vertical line $ x = -5 $ in the coordinate plane

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✔ Summary of Correct Answers:

| Q | Answer |
|---|--------|
| 1 | $ m = -\frac{9}{13} $ (not $ -\frac{24}{47} $) |
| 2 | $ a = 1 $ |
| 3 | $ a = 1, b = -\sqrt{3}, c = -4 $ |
| 4 | $ y = -2.5 $ or $ y + \frac{5}{2} = 0 $ |
| 5 | $ y = 8x + 7 $ |
| 6 | $ a = -\frac{1}{4}, b = -\frac{1}{6} $ (not $ \frac{1}{4}, \frac{1}{6} $) |
| 7 | X-axis: $ (\frac{4}{7}, 0) $, Y-axis: $ (0, -\frac{4}{3}) $ |
| 8 | $ x = -\frac{16}{59} $ (not $ x = 4 $) |
| 9 | Lines intersect at origin, are perpendicular |
| 10 | Both points lie on the graph |
| 11 | (i) Point at $ x = -5 $; (ii) Vertical line $ x = -5 $ |

> ⚠️ Note: Several answers in the key seem incorrect. Likely typos in the answer sheet.

Let me know if you'd like diagrams for Q9 or Q10.