Solve {c}{y=2x^3-3x^2-12x+5}{x=2}

$$\left. \begin{array} { c } { y = 2 x ^ { 3 } - 3 x ^ { 2 } - 12 x + 5 } \\ { x = 2 } \end{array} \right.$$

Solve for y, x

$$x=2$$

$$y=-15$$

Solution Steps

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

$$y=2\times 2^{3}-3\times 2^{2}-12\times 2+5$$

To multiply powers of the same base, add their exponents. Add 1 and 3 to get 4.

$$y=2^{4}-3\times 2^{2}-12\times 2+5$$

Calculate 2 to the power of 4 and get 16.

$$y=16-3\times 2^{2}-12\times 2+5$$

Calculate 2 to the power of 2 and get 4.

$$y=16-3\times 4-12\times 2+5$$

Multiply -3 and 4 to get -12.

$$y=16-12-12\times 2+5$$

Subtract 12 from 16 to get 4.

$$y=4-12\times 2+5$$

Multiply -12 and 2 to get -24.

$$y=4-24+5$$

Subtract 24 from 4 to get -20.

$$y=-20+5$$

Add -20 and 5 to get -15.

$$y=-15$$

The system is now solved.

$$y=-15 x=2$$