1964 AHSME Problems/Problem 19

Problem 19

If $2x-3y-z=0$ and $x+3y-14z=0, z \neq 0$ , the numerical value of $\frac{x^2+3xy}{y^2+z^2}$ is:

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ -20/17\qquad \textbf{(E)}\ -2$

Solution 1

If the value of $\frac{x^2+3xy}{y^2+z^2}$ is constant, as the answers imply, we can pick a value of $z$ , and then solve the two linear equations for the corresponding $(x, y)$ . We can then plug in $(x,y, z)$ into the expression to get the answer.

If $z=1$ , then $2x - 3y = 1$ and $x + 3y = 14$ . We can solve each equation for $3y$ and set them equal, which leads to $2x - 1 = 14 - x$ . This leads to $x = 5$ . Plugging in $x=5$ into $x + 3y = 14$ gives $y = 3$ . Thus, $(5, 3, 1)$ is one solution to the intersection of the two planes given.

Plugging $(5, 3, -1)$ into the expression gives $\frac{x^2+3xy}{y^2+z^2}$ gives $\frac{25 + 45}{10}$ , or $7$ , which is answer $\boxed{\textbf{(A)}}$

Solution 2

If we think of $z$ as a parameter, we get $2x - 3y = z$ and $x + 3y = 14z$ . Adding the equations leads to $3x = 15z$ , or $x = 5z$ . Plugging that into $x + 3y = 14z$ gives $5z + 3y = 14z$ , or $y = 3z$ . Thus, the intersection of the two planes is given by the parametric line $(5z, 3z, z)$ , where $z$ varies along all real numbers.

We plug this in to $\frac{x^2+3xy}{y^2+z^2}$ to get $\frac{25z^2 + 45z^2}{10z^2}$ , or $7$ , which is answer $\boxed{\textbf{(A)}}$ .

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1964 AHSME Problems/Problem 19

Contents

Problem 19

Solution 1

Solution 2

See Also