Re: Which points lie on the graph of
[#permalink]
31 Aug 2025, 10:04
Given the equation:
$$
\(\frac{3 y+x y}{x+3}=x-2\)
$$
Simplify the left-hand side:
$$
\(\frac{3 y+x y}{x+3}=\frac{y(3+x)}{x+3}=y\)
$$
So the equation reduces to:
$$
\(y=x-2\)
$$
Now check which of the given points satisfy $\(y=x-2\)$ :
A. $\((2,0)\)$ :
$$
\(y=2-2=0\)
$$
B. \((-2, 1)\):
$$
\(y=-2-2=-4 \neq 1\)
$$
C. $\((-1,-3)\)$ :
$$
\(y=-1-2=-3\)
$$
D. $\(\left(\frac{4}{3}, \frac{1}{3}\right)\)$ :
$$
\(y=\frac{4}{3}-2=\frac{4}{3}-\frac{6}{3}=-\frac{2}{3} \neq \frac{1}{3}\)
$$
E. $\((1,3)\)$ :
$$
\(y=1-2=-1 \neq 3\)
$$
Final points that lie on the graph:
$\(A(2,0)\)$ and $\(C(-1,-3)\)$