Solve 0.8left(8y+6left(50-yright)right)leq400-10*11+17

$$0.8 \left( 8y+6 \left( 50-y \right) \right) \leq 400-10 \cdot 11+17$$

Solve for y

$$y\leq 41.875$$

Solution Steps

Use the distributive property to multiply 6 by 50-y.

$$0.8\left(8y+300-6y\right)\leq 400-10\times 11+17$$

Combine 8y and -6y to get 2y.

$$0.8\left(2y+300\right)\leq 400-10\times 11+17$$

Use the distributive property to multiply 0.8 by 2y+300.

$$1.6y+240\leq 400-10\times 11+17$$

Multiply 10 and 11 to get 110.

$$1.6y+240\leq 400-110+17$$

Subtract 110 from 400 to get 290.

$$1.6y+240\leq 290+17$$

Add 290 and 17 to get 307.

$$1.6y+240\leq 307$$

Subtract 240 from both sides.

$$1.6y\leq 307-240$$

Subtract 240 from 307 to get 67.

$$1.6y\leq 67$$

Divide both sides by 1.6. Since 1.6 is positive, the inequality direction remains the same.

$$y\leq \frac{67}{1.6}$$

Expand \frac{67}{1.6} by multiplying both numerator and the denominator by 10.

$$y\leq \frac{670}{16}$$

Reduce the fraction \frac{670}{16} to lowest terms by extracting and canceling out 2.

$$y\leq \frac{335}{8}$$