https://file.aiquickdraw.com/mathbot/file/1726151456349b4x8gwlb.jpg

Let's work through each of the questions in the image.

### Question 43:
Regina is 34 years old, and her daughter is 5 years old. In \( n \) years, Regina will be twice as old as her daughter.

Let Regina's age in \( n \) years be \( 34 + n \) and her daughter's age be \( 5 + n \). The condition says that Regina will be twice her daughter's age:
\[
34 + n = 2(5 + n)
\]
Expanding the right-hand side:
\[
34 + n = 10 + 2n
\]
Subtract \( n \) from both sides:
\[
34 = 10 + n
\]
Subtract 10 from both sides:
\[
n = 24
\]
So, \( n = 24 \). The answer is **B. 24**.

---

### Question 44:
In an examination with 120 students, 90 passed Mathematics, 40 passed Science, and 5 failed both subjects. We need to find how many students passed Science only.

Using the inclusion-exclusion principle:
\[
\text{Total students} = 120
\]
\[
\text{Students who passed Mathematics} = 90
\]
\[
\text{Students who passed Science} = 40
\]
\[
\text{Students who failed both} = 5
\]

Let \( x \) be the number of students who passed both Mathematics and Science. The total number of students who passed at least one subject is:
\[
120 - 5 = 115
\]
Now, applying the formula for the union of two sets:
\[
90 + 40 - x = 115
\]
Solving for \( x \):
\[
130 - x = 115 \quad \Rightarrow \quad x = 15
\]
So, 15 students passed both subjects. Therefore, the number of students who passed Science only is:
\[
40 - 15 = 25
\]
The answer is **C. 25**.

---

### Question 45:
We need to find the probability that a randomly selected student passed only one subject.

The total number of students is 120, and we know:
- 25 students passed only Science.
- 75 students passed only Mathematics (from \( 90 - 15 = 75 \)).

So, the number of students who passed only one subject is:
\[
25 + 75 = 100
\]
The probability of selecting a student who passed only one subject is:
\[
\frac{100}{120} = \frac{5}{6}
\]
The answer is **C. \( \frac{5}{6} \)**.

---

### Question 46:
We need to solve the equation \( x(3x + 4) = 4 \).

First, expand the left-hand side:
\[
3x^2 + 4x = 4
\]
Rearrange the equation:
\[
3x^2 + 4x - 4 = 0
\]
Now, use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 3 \), \( b = 4 \), and \( c = -4 \):
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(3)(-4)}}{2(3)} = \frac{-4 \pm \sqrt{16 + 48}}{6} = \frac{-4 \pm \sqrt{64}}{6} = \frac{-4 \pm 8}{6}
\]
So, we have two solutions:
\[
x = \frac{-4 + 8}{6} = \frac{4}{6} = \frac{2}{3}, \quad x = \frac{-4 - 8}{6} = \frac{-12}{6} = -2
\]
The solutions are \( x = \frac{2}{3} \) and \( x = -2 \).

The answer is **D**.

---

Would you like further details or explanations on any of the questions? Here are five related questions you could consider:

1. What is the inclusion-exclusion principle, and how is it used in set theory?
2. How would you solve a similar quadratic equation using the quadratic formula?
3. Can we solve the same system using Venn diagrams in probability?
4. What are other methods to solve word problems in algebra, like question 43?
5. How is probability calculated when dealing with mutually exclusive events?

**Tip**: Always cross-check your answer by plugging values back into the original equations to ensure they satisfy the conditions.

Questions 43 to 46: Age problem, examination results probability, and quadratic equation.

Solve algebraic problems including an age-based word problem, exam results with probability using the inclusion-exclusion principle, and a quadratic equation. This detailed explanation covers key concepts suitable for students in Grades 8-10.

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