Solve {l}{1=4m_{11}+2sin21}{2=3m_{12}+1min1}

$$\left. \begin{array} { l } { 1 = 4 m _ { 11 } + 2 \sin 21 } \\ { 2 = 3 m _ { 12 } + 1 \min 1 } \end{array} \right.$$

Solve for m_11, m_12

$$m_{11}=0.070816025227349865$$

$$m_{12}=\frac{1}{3}\approx 0.333333333$$

Solution Steps

Evaluate trigonometric functions in the problem

$$\left. \begin{array} { l } { 1 = 4 m_{11} + 2 \cdot 0.35836794954530027 }\\ { 2 = 3 m_{12} + 1 \min 1 } \end{array} \right.$$

Consider the first equation. Multiply 2 and 0.35836794954530027 to get 0.71673589909060054.

$$1=4m_{11}+0.71673589909060054$$

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

$$4m_{11}+0.71673589909060054=1$$

Subtract 0.71673589909060054 from both sides.

$$4m_{11}=1-0.71673589909060054$$

Subtract 0.71673589909060054 from 1 to get 0.28326410090939946.

$$4m_{11}=0.28326410090939946$$

Divide both sides by 4.

$$m_{11}=\frac{0.28326410090939946}{4}$$

Expand \frac{0.28326410090939946}{4} by multiplying both numerator and the denominator by 100000000000000000.

$$m_{11}=\frac{28326410090939946}{400000000000000000}$$

Reduce the fraction \frac{28326410090939946}{400000000000000000} to lowest terms by extracting and canceling out 2.

$$m_{11}=\frac{14163205045469973}{200000000000000000}$$

Consider the second equation. To find the minimum of 1, first put the numbers in order from least to greatest. This set is already in order.

$$1$$

The minimum is 1, the leftmost value in the set ordered from least to greatest.

$$2=3m_{12}+1\times 1$$

Multiply 1 and 1 to get 1.

$$2=3m_{12}+1$$

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

$$3m_{12}+1=2$$

Subtract 1 from both sides.

$$3m_{12}=2-1$$

Subtract 1 from 2 to get 1.

$$3m_{12}=1$$

Divide both sides by 3.

$$m_{12}=\frac{1}{3}$$

The system is now solved.

$$m_{11}=\frac{14163205045469973}{200000000000000000} m_{12}=\frac{1}{3}$$