Question

Given, $y=x$ and ${(y-2)}^2-4=-x$
The system of equations above intersects at two points. Find the sum of the $x$ and $y$ co-ordinates of the point of intersection of the pair of equations in Quadrant $1$.

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The correct option is C 6

Substitute $x$ for $y$ in the second equation to get $(x-2{)}^2-4=-x$.

Expand the left side of the equation to get $(x-2)(x-2)-4=-x$ or $x^2-4x+4-4=-x$.

Simplify the equation to get $x^2-4x=-x$.

Set the equation to $0$ to get $x^2-3x=0$.

Factor an $x$ out of the equation to get $x(x-3)=0$.

Therefore, either $x=0orx-3=0$ and $x=3$.

According to the question, the point of intersection is in quadrant I, where the $x$ and $y$ values are both positive.

Therefore, $x=3$ and $y=3$.

The sum of $3+3=6$.

The correct answer is $6$.