Solve 1/x+3/2(x-3)=1/7

$$\frac { 1 } { x } + \frac { 3 } { 2 ( x - 3 ) } = \frac { 1 } { 7 }$$

Solve for x

$$x = \frac{\sqrt{1345} + 41}{4} \approx 19.41856041$$

$$x = \frac{41 - \sqrt{1345}}{4} \approx 1.08143959$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 14x\left(x-3\right), the least common multiple of x,2\left(x-3\right),7.

$$14x-42+7x\times 3=2x\left(x-3\right)$$

Multiply 7 and 3 to get 21.

$$14x-42+21x=2x\left(x-3\right)$$

Combine 14x and 21x to get 35x.

$$35x-42=2x\left(x-3\right)$$

Use the distributive property to multiply 2x by x-3.

$$35x-42=2x^{2}-6x$$

Subtract 2x^{2} from both sides.

$$35x-42-2x^{2}=-6x$$

Add 6x to both sides.

$$35x-42-2x^{2}+6x=0$$

Combine 35x and 6x to get 41x.

$$41x-42-2x^{2}=0$$

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.

$$-2x^{2}+41x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 41 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}-4\left(-2\right)\left(-42\right)}}{2\left(-2\right)}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\left(-2\right)\left(-42\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-41±\sqrt{1681+8\left(-42\right)}}{2\left(-2\right)}$$

Multiply 8 times -42.

$$x=\frac{-41±\sqrt{1681-336}}{2\left(-2\right)}$$

Add 1681 to -336.

$$x=\frac{-41±\sqrt{1345}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-41±\sqrt{1345}}{-4}$$

Now solve the equation x=\frac{-41±\sqrt{1345}}{-4} when ± is plus. Add -41 to \sqrt{1345}.

$$x=\frac{\sqrt{1345}-41}{-4}$$

Divide -41+\sqrt{1345} by -4.

$$x=\frac{41-\sqrt{1345}}{4}$$

Now solve the equation x=\frac{-41±\sqrt{1345}}{-4} when ± is minus. Subtract \sqrt{1345} from -41.

$$x=\frac{-\sqrt{1345}-41}{-4}$$

Divide -41-\sqrt{1345} by -4.

$$x=\frac{\sqrt{1345}+41}{4}$$

The equation is now solved.

$$x=\frac{41-\sqrt{1345}}{4} x=\frac{\sqrt{1345}+41}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by 14x\left(x-3\right), the least common multiple of x,2\left(x-3\right),7.

$$14x-42+7x\times 3=2x\left(x-3\right)$$

Multiply 7 and 3 to get 21.

$$14x-42+21x=2x\left(x-3\right)$$

Combine 14x and 21x to get 35x.

$$35x-42=2x\left(x-3\right)$$

Use the distributive property to multiply 2x by x-3.

$$35x-42=2x^{2}-6x$$

Subtract 2x^{2} from both sides.

$$35x-42-2x^{2}=-6x$$

Add 6x to both sides.

$$35x-42-2x^{2}+6x=0$$

Combine 35x and 6x to get 41x.

$$41x-42-2x^{2}=0$$

Add 42 to both sides. Anything plus zero gives itself.

$$41x-2x^{2}=42$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}+41x=42$$

Divide both sides by -2.

$$\frac{-2x^{2}+41x}{-2}=\frac{42}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{41}{-2}x=\frac{42}{-2}$$

Divide 41 by -2.

$$x^{2}-\frac{41}{2}x=\frac{42}{-2}$$

Divide 42 by -2.

$$x^{2}-\frac{41}{2}x=-21$$

Divide -\frac{41}{2}, the coefficient of the x term, by 2 to get -\frac{41}{4}. Then add the square of -\frac{41}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{41}{2}x+\left(-\frac{41}{4}\right)^{2}=-21+\left(-\frac{41}{4}\right)^{2}$$

Square -\frac{41}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{41}{2}x+\frac{1681}{16}=-21+\frac{1681}{16}$$

Add -21 to \frac{1681}{16}.

$$x^{2}-\frac{41}{2}x+\frac{1681}{16}=\frac{1345}{16}$$

Factor x^{2}-\frac{41}{2}x+\frac{1681}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{41}{4}\right)^{2}=\frac{1345}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{4}\right)^{2}}=\sqrt{\frac{1345}{16}}$$

Simplify.

$$x-\frac{41}{4}=\frac{\sqrt{1345}}{4} x-\frac{41}{4}=-\frac{\sqrt{1345}}{4}$$

Add \frac{41}{4} to both sides of the equation.

$$x=\frac{\sqrt{1345}+41}{4} x=\frac{41-\sqrt{1345}}{4}$$