Question 2

Given two real numbers with the following conditions:

Twice the first number minus three times the second number equals $14$ .
If the first number is multiplied by $k$ plus twice the second number, the result is $4$ , where $k$ is a real number.

If the value of the first number is $2$ more than $k$ , the largest possible value of the second number is ....

Let the first number be $A$ and the second number be $B$ .

From the problem, we get the following information:

Twice the first number minus three times the second number equals $14$ :

$2A - 3B = 14$
The first number multiplied by $k$ plus twice the second number equals $4$ :

$Ak + 2B = 4$
The value of the first number is $2$ more than $k$ (meaning $A = k + 2$ or $k = A - 2$ ).

Substitute $k = A - 2$ into equation $(2)$ :

A(A - 2) + 2B = 4

A^2 - 2A + 2B = 4

Rearrange the equation to solve for $B$ :

2B = 4 + 2A - A^2

B = \frac{-A^2 + 2A + 4}{2}

Next, substitute this value of $B$ into equation $(1)$ :

2A - 3\left(\frac{-A^2 + 2A + 4}{2}\right) = 14

Multiply both sides by $2$ to eliminate the denominator:

4A - 3(-A^2 + 2A + 4) = 28

4A + 3A^2 - 6A - 12 = 28

3A^2 - 2A - 12 = 28

3A^2 - 2A - 40 = 0

Factor the quadratic equation:

(3A - 12)(3A + 10) = 0

3(A - 4)(3A + 10) = 0

This gives two possible values for $A$ :

A = 4 \quad \text{or} \quad A = -\frac{10}{3}

We find the value of $B$ for each value of $A$ using equation $2A - 3B = 14 \Rightarrow 3B = 2A - 14 \Rightarrow B = \frac{2A - 14}{3}$ .

Case 1: For $A = 4$

B = \frac{2(4) - 14}{3} = \frac{8 - 14}{3} = \frac{-6}{3} = -2

Case 2: For $A = -\frac{10}{3}$

B = \frac{2(-\frac{10}{3}) - 14}{3} = \frac{-\frac{20}{3} - \frac{42}{3}}{3} = \frac{-\frac{62}{3}}{3} = -\frac{62}{9} \approx -6.88

Compare the two obtained values of $B$ : $-2$ and $-\frac{62}{9}$ .

The value $-2$ is larger than $-\frac{62}{9}$ .

Thus, the largest possible value of the second number is $-2$ .

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