# Math Help - stuck on this one - first post

1. ## stuck on this one - first post

just joined so don't let me down

Can't sort this one:

y = 3x
2y(squared) - xy = 15

Tried substituting y = 3x and x = y/3. I only get half of the right answer using x = y/3 as y = -3 and x= -1. Can't seem to get the full right answer.

Any ideas

Becca.

2. Originally Posted by becca999
just joined so don't let me down

Can't sort this one:

y = 3x
2y(squared) - xy = 15

Tried substituting y = 3x and x = y/3. I only get half of the right answer using x = y/3 as y = -3 and x= -1. Can't seem to get the full right answer.

Any ideas

Becca.
$\begin{array}{l}
y = 3x \\
2y^2 - xy = 15 \\
\end{array}
$

$\begin{array}{l}
2(3x)^2 - x(3x) = 15 \\
2(9x^2 ) - 3x^2 = 15 \\
18x^2 - 3x^2 = 15 \\
15x^2 = 15 \\
x^2 = 1 \\
x = 1 \\
\\
y = 3 \\
\end{array}
$

3. Originally Posted by OReilly
$\begin{array}{l}
y = 3x \\
2y^2 - xy = 15 \\
\end{array}
$

$\begin{array}{l}
2(3x)^2 - x(3x) = 15 \\
2(9x^2 ) - 3x^2 = 15 \\
18x^2 - 3x^2 = 15 \\
15x^2 = 15 \\
x^2 = 1 \\
x = 1 \\
\\
y = 3 \\
\end{array}
$
Of course, $x^2 = 1$ also has the solution x = -1, but substitution shows that this x value does not provide a new intersection point.

-Dan

4. Originally Posted by topsquark
Of course, $x^2 = 1$ also has the solution x = -1, but substitution shows that this x value does not provide a new intersection point.

-Dan
Both solutions (1,3) and (-1,-3) are correct and they providing intersection points.

Graph is showing that.

5. Originally Posted by topsquark
Of course, $x^2 = 1$ also has the solution x = -1, but substitution shows that this x value does not provide a new intersection point.
-Dan
Hi,

actually you calculate the intersection between a hyperbola and a straight line. And of course there are two intersection points.
I've attached a diagram to show you where these points are:

6. Pffl on me. The second equation is $2y^2 - xy = 15$. When I did the problem in my head for x = -1 I was using "+xy" Ah well. You win some, you lose some!

-Dan