Solve {l}{2x^2+yx+2=0}{x=2}

$$\left. \begin{array} { l } { 2 x ^ { 2 } + y x + 2 = 0 } \\ { x = 2 } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=-5$$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$2\times 2^{2}+y\times 2+2=0$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$2^{3}+y\times 2+2=0$$

Calculate 2 to the power of 3 and get 8.

$$8+y\times 2+2=0$$

Add 8 and 2 to get 10.

$$10+y\times 2=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$y\times 2=-10$$

Divide both sides by 2.

$$y=\frac{-10}{2}$$

Divide -10 by 2 to get -5.

$$y=-5$$

The system is now solved.

$$x=2 y=-5$$