Solve {l}{y=1}{y=-2x+6}{z=5x-1}{a=z}{b=a}{c=b}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { y = 1 }\\ { y = -2 x + 6 }\\ { z = 5 x - 1 }\\ { a = z }\\ { b = a }\\ { c = b }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for y, x, z, a, b, c, d

$$d = \frac{23}{2} = 11\frac{1}{2} = 11.5$$

Solution Steps

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

$$1=-2x+6$$

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

$$-2x+6=1$$

Subtract 6 from both sides.

$$-2x=1-6$$

Subtract 6 from 1 to get -5.

$$-2x=-5$$

Divide both sides by -2.

$$x=\frac{-5}{-2}$$

Fraction \frac{-5}{-2} can be simplified to \frac{5}{2} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{5}{2}$$

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

$$z=5\times \frac{5}{2}-1$$

Multiply 5 and \frac{5}{2} to get \frac{25}{2}.

$$z=\frac{25}{2}-1$$

Subtract 1 from \frac{25}{2} to get \frac{23}{2}.

$$z=\frac{23}{2}$$

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

$$a=\frac{23}{2}$$

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

$$b=\frac{23}{2}$$

Consider the equation (6). Insert the known values of variables into the equation.

$$c=\frac{23}{2}$$

Consider the equation (7). Insert the known values of variables into the equation.

$$d=\frac{23}{2}$$

The system is now solved.

$$y=1 x=\frac{5}{2} z=\frac{23}{2} a=\frac{23}{2} b=\frac{23}{2} c=\frac{23}{2} d=\frac{23}{2}$$