# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-6/7

Exercises: Try Out - Set 6: Real exam simulation to sharpen your skills and build confidence. - Question 7

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## Exercise 7

### Question

export const metadata = {
  title: "Question 7",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

The value of <InlineMath math="x" /> that satisfies <InlineMath math="\begin{vmatrix} 3x & x \\ 2 & x \end{vmatrix} = \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}" /> is ....


### Choices

- [ ] $$5 \text{ and } -3$$
- [ ] $$-5 \text{ and } 3$$
- [x] $$\frac{5}{3} \text{ and } -1$$
- [ ] $$-\frac{5}{3} \text{ and } 1$$
- [ ] $$5 \text{ and } -1$$

### Answer & Explanation

export const metadata = {
  title: "Solution for Question 7",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

We are given the following determinant equation:

<MathContainer>
  <BlockMath math="\begin{vmatrix} 3x & x \\ 2 & x \end{vmatrix} = \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}" />
</MathContainer>

The first step is to calculate the determinant of the matrices on the left and right sides. Recall that the determinant of a matrix <InlineMath math="\begin{pmatrix} a & b \\ c & d \end{pmatrix}" /> is <InlineMath math="ad - bc" />.

Determinant of the left side:

<MathContainer>
  <BlockMath math="\begin{vmatrix} 3x & x \\ 2 & x \end{vmatrix} = (3x)(x) - (x)(2) = 3x^2 - 2x" />
</MathContainer>

Determinant of the right side:

<MathContainer>
  <BlockMath math="\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = (3)(2) - (1)(1) = 6 - 1 = 5" />
</MathContainer>

Now we equate both results:

<MathContainer>
  <BlockMath math="3x^2 - 2x = 5" />
</MathContainer>

Move <InlineMath math="5" /> to the left side to form a quadratic equation:

<MathContainer>
  <BlockMath math="3x^2 - 2x - 5 = 0" />
</MathContainer>

We factor this quadratic equation. We look for two numbers that multiply to <InlineMath math="3 \times (-5) = -15" /> and sum to <InlineMath math="-2" />. These numbers are <InlineMath math="-5" /> and <InlineMath math="3" />.

<MathContainer>
  <BlockMath math="3x^2 - 5x + 3x - 5 = 0" />
  <BlockMath math="x(3x - 5) + 1(3x - 5) = 0" />
  <BlockMath math="(x + 1)(3x - 5) = 0" />
</MathContainer>

Thus, the solutions are:

<MathContainer>
  <BlockMath math="x + 1 = 0 \Rightarrow x = -1" />
  <BlockMath math="3x - 5 = 0 \Rightarrow x = \frac{5}{3}" />
</MathContainer>

So, the values of <InlineMath math="x" /> that satisfy the equation are <InlineMath math="\frac{5}{3}" /> and <InlineMath math="-1" />.


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