Solve 5/2x+1geq3/2-2/9x

$$\frac { 5 } { 2 } x + 1 \geq \frac { 3 } { 2 } - \frac { 2 } { 9 } x$$

Solve for x

$$x\geq \frac{9}{49}$$

Solution Steps

Add \frac{2}{9}x to both sides.

$$\frac{5}{2}x+1+\frac{2}{9}x\geq \frac{3}{2}$$

Combine \frac{5}{2}x and \frac{2}{9}x to get \frac{49}{18}x.

$$\frac{49}{18}x+1\geq \frac{3}{2}$$

Subtract 1 from both sides.

$$\frac{49}{18}x\geq \frac{3}{2}-1$$

Convert 1 to fraction \frac{2}{2}.

$$\frac{49}{18}x\geq \frac{3}{2}-\frac{2}{2}$$

Since \frac{3}{2} and \frac{2}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{49}{18}x\geq \frac{3-2}{2}$$

Subtract 2 from 3 to get 1.

$$\frac{49}{18}x\geq \frac{1}{2}$$

Multiply both sides by \frac{18}{49}, the reciprocal of \frac{49}{18}. Since \frac{49}{18} is positive, the inequality direction remains the same.

$$x\geq \frac{1}{2}\times \frac{18}{49}$$

Multiply \frac{1}{2} times \frac{18}{49} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{1\times 18}{2\times 49}$$

Do the multiplications in the fraction \frac{1\times 18}{2\times 49}.

$$x\geq \frac{18}{98}$$

Reduce the fraction \frac{18}{98} to lowest terms by extracting and canceling out 2.

$$x\geq \frac{9}{49}$$