Solve the following linear equations
Ejercicio 1
\[- (-3x + 6) = - ( -4x - 2 ) - 13\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 2
\[(2x + 5) = - ( -4x - 1 ) + 16\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 3
\[- (4x + 4) = ( 4x + 4 ) + 32\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 4
\[2 ( x - 4 ) = -3 ( x - 1 ) - 36\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 5
\[-3 ( x - 2 ) = 5 ( 5x + 4 ) + 14\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 6
\[1 ( 3x - 4 ) = -4 ( -x + 6 ) + 25\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 7
\[\frac{4}{3} ( \frac{-1}{3}x + \frac{1}{2} ) = 2 ( -x + 2 ) + \frac{4}{3}\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 8
\[\frac{5}{3} ( \frac{1}{2}x + 3 ) = \frac{-1}{2} ( x + 1 ) - \frac{7}{6}\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
Ejercicio 9
\[-1 ( \frac{-5}{4}x + 2 ) = 1 ( \frac{-3}{2}x + 2 ) + \frac{1}{8}\]
\[x = \mathbf{?}\]
\[x = \mathbf{?}\]
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