If 2x-3y-z=0 and x+3y-14z=0, z≠0, the numerical value of (x^2+3xy)/(y^2+z^2) is:

1964 AMC 12 Problem 19

Attempt Problem 19 of the 1964 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1964 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

2x-3y-z=0

and

x+3y-14z=0,

z\ne0,

the numerical value of

\dfrac{x^2+3xy}{y^2+z^2}

is:

7

2

0

-\dfrac{20}{17}

-2

Solve both linear equations for

x

and

y

in terms of

z

The equations give

y=3z

and

x=5z

Solving the equations gives

y=3z

and

x=5z.

Therefore

\frac{x^2+3xy}{y^2+z^2}=\frac{25z^2+45z^2}{9z^2+z^2}=7.

Thus, the correct answer is A.