Solve 9c(c-11)+10(5c-3)=3c(c+5)+c(6c-3)-30

$$9 c ( c - 11 ) + 10 ( 5 c - 3 ) = 3 c ( c + 5 ) + c ( 6 c - 3 ) - 30$$

Solve for c

$$c=0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 9c by c-11.

$$9c^{2}-99c+10\left(5c-3\right)=3c\left(c+5\right)+c\left(6c-3\right)-30$$

Use the distributive property to multiply 10 by 5c-3.

$$9c^{2}-99c+50c-30=3c\left(c+5\right)+c\left(6c-3\right)-30$$

Combine -99c and 50c to get -49c.

$$9c^{2}-49c-30=3c\left(c+5\right)+c\left(6c-3\right)-30$$

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

$$9c^{2}-49c-30=3c^{2}+15c+c\left(6c-3\right)-30$$

Use the distributive property to multiply c by 6c-3.

$$9c^{2}-49c-30=3c^{2}+15c+6c^{2}-3c-30$$

Combine 3c^{2} and 6c^{2} to get 9c^{2}.

$$9c^{2}-49c-30=9c^{2}+15c-3c-30$$

Combine 15c and -3c to get 12c.

$$9c^{2}-49c-30=9c^{2}+12c-30$$

Subtract 9c^{2} from both sides.

$$9c^{2}-49c-30-9c^{2}=12c-30$$

Combine 9c^{2} and -9c^{2} to get 0.

$$-49c-30=12c-30$$

Subtract 12c from both sides.

$$-49c-30-12c=-30$$

Combine -49c and -12c to get -61c.

$$-61c-30=-30$$

Add 30 to both sides.

$$-61c=-30+30$$

Add -30 and 30 to get 0.

$$-61c=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since -61 is not equal to 0, c must be equal to 0.

$$c=0$$