Question

How many solutions does the system of equation have?

$\begin{array}{c}{x}^2+y^2=17\\ 2x^2+y=16\\ 3x-2y=-5\end{array}.$

• A. None
• B. One
• C. Two
• D. Three

Answer 1

The correct option is A None

Let $x^2+y^2=17...\left(1\right)$

$2x^2+y=16...\left(2\right)$

and $3x-2y=-5...\left(3\right)$

Substituting value of $y$ from equation $\left(2\right)$ into equation $\left(3\right)$, we get

$3x-2(16-2x^2)=-5$

$\Rightarrow 3x-32+4x^2=-5$

$\therefore 4x^2+3x-27=0$

$\therefore 4x^2+12x-9x-27=0$

$\therefore (4x-9)(x+3)=0$

$\therefore x=\frac{9}{4},-3$

Substituting $x=-3$ in equations $\left(1\right),\left(2\right),\left(3\right)$, we get

$y^2=8$ or $y=\pm \sqrt{8}$

$y=16-18=-2$

$-2y=4$ or $y=-2$

Since all the values of $y$ aren't same, $x=-3$ cannot be a solution.

Substituting $x=\frac{9}{4}$ in equation $\left(1\right)$, we get

$y^2=17-\frac{81}{16}$ or $y=\pm \sqrt{\frac{191}{16}}$, but from equation $\left(3\right)$, we get

$2\left(\frac{81}{16}\right)+y=16\Rightarrow y=\frac{47}{8}$

Since, the value of $y$ is not same, $x=\frac{9}{4}$ is also not a solution. Hence, the given set of equations have no solution.