Solve 10c+3p+0.5H=100lfloor24

$$10 c + 3 p + 0.5 H = 100 \lfloor 24$$

Solve for H

$$H=4800-20c-6p$$

Steps for Solving Linear Equation

The floor of a real number a is the largest integer number less than or equal to a. The floor of 24 is 24.

$$10c+3p+0.5H=100\times 24$$

Multiply 100 and 24 to get 2400.

$$10c+3p+0.5H=2400$$

Subtract 10c from both sides.

$$3p+0.5H=2400-10c$$

Subtract 3p from both sides.

$$0.5H=2400-10c-3p$$

Multiply both sides by 2.

$$\frac{0.5H}{0.5}=\frac{2400-10c-3p}{0.5}$$

Dividing by 0.5 undoes the multiplication by 0.5.

$$H=\frac{2400-10c-3p}{0.5}$$

Divide 2400-10c-3p by 0.5 by multiplying 2400-10c-3p by the reciprocal of 0.5.

$$H=4800-20c-6p$$

Solve for c

$$c=-\frac{H}{20}-\frac{3p}{10}+240$$

Steps for Solving Linear Equation

The floor of a real number a is the largest integer number less than or equal to a. The floor of 24 is 24.

$$10c+3p+0.5H=100\times 24$$

Multiply 100 and 24 to get 2400.

$$10c+3p+0.5H=2400$$

Subtract 3p from both sides.

$$10c+0.5H=2400-3p$$

Subtract 0.5H from both sides.

$$10c=2400-3p-0.5H$$

The equation is in standard form.

$$10c=-\frac{H}{2}-3p+2400$$

Divide both sides by 10.

$$\frac{10c}{10}=\frac{-\frac{H}{2}-3p+2400}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$c=\frac{-\frac{H}{2}-3p+2400}{10}$$

Divide 2400-3p-\frac{H}{2} by 10.

$$c=-\frac{H}{20}-\frac{3p}{10}+240$$