Solve 3/5y+1/4geq3/7

$$\frac { 3 } { 5 } y + \frac { 1 } { 4 } \geq \frac { 3 } { 7 }$$

Solve for y

$$y\geq \frac{25}{84}$$

Solution Steps

Subtract \frac{1}{4} from both sides.

$$\frac{3}{5}y\geq \frac{3}{7}-\frac{1}{4}$$

Least common multiple of 7 and 4 is 28. Convert \frac{3}{7} and \frac{1}{4} to fractions with denominator 28.

$$\frac{3}{5}y\geq \frac{12}{28}-\frac{7}{28}$$

Since \frac{12}{28} and \frac{7}{28} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{5}y\geq \frac{12-7}{28}$$

Subtract 7 from 12 to get 5.

$$\frac{3}{5}y\geq \frac{5}{28}$$

Multiply both sides by \frac{5}{3}, the reciprocal of \frac{3}{5}. Since \frac{3}{5} is positive, the inequality direction remains the same.

$$y\geq \frac{5}{28}\times \frac{5}{3}$$

Multiply \frac{5}{28} times \frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$y\geq \frac{5\times 5}{28\times 3}$$

Do the multiplications in the fraction \frac{5\times 5}{28\times 3}.

$$y\geq \frac{25}{84}$$