Solve {l}{x=-3}{y=|x^2-5x+10|}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { x = -3 }\\ { y = | x ^ {2} - 5 x + 10 | }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b=34$$

Solution Steps

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

$$y=|\left(-3\right)^{2}-5\left(-3\right)+10|$$

Calculate -3 to the power of 2 and get 9.

$$y=|9-5\left(-3\right)+10|$$

Multiply -5 and -3 to get 15.

$$y=|9+15+10|$$

Add 9 and 15 to get 24.

$$y=|24+10|$$

Add 24 and 10 to get 34.

$$y=|34|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 34 is 34.

$$y=34$$

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

$$z=34$$

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

$$a=34$$

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

$$b=34$$

The system is now solved.

$$x=-3 y=34 z=34 a=34 b=34$$