Solve 3*2x-3+2sqrt{7-x}=11

$$3 \cdot 2 x - 3 + 2 \sqrt { 7 - x } = 11$$

Solve for x

$$x = \frac{14}{9} = 1\frac{5}{9} \approx 1.555555556$$

Solution Steps

Subtract 3\times 2x-3 from both sides of the equation.

$$2\sqrt{7-x}=11-\left(3\times 2x-3\right)$$

Multiply 3 and 2 to get 6.

$$2\sqrt{7-x}=11-\left(6x-3\right)$$

To find the opposite of 6x-3, find the opposite of each term.

$$2\sqrt{7-x}=11-6x-\left(-3\right)$$

The opposite of -3 is 3.

$$2\sqrt{7-x}=11-6x+3$$

Add 11 and 3 to get 14.

$$2\sqrt{7-x}=14-6x$$

Square both sides of the equation.

$$\left(2\sqrt{7-x}\right)^{2}=\left(14-6x\right)^{2}$$

Expand \left(2\sqrt{7-x}\right)^{2}.

$$2^{2}\left(\sqrt{7-x}\right)^{2}=\left(14-6x\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{7-x}\right)^{2}=\left(14-6x\right)^{2}$$

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

$$4\left(7-x\right)=\left(14-6x\right)^{2}$$

Use the distributive property to multiply 4 by 7-x.

$$28-4x=\left(14-6x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-6x\right)^{2}.

$$28-4x=196-168x+36x^{2}$$

Subtract 196 from both sides.

$$28-4x-196=-168x+36x^{2}$$

Subtract 196 from 28 to get -168.

$$-168-4x=-168x+36x^{2}$$

Add 168x to both sides.

$$-168-4x+168x=36x^{2}$$

Combine -4x and 168x to get 164x.

$$-168+164x=36x^{2}$$

Subtract 36x^{2} from both sides.

$$-168+164x-36x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-36x^{2}+164x-168=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -36 for a, 164 for b, and -168 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-164±\sqrt{164^{2}-4\left(-36\right)\left(-168\right)}}{2\left(-36\right)}$$

Square 164.

$$x=\frac{-164±\sqrt{26896-4\left(-36\right)\left(-168\right)}}{2\left(-36\right)}$$

Multiply -4 times -36.

$$x=\frac{-164±\sqrt{26896+144\left(-168\right)}}{2\left(-36\right)}$$

Multiply 144 times -168.

$$x=\frac{-164±\sqrt{26896-24192}}{2\left(-36\right)}$$

Add 26896 to -24192.

$$x=\frac{-164±\sqrt{2704}}{2\left(-36\right)}$$

Take the square root of 2704.

$$x=\frac{-164±52}{2\left(-36\right)}$$

Multiply 2 times -36.

$$x=\frac{-164±52}{-72}$$

Now solve the equation x=\frac{-164±52}{-72} when ± is plus. Add -164 to 52.

$$x=-\frac{112}{-72}$$

Reduce the fraction \frac{-112}{-72} to lowest terms by extracting and canceling out 8.

$$x=\frac{14}{9}$$

Now solve the equation x=\frac{-164±52}{-72} when ± is minus. Subtract 52 from -164.

$$x=-\frac{216}{-72}$$

Divide -216 by -72.

$$x=3$$

The equation is now solved.

$$x=\frac{14}{9} x=3$$

Substitute \frac{14}{9} for x in the equation 3\times 2x-3+2\sqrt{7-x}=11.

$$3\times 2\times \frac{14}{9}-3+2\sqrt{7-\frac{14}{9}}=11$$

Simplify. The value x=\frac{14}{9} satisfies the equation.

$$11=11$$

Substitute 3 for x in the equation 3\times 2x-3+2\sqrt{7-x}=11.

$$3\times 2\times 3-3+2\sqrt{7-3}=11$$

Simplify. The value x=3 does not satisfy the equation.

$$19=11$$

Equation 2\sqrt{7-x}=14-6x has a unique solution.

$$x=\frac{14}{9}$$