Solve sqrt{3x+7}=4-sqrt{3x-1}

$$\sqrt{ 3x+7 } =4- \sqrt{ 3x-1 }$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x+7}\right)^{2}=\left(4-\sqrt{3x-1}\right)^{2}$$

Calculate \sqrt{3x+7} to the power of 2 and get 3x+7.

$$3x+7=\left(4-\sqrt{3x-1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{3x-1}\right)^{2}.

$$3x+7=16-8\sqrt{3x-1}+\left(\sqrt{3x-1}\right)^{2}$$

Calculate \sqrt{3x-1} to the power of 2 and get 3x-1.

$$3x+7=16-8\sqrt{3x-1}+3x-1$$

Subtract 1 from 16 to get 15.

$$3x+7=15-8\sqrt{3x-1}+3x$$

Add 8\sqrt{3x-1} to both sides.

$$3x+7+8\sqrt{3x-1}=15+3x$$

Subtract 3x from both sides.

$$3x+7+8\sqrt{3x-1}-3x=15$$

Combine 3x and -3x to get 0.

$$7+8\sqrt{3x-1}=15$$

Subtract 7 from both sides.

$$8\sqrt{3x-1}=15-7$$

Subtract 7 from 15 to get 8.

$$8\sqrt{3x-1}=8$$

Divide both sides by 8.

$$\sqrt{3x-1}=\frac{8}{8}$$

Divide 8 by 8 to get 1.

$$\sqrt{3x-1}=1$$

Square both sides of the equation.

$$3x-1=1$$

Add 1 to both sides of the equation.

$$3x-1-\left(-1\right)=1-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$3x=1-\left(-1\right)$$

Subtract -1 from 1.

$$3x=2$$

Divide both sides by 3.

$$\frac{3x}{3}=\frac{2}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x=\frac{2}{3}$$

Substitute \frac{2}{3} for x in the equation \sqrt{3x+7}=4-\sqrt{3x-1}.

$$\sqrt{3\times \frac{2}{3}+7}=4-\sqrt{3\times \frac{2}{3}-1}$$

Simplify. The value x=\frac{2}{3} satisfies the equation.

$$3=3$$

Equation \sqrt{3x+7}=-\sqrt{3x-1}+4 has a unique solution.

$$x=\frac{2}{3}$$