Solve {l}{y=3}{x=5}{y=x^2+cx+d}{c=-10}

$$\left. \begin{array} { l } { y = 3 } \\ { x = 5 } \\ { y = x ^ { 2 } + c x + d } \\ { c = - 10 } \end{array} \right.$$

Solve for y, x, c, d

$$x=5$$

$$y=3$$

$$c=-10$$

$$d=28$$

Solution Steps

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

$$3=5^{2}-10\times 5+d$$

Calculate 5 to the power of 2 and get 25.

$$3=25-10\times 5+d$$

Multiply -10 and 5 to get -50.

$$3=25-50+d$$

Subtract 50 from 25 to get -25.

$$3=-25+d$$

Swap sides so that all variable terms are on the left hand side.

$$-25+d=3$$

Add 25 to both sides.

$$d=3+25$$

Add 3 and 25 to get 28.

$$d=28$$

The system is now solved.

$$y=3 x=5 c=-10 d=28$$