Solve 4+x/2+1geq4(x+1)/3

$$\frac { 4 + x } { 2 } + 1 \geq \frac { 4 ( x + 1 ) } { 3 }$$

Solve for x

$$x\leq 2$$

Solution Steps

Multiply both sides of the equation by 6, the least common multiple of 2,3. Since 6 is positive, the inequality direction remains the same.

$$3\left(4+x\right)+6\geq 2\times 4\left(x+1\right)$$

Use the distributive property to multiply 3 by 4+x.

$$12+3x+6\geq 2\times 4\left(x+1\right)$$

Add 12 and 6 to get 18.

$$18+3x\geq 2\times 4\left(x+1\right)$$

Multiply 2 and 4 to get 8.

$$18+3x\geq 8\left(x+1\right)$$

Use the distributive property to multiply 8 by x+1.

$$18+3x\geq 8x+8$$

Subtract 8x from both sides.

$$18+3x-8x\geq 8$$

Combine 3x and -8x to get -5x.

$$18-5x\geq 8$$

Subtract 18 from both sides.

$$-5x\geq 8-18$$

Subtract 18 from 8 to get -10.

$$-5x\geq -10$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x\leq \frac{-10}{-5}$$

Divide -10 by -5 to get 2.

$$x\leq 2$$