Question

If $2x + 3y = 15$ and $xy = 6$, find $x^2 + y^2$.

• A. 37
• B. 24
• C. 27
• D. 30

Hint:

Use $x^2 + y^2 = (x + y)^2 - 2xy$ with given $xy$ and linear equation.

Answer 1

The Correct Option is A

Step 1: We know that: \[ 2x + 3y = 15 \quad \text{(1)}, \quad xy = 6 \quad \text{(2)} \] We want $x^2 + y^2$. Recall the identity: \[ x^2 + y^2 = (x + y)^2 - 2xy \] So, if we find $x + y$, we can get $x^2 + y^2$ easily. Step 2: Find $x + y$ and $x - y$ relation.
From (1): \[ 2x + 3y = 15 \] Divide through by $1$ is not useful, so instead we try to get $x + y$. Let us write $x$ from (1): \[ 2x = 15 - 3y \quad \Rightarrow \quad x = \frac{15 - 3y}{2} \] Step 3: Use $xy = 6$ to get $y$.
Substitute $x = \frac{15 - 3y}{2}$ into $xy = 6$: \[ \frac{15 - 3y}{2} \cdot y = 6 \] \[ 15y - 3y^2 = 12 \] \[ -3y^2 + 15y - 12 = 0 \] Multiply by $-1$: \[ 3y^2 - 15y + 12 = 0 \] Divide through by $3$: \[ y^2 - 5y + 4 = 0 \] Factorise: \[ (y - 4)(y - 1) = 0 \] So: \[ y = 4 \quad \text{or} \quad y = 1 \] Step 4: Find $x$ for each case.
If $y = 4$: From (1): $2x + 3(4) = 15$ $2x + 12 = 15 \quad \Rightarrow \quad 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2}$ If $y = 1$: From (1): $2x + 3(1) = 15$ $2x + 3 = 15 \quad \Rightarrow \quad 2x = 12 \quad \Rightarrow \quad x = 6$ Step 5: Find $x^2 + y^2$.
Case 1: $(x, y) = \left(\frac{3}{2}, 4\right)$ \[ x^2 + y^2 = \left(\frac{3}{2}\right)^2 + (4)^2 = \frac{9}{4} + \frac{64}{4} = \frac{73}{4} \] Case 2: $(x, y) = (6, 1)$ \[ x^2 + y^2 = (6)^2 + (1)^2 = 36 + 1 = 37 \] Final Answer: \[ \boxed{\frac{73}{4} \ \text{or} \ 37} \] Both answers are possible depending on which pair $(x, y)$ satisfies the given equations.