# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-6/1

Exercises: Try Out - Set 6: Real exam simulation to sharpen your skills and build confidence. - Problem 1

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## Exercise 1

### Question

export const metadata = {
  title: "Problem 1",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

Given two lines <InlineMath math="y = 2x + 3" /> and <InlineMath math="y = ax + b" /> have an intersection point. Determine the intersection point!

Decide whether statements <InlineMath math="(1)" /> and <InlineMath math="(2)" /> below are sufficient to answer the question.

1. <InlineMath math="a \neq 2" />
2. <InlineMath math="b = 3" />


### Choices

- [ ] Statement $$(1)$$ ALONE is sufficient, but statement $$(2)$$ alone is not sufficient.
- [ ] Statement $$(2)$$ ALONE is sufficient, but statement $$(1)$$ alone is not sufficient.
- [x] BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
- [ ] EACH statement ALONE is sufficient.
- [ ] Statements $$(1)$$ and $$(2)$$ TOGETHER are NOT sufficient.

### Answer & Explanation

export const metadata = {
  title: "Problem 1 Solution",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/23/2025",
};

We are asked to determine the intersection point of two lines:

1. <InlineMath math="y = 2x + 3" />
2. <InlineMath math="y = ax + b" />

To determine a unique intersection point <InlineMath math="(x, y)" />, we must be able to solve this system of equations. Let's equate the two equations:

<MathContainer>
  <BlockMath math="2x + 3 = ax + b" />
  <BlockMath math="2x - ax = b - 3" />
  <BlockMath math="(2 - a)x = b - 3" />
</MathContainer>

From the equation above, the value of <InlineMath math="x" /> can be determined if and only if the coefficient of <InlineMath math="x" /> is not zero, i.e., <InlineMath math="2 - a \neq 0" /> or <InlineMath math="a \neq 2" />. If <InlineMath math="a \neq 2" />, then:

<MathContainer>
  <BlockMath math="x = \frac{b - 3}{2 - a}" />
</MathContainer>

After finding <InlineMath math="x" />, we can find <InlineMath math="y" /> by substituting it into one of the equations. So, to determine the intersection point specifically, we need to know:

1. That <InlineMath math="a \neq 2" /> (so the lines intersect at one point).
2. The value of <InlineMath math="b" /> (to calculate the coordinate values).

#### Analysis of Statement 1

Statement <InlineMath math="(1)" /> provides the information <InlineMath math="a \neq 2" />. This guarantees that the two lines have different gradients (<InlineMath math="m_1 = 2" /> and <InlineMath math="m_2 = a \neq 2" />), so they definitely intersect at one point. However, we do not know the value of <InlineMath math="b" />. Without the value of <InlineMath math="b" />, we cannot determine the coordinates of the intersection point (values of <InlineMath math="x" /> and <InlineMath math="y" /> still depend on <InlineMath math="b" />). Therefore, statement <InlineMath math="(1)" /> alone is **not sufficient**.

#### Analysis of Statement 2

Statement <InlineMath math="(2)" /> provides the information <InlineMath math="b = 3" />. The second equation becomes <InlineMath math="y = ax + 3" />. We do not know the value of <InlineMath math="a" />.

- If <InlineMath math="a = 2" />, then the line becomes <InlineMath math="y = 2x + 3" />, which coincides with the first line (infinitely many intersection points).
- If <InlineMath math="a \neq 2" />, the lines intersect at one point.

Since <InlineMath math="a" /> is unknown, we cannot ensure there is a single intersection point, nor can we determine its coordinates. Therefore, statement <InlineMath math="(2)" /> alone is **not sufficient**.

#### Analysis of Both Statements Together

If we combine both statements:

1. <InlineMath math="a \neq 2" />
2. <InlineMath math="b = 3" />

We substitute into the equation to find <InlineMath math="x" />:

<MathContainer>
  <BlockMath math="x = \frac{3 - 3}{2 - a} = \frac{0}{2 - a} = 0" />
</MathContainer>

(Division by <InlineMath math="2 - a" /> is valid because <InlineMath math="a \neq 2" />). Then we find <InlineMath math="y" />:

<MathContainer>
  <BlockMath math="y = 2(0) + 3 = 3" />
</MathContainer>

We get a unique intersection point <InlineMath math="(0, 3)" />. Since we can determine the intersection point, both statements together are **sufficient**.

Therefore, **BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient**.


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