Calculus 10th Edition

The points of intersection are $(2,1)$ and $(-1,-2)$
$x^{2}+y^{2}=5$ $;$ $x-y=1$ 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 $(-1,-2)$