What is the answer to solvefor x,m_{1}(x)=m_{2}(3-x) ?

The answer to solvefor x,m_{1}(x)=m_{2}(3-x) is x=(3m_{2})/(m_{1)+m_{2}};\quad m_{1}\ne-m_{2}

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solvefor x,m_{1}(x)=m_{2}(3-x)

solve for

x, m_{1} (x) = m_{2} (3 - x)

x = \frac{3 m _{2}}{m _{1} + m _{2}}; m_{1} \neq = - m_{2}

Solution steps

m_{1} (x) = m_{2} (3 - x)

Expand

m_{1} (x) : m_{1} x

Expand

m_{2} (3 - x) : 3 m_{2} - m_{2} x

m_{1} x = 3 m_{2} - m_{2} x

Move

m_{2} x

to the left side

m_{1} x + m_{2} x = 3 m_{2}

Factor

m_{1} x + m_{2} x : x (m_{1} + m_{2})

x (m_{1} + m_{2}) = 3 m_{2}

Divide both sides by

m_{1} + m_{2}; m_{1} \neq = - m_{2}

x = \frac{3 m _{2}}{m _{1} + m _{2}}; m_{1} \neq = - m_{2}

3 x - 5 = 2.5 + 3 - (x - 4)

3 (x - 4) = 5 (x + 6)

Q = 20 - 5 P

2 (4 s - 16) - 5 s = - 5

10 + 14 = - 3 (2 x - 8)

What is the answer to solvefor x,m_{1}(x)=m_{2}(3-x) ?
The answer to solvefor x,m_{1}(x)=m_{2}(3-x) is x=(3m_{2})/(m_{1)+m_{2}};\quad m_{1}\ne-m_{2}