Solve 2w/w-3=3/w^2-3w+5/w-3

$$\frac { 2 w } { w - 3 } = \frac { 3 } { w ^ { 2 } - 3 w } + \frac { 5 } { w - 3 }$$

Solve for w

$$w=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

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

$$w\times 2w=3+w\times 5$$

Multiply w and w to get w^{2}.

$$w^{2}\times 2=3+w\times 5$$

Subtract 3 from both sides.

$$w^{2}\times 2-3=w\times 5$$

Subtract w\times 5 from both sides.

$$w^{2}\times 2-3-w\times 5=0$$

Multiply -1 and 5 to get -5.

$$w^{2}\times 2-3-5w=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2w^{2}-5w-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2w^{2}+aw+bw-3. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\left(-3\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite 2w^{2}-5w-3 as \left(2w^{2}-6w\right)+\left(w-3\right).

$$\left(2w^{2}-6w\right)+\left(w-3\right)$$

Factor out 2w in 2w^{2}-6w.

$$2w\left(w-3\right)+w-3$$

Factor out common term w-3 by using distributive property.

$$\left(w-3\right)\left(2w+1\right)$$

To find equation solutions, solve w-3=0 and 2w+1=0.

$$w=3 w=-\frac{1}{2}$$

Variable w cannot be equal to 3.

$$w=-\frac{1}{2}$$

Steps Using the Quadratic Formula

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

$$w\times 2w=3+w\times 5$$

Multiply w and w to get w^{2}.

$$w^{2}\times 2=3+w\times 5$$

Subtract 3 from both sides.

$$w^{2}\times 2-3=w\times 5$$

Subtract w\times 5 from both sides.

$$w^{2}\times 2-3-w\times 5=0$$

Multiply -1 and 5 to get -5.

$$w^{2}\times 2-3-5w=0$$

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.

$$2w^{2}-5w-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

Square -5.

$$w=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

Multiply -4 times 2.

$$w=\frac{-\left(-5\right)±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$w=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 2}$$

Add 25 to 24.

$$w=\frac{-\left(-5\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$w=\frac{-\left(-5\right)±7}{2\times 2}$$

The opposite of -5 is 5.

$$w=\frac{5±7}{2\times 2}$$

Multiply 2 times 2.

$$w=\frac{5±7}{4}$$

Now solve the equation w=\frac{5±7}{4} when ± is plus. Add 5 to 7.

$$w=\frac{12}{4}$$

Divide 12 by 4.

$$w=3$$

Now solve the equation w=\frac{5±7}{4} when ± is minus. Subtract 7 from 5.

$$w=-\frac{2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$w=-\frac{1}{2}$$

The equation is now solved.

$$w=3 w=-\frac{1}{2}$$

Variable w cannot be equal to 3.

$$w=-\frac{1}{2}$$

Steps for Completing the Square

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

$$w\times 2w=3+w\times 5$$

Multiply w and w to get w^{2}.

$$w^{2}\times 2=3+w\times 5$$

Subtract w\times 5 from both sides.

$$w^{2}\times 2-w\times 5=3$$

Multiply -1 and 5 to get -5.

$$w^{2}\times 2-5w=3$$

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.

$$2w^{2}-5w=3$$

Divide both sides by 2.

$$\frac{2w^{2}-5w}{2}=\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$w^{2}-\frac{5}{2}w=\frac{3}{2}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$w^{2}-\frac{5}{2}w+\left(-\frac{5}{4}\right)^{2}=\frac{3}{2}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$w^{2}-\frac{5}{2}w+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Add \frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$w^{2}-\frac{5}{2}w+\frac{25}{16}=\frac{49}{16}$$

Factor w^{2}-\frac{5}{2}w+\frac{25}{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}.

$$\left(w-\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w-\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$w-\frac{5}{4}=\frac{7}{4} w-\frac{5}{4}=-\frac{7}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$w=3 w=-\frac{1}{2}$$

Variable w cannot be equal to 3.

$$w=-\frac{1}{2}$$