Question

If $(a,b)$ satisfies the given system of equations

$\{\begin{array}{c}y=\frac{1}{2}x-2\\ y=-x^2+1\end{array}$, then the value of $b$ is

• A. -3
• B. -2
• C. 1
• D. 2

Answer 1

Answer: (A)

-3

Since $(a,b)$ satisfies the equations $y=\frac{x}{2}-2$ and $y=-x^2+1$, we have

$\frac{a}{2}-2=-a^2+1$

$\therefore a-4=-2a^2+2$

$\therefore 2a^2+a-6=0$

$\therefore 2a^2+4a-3a-6=0$

$\therefore (a+2)(2a-3)=0$

$\therefore a=-2$ or $\frac{3}{2}$

Substituting $a=-2$ in two equations we have $b=-1-2=-3$ and $b=-(-2{)}^2+1=-3$