Answer
The points of intersection are $(2,1)$ and $(-1,-2)$
Work Step by Step
$x^{2}+y^{2}=5$ $;$ $x-y=1$
From the graph (shown below), it can be seen that the points of intersection are $(-1,-2)$ and $(2,1)$
Check these results analytically. Start by solving $x-y=1$ for $y$:
$x-y=1$
$x-1=y$
$y=x-1$
Substitute $y$ by $x-1$ in $x^{2}+y^{2}=5$ and simplify the equation:
$x^{2}+(x-1)^{2}=5$
$x^{2}+x^{2}-2x+1=5$
$2x^{2}-2x+1-5=0$
$2x^{2}-2x-4=0$
Solve this equation to find the points of intersection. Start by taking out common factor $2$ and taking it to divide the right side:
$2(x^{2}-x-2)=0$
$x^{2}-x-2=\dfrac{0}{2}$
$x^{2}-x-2=0$
Solve by factoring:
$(x-2)(x+1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-2=0$
$x=2$
$x+1=0$
$x=-1$
The $x$-coordinates of the points of intersection are $x=2$ and $x=-1$. Substitute these values into $x-y=1$ and solve for $y$ to obtain the $y$-coordinates of the points:
$x=2$
$2-y=1$
$-y=1-2$
$-y=-1$
$y=1$
$x=-1$
$-1-y=1$
$-y=1+1$
$-y=2$
$y=-2$
It is confirmed that the points of intersection are $(2,1)$ and $(-1,-2)$
