Solve 6*(x-1)+7>3*(5x+3)-8

$$6 \cdot ( x - 1 ) + 7 > 3 \cdot ( 5 x + 3 ) - 8$$

Solve for x

$$x<0$$

Solution Steps

Use the distributive property to multiply 6 by x-1.

$$6x-6+7>3\left(5x+3\right)-8$$

Add -6 and 7 to get 1.

$$6x+1>3\left(5x+3\right)-8$$

Use the distributive property to multiply 3 by 5x+3.

$$6x+1>15x+9-8$$

Subtract 8 from 9 to get 1.

$$6x+1>15x+1$$

Subtract 15x from both sides.

$$6x+1-15x>1$$

Combine 6x and -15x to get -9x.

$$-9x+1>1$$

Subtract 1 from both sides.

$$-9x>1-1$$

Subtract 1 from 1 to get 0.

$$-9x>0$$

Product of two numbers is >0 if both are >0 or both are <0. Since -9<0, x must be <0.

$$x<0$$