Given x/y = 2/3, find the value of (3x² − 2xy + 5y²)/(5x² − 3xy + 3y²).

Step-by-step solution

1. 1

   Notice the expression is homogeneous of degree 2. Every term on the top and bottom has total degree $2$ in $x$ and $y$. That means the value depends only on the ratio $x:y$, not on the individual sizes — which is exactly why a single ratio is enough information.

2. 2

   Parametrise the ratio. From $\dfrac xy=\dfrac23$, write

   $x=2k,\qquad y=3k\qquad(k\neq 0)$

   Using $x=2$, $y=3$ directly also works; the parameter $k$ just makes the cancellation visible.

3. 3

   Compute the numerator.

   $3(2k)^{2}-2(2k)(3k)+5(3k)^{2}=12k^{2}-12k^{2}+45k^{2}=45k^{2}$

4. 4

   Compute the denominator.

   $5(2k)^{2}-3(2k)(3k)+3(3k)^{2}=20k^{2}-18k^{2}+27k^{2}=29k^{2}$

5. 5

   Divide and cancel.

   $\frac{45k^{2}}{29k^{2}}=\frac{45}{29}$

   $\boxed{\dfrac{45}{29}}$

6. 6

   Check with a second pair in the same ratio. Take $x=4$, $y=6$ (still $2:3$): numerator $=48-48+180=180$, denominator $=80-72+108=116$, and $\tfrac{180}{116}=\tfrac{45}{29}$ ✓ — confirming the answer really depends only on the ratio.

7. 7

   Alternative route: divide through by y². Writing $r=\tfrac xy=\tfrac23$, the expression equals $\dfrac{3r^{2}-2r+5}{5r^{2}-3r+3}=\dfrac{\tfrac{12}{9}-\tfrac43+5}{\tfrac{20}{9}-2+3}=\dfrac{5}{\tfrac{29}{9}}=\dfrac{45}{29}$ — the same value by a shorter path.