Math worksheet for Grade 8 students covering linear equations and word problems.

It looks like you've uploaded a worksheet on Linear Equations for 8th-grade mathematics from the International Indian School, Riyadh. However, some parts of the questions are incomplete or missing, particularly in problems 6, 9, and 10, where key values (like fractions or ratios) are not visible.

Despite that, I’ll solve all the complete problems step-by-step and point out where information is missing so you can fix them.

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Problem 1: Solve for x

#### i) $ 4x - 7 - (x + 4) = 3x + 4 - (2x - 1) $

Step 1: Expand both sides
Left side:
$ 4x - 7 - x - 4 = 3x - 11 $

Right side:
$ 3x + 4 - 2x + 1 = x + 5 $

Now equation becomes:
$$
3x - 11 = x + 5
$$

Step 2: Subtract $x$ from both sides:
$$
2x - 11 = 5
$$

Step 3: Add 11 to both sides:
$$
2x = 16
$$

Step 4: Divide by 2:
$$
x = 8
$$

✔ Answer: $ x = 8 $

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#### ii) $ \frac{17(2 - x) - 5(x + 12)}{1 - 7x} = 8 $

Step 1: Simplify numerator:
$$
17(2 - x) = 34 - 17x \\
-5(x + 12) = -5x - 60 \\
\text{Numerator: } 34 - 17x - 5x - 60 = -22x - 26
$$

So the equation becomes:
$$
\frac{-22x - 26}{1 - 7x} = 8
$$

Step 2: Multiply both sides by $1 - 7x$ (assuming $1 - 7x \neq 0$):
$$
-22x - 26 = 8(1 - 7x)
$$

Step 3: Expand right side:
$$
-22x - 26 = 8 - 56x
$$

Step 4: Add $56x$ to both sides:
$$
34x - 26 = 8
$$

Step 5: Add 26:
$$
34x = 34
$$

Step 6: Divide:
$$
x = 1
$$

✔ Check if denominator ≠ 0: $1 - 7(1) = -6 \neq 0$, so valid.

✔ Answer: $ x = 1 $

---

Problem 2: Sum of three consecutive even numbers is 30. Find the numbers.

Let the numbers be:
$ x, x+2, x+4 $

Sum:
$$
x + (x+2) + (x+4) = 30 \\
3x + 6 = 30 \\
3x = 24 \\
x = 8
$$

Numbers: $8, 10, 12$

✔ Answer: 8, 10, 12

---

Problem 3: Sum of three consecutive odd numbers is 63. Find the numbers.

Let numbers be: $ x, x+2, x+4 $

Sum:
$$
x + x+2 + x+4 = 63 \\
3x + 6 = 63 \\
3x = 57 \\
x = 19
$$

Numbers: $19, 21, 23$

✔ Answer: 19, 21, 23

---

Problem 4: The sum of two twin primes is 60. Find the two prime numbers.

Twin primes are pairs of primes differing by 2.

We need two primes $p$ and $p+2$ such that:
$$
p + (p+2) = 60 \\
2p + 2 = 60 \\
2p = 58 \\
p = 29
$$

So numbers: $29$ and $31$

Check: both are prime? Yes. Difference = 2 → twin primes.

✔ Answer: 29 and 31

---

Problem 5: Angles of triangle in ratio 1:2:3. Find angles.

Let angles be: $x, 2x, 3x$

Sum of angles in triangle = $180^\circ$:
$$
x + 2x + 3x = 180 \\
6x = 180 \\
x = 30
$$

Angles: $30^\circ, 60^\circ, 90^\circ$

✔ Answer: $30^\circ, 60^\circ, 90^\circ$

---

Problem 6: Numerator is 3 less than denominator. If we add 1 to both, it becomes equal to ___

> ✘ Incomplete: The result after adding 1 is missing (e.g., equals what fraction?)

But let’s assume it says: "it becomes equal to $ \frac{1}{2} $" — this is a common type.

Let denominator = $x$, then numerator = $x - 3$

After adding 1 to both:
$$
\frac{x - 3 + 1}{x + 1} = \frac{x - 2}{x + 1}
$$

Suppose this equals $ \frac{1}{2} $:

$$
\frac{x - 2}{x + 1} = \frac{1}{2}
$$

Cross-multiply:
$$
2(x - 2) = 1(x + 1) \\
2x - 4 = x + 1 \\
x = 5
$$

So numerator = $5 - 3 = 2$, denominator = 5 → Fraction = $ \frac{2}{5} $

✔ Answer (if result is $ \frac{1}{2} $): $ \frac{2}{5} $

⚠️ But since the original problem is missing the value, please confirm the missing part.

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Problem 7: Renu’s mother is four times as old as Renu. After 5 years, mother will be three times as old as Renu. Find their present ages.

Let Renu’s age = $x$, mother’s age = $4x$

After 5 years:
- Renu: $x + 5$
- Mother: $4x + 5$

Given:
$$
4x + 5 = 3(x + 5)
$$

Solve:
$$
4x + 5 = 3x + 15 \\
x = 10
$$

So Renu = 10 years, Mother = $4 \times 10 = 40$ years

✔ Answer: Renu = 10 years, Mother = 40 years

---

Problem 8: Sum of four consecutive multiples of 7 is 70. Find these multiples.

Let the multiples be: $7x, 7(x+1), 7(x+2), 7(x+3)$

Sum:
$$
7x + 7(x+1) + 7(x+2) + 7(x+3) = 70 \\
7[x + (x+1) + (x+2) + (x+3)] = 70 \\
7[4x + 6] = 70 \\
28x + 42 = 70 \\
28x = 28 \\
x = 1
$$

So multiples: $7(1) = 7$, $14$, $21$, $28$

✔ Answer: 7, 14, 21, 28

---

Problem 9: Sum of two numbers is 50. Larger divided by smaller gives ___

> ✘ Incomplete: The quotient is missing (e.g., “we get 3” or “we get $ \frac{3}{2} $”)

Let’s suppose it says: “If the larger number is divided by the smaller, we get 3”

Let smaller number = $x$, larger = $50 - x$

Then:
$$
\frac{50 - x}{x} = 3 \\
50 - x = 3x \\
50 = 4x \\
x = 12.5
$$

Larger = $50 - 12.5 = 37.5$

But these are decimals — possible, but maybe intended to be integers.

Try if ratio is $ \frac{3}{2} $:

$$
\frac{50 - x}{x} = \frac{3}{2} \\
2(50 - x) = 3x \\
100 - 2x = 3x \\
100 = 5x \\
x = 20
$$

Then numbers: 20 and 30

✔ Answer (if ratio is $ \frac{3}{2} $): 20 and 30

⚠️ Please provide the missing value.

---

Problem 10: Perimeter of triangle is 49 cm. One side is 7 cm longer than another and 5 cm shorter than the third. Find the sides.

Let the sides be:
Let middle side = $x$
Then:
- One side = $x - 7$ (7 cm shorter than $x$)
- Third side = $x + 5$ (since one side is 5 cm shorter than third → third is 5 cm longer than that side)

Wait — rephrase carefully:

Let’s define based on description:

Let the shortest side be $a$.
Then one side is $a + 7$ (7 cm longer than $a$)
And that same side is 5 cm shorter than the third side → so third side = $(a + 7) + 5 = a + 12$

So sides:
$a$, $a + 7$, $a + 12$

Perimeter:
$$
a + (a + 7) + (a + 12) = 49 \\
3a + 19 = 49 \\
3a = 30 \\
a = 10
$$

Sides: $10$, $17$, $22$

✔ Check: $10 + 17 + 22 = 49$ ✓
Check differences:
- $17 - 10 = 7$ ✓
- $22 - 17 = 5$ ✓

✔ Answer: Sides are 10 cm, 17 cm, 22 cm

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1i | $ x = 8 $ |
| 1ii | $ x = 1 $ |
| 2 | 8, 10, 12 |
| 3 | 19, 21, 23 |
| 4 | 29 and 31 |
| 5 | $30^\circ, 60^\circ, 90^\circ$ |
| 6 | Missing value — assuming $ \frac{1}{2} $, answer is $ \frac{2}{5} $ |
| 7 | Renu = 10, Mother = 40 |
| 8 | 7, 14, 21, 28 |
| 9 | Missing ratio — assuming $ \frac{3}{2} $, answer is 20 and 30 |
| 10 | 10 cm, 17 cm, 22 cm |

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📌 Please check Problems 6, 9, and possibly others for missing values (likely due to image quality). If you can provide the missing parts (e.g., "becomes equal to $ \frac{1}{2} $" or "we get 3"), I can revise accordingly.

Let me know!