Solve {l}{0.52*4=2.08{(t)}}{u=0.25*0.05+0.4*0.01}{text{Solvefor}vtext{where}}{v=u}

$$\left. \begin{array} { l } { 0.52 \cdot 4 = 2.08 {(t)} }\\ { u = 0.25 \cdot 0.05 + 0.4 \cdot 0.01 }\\ { \text{Solve for } v \text{ where} } \\ { v = u } \end{array} \right.$$

Solve for t, u, v

$$v=0.0165$$

Solution Steps

Consider the first equation. Multiply 0.52 and 4 to get 2.08.

$$2.08=2.08t$$

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

$$2.08t=2.08$$

Divide both sides by 2.08.

$$t=\frac{2.08}{2.08}$$

Divide 2.08 by 2.08 to get 1.

$$t=1$$

Consider the second equation. Multiply 0.25 and 0.05 to get 0.0125.

$$u=0.0125+0.4\times 0.01$$

Multiply 0.4 and 0.01 to get 0.004.

$$u=0.0125+0.004$$

Add 0.0125 and 0.004 to get 0.0165.

$$u=0.0165$$

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

$$v=0.0165$$

The system is now solved.

$$t=1 u=0.0165 v=0.0165$$