What is the intersection of the lines given by $2y=-x+3$ and $-y=5x+1$?  Enter the answer as an ordered pair.
To find the intersection, we must find the point satisfying both equations. Hence we must solve the system \begin{align*}
2y&=-x+3, \\
-y&=5x+1.
\end{align*}Adding two times the second equation to the first, we get $2y+2(-y)=-x+3+2(5x+1)$, which simplifies to $0=9x+5$. Solving for $x$, we find that $x=-\frac{5}{9}$. Plugging this into the second equation above, we obtain $-y=5\cdot -\frac{5}{9}+1=-\frac{16}{9}$. So the intersection is $\boxed{\left(-\frac{5}{9}, \frac{16}{9}\right)}$.
\le(-\frac{5}{9},\frac{16}{9}\right)
Level 5
Algebra
Problem: What is the intersection of the lines given by $2y=-x+3$ and $-y=5x+1$?  Enter the answer as an ordered pair.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the intersection of the two lines, we first need to rewrite both equations in slope-intercept form (y = mx + b). 

Starting with the first equation, $2y = -x + 3$, we divide both sides by 2 to isolate y:
y = -\frac{1}{2}x + \frac{3}{2}.

Next, we take the second equation, $-y = 5x + 1$, and multiply through by -1 to solve for y:
y = -5x - 1.

Now we have the two equations:
1. y = -\frac{1}{2}x + \frac{3}{2}
2. y = -5x - 1.

To find the intersection, we set the two expressions for y equal to each other:
-\frac{1}{2}x + \frac{3}{2} = -5x - 1.

Next, we will eliminate the fractions by multiplying the entire equation by 2:
-x + 3 = -10x - 2.

Now, we rearrange the equation to isolate x. Adding 10x to both sides gives:
9x + 3 = -2.

Subtracting 3 from both sides results in:
9x = -5.

Dividing both sides by 9 yields:
x = -\frac{5}{9}.

Now that we have the x-coordinate of the intersection, we substitute x back into one of the original equations to find the corresponding y-coordinate. Using the first equation:
y = -\frac{1}{2}(-\frac{5}{9}) + \frac{3}{2}.

Calculating this gives:
y = \frac{5}{18} + \frac{27}{18} = \frac{32}{18} = \frac{16}{9}.

Thus, the intersection point of the two lines is the ordered pair:
\left(-\frac{5}{9}, \frac{16}{9}\right).<|END_OF_LATENT|> The final answer is $\le(-\frac{5}{9},\frac{16}{9}\right)$.