Solve 2k+5/k=k/k-6

$$\frac { 2 k + 5 } { k } = \frac { k } { k - 6 }$$

Solve for k

$$k=-3$$

$$k=10$$

Steps Using Factoring

Variable k cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by k\left(k-6\right), the least common multiple of k,k-6.

$$\left(k-6\right)\left(2k+5\right)=kk$$

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

$$\left(k-6\right)\left(2k+5\right)=k^{2}$$

Use the distributive property to multiply k-6 by 2k+5 and combine like terms.

$$2k^{2}-7k-30=k^{2}$$

Subtract k^{2} from both sides.

$$2k^{2}-7k-30-k^{2}=0$$

Combine 2k^{2} and -k^{2} to get k^{2}.

$$k^{2}-7k-30=0$$

To solve the equation, factor k^{2}-7k-30 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=-7 ab=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=-10 b=3$$

Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.

$$\left(k-10\right)\left(k+3\right)$$

To find equation solutions, solve k-10=0 and k+3=0.

$$k=10 k=-3$$

Steps Using Factoring By Grouping

Variable k cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by k\left(k-6\right), the least common multiple of k,k-6.

$$\left(k-6\right)\left(2k+5\right)=kk$$

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

$$\left(k-6\right)\left(2k+5\right)=k^{2}$$

Use the distributive property to multiply k-6 by 2k+5 and combine like terms.

$$2k^{2}-7k-30=k^{2}$$

Subtract k^{2} from both sides.

$$2k^{2}-7k-30-k^{2}=0$$

Combine 2k^{2} and -k^{2} to get k^{2}.

$$k^{2}-7k-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk-30. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\left(-30\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=-10 b=3$$

Rewrite k^{2}-7k-30 as \left(k^{2}-10k\right)+\left(3k-30\right).

$$\left(k^{2}-10k\right)+\left(3k-30\right)$$

Factor out k in the first and 3 in the second group.

$$k\left(k-10\right)+3\left(k-10\right)$$

Factor out common term k-10 by using distributive property.

$$\left(k-10\right)\left(k+3\right)$$

To find equation solutions, solve k-10=0 and k+3=0.

$$k=10 k=-3$$

Steps Using the Quadratic Formula

Variable k cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by k\left(k-6\right), the least common multiple of k,k-6.

$$\left(k-6\right)\left(2k+5\right)=kk$$

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

$$\left(k-6\right)\left(2k+5\right)=k^{2}$$

Use the distributive property to multiply k-6 by 2k+5 and combine like terms.

$$2k^{2}-7k-30=k^{2}$$

Subtract k^{2} from both sides.

$$2k^{2}-7k-30-k^{2}=0$$

Combine 2k^{2} and -k^{2} to get k^{2}.

$$k^{2}-7k-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-30\right)}}{2}$$

Square -7.

$$k=\frac{-\left(-7\right)±\sqrt{49-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$k=\frac{-\left(-7\right)±\sqrt{49+120}}{2}$$

Add 49 to 120.

$$k=\frac{-\left(-7\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$k=\frac{-\left(-7\right)±13}{2}$$

The opposite of -7 is 7.

$$k=\frac{7±13}{2}$$

Now solve the equation k=\frac{7±13}{2} when ± is plus. Add 7 to 13.

$$k=\frac{20}{2}$$

Divide 20 by 2.

$$k=10$$

Now solve the equation k=\frac{7±13}{2} when ± is minus. Subtract 13 from 7.

$$k=-\frac{6}{2}$$

Divide -6 by 2.

$$k=-3$$

The equation is now solved.

$$k=10 k=-3$$

Steps for Completing the Square

Variable k cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by k\left(k-6\right), the least common multiple of k,k-6.

$$\left(k-6\right)\left(2k+5\right)=kk$$

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

$$\left(k-6\right)\left(2k+5\right)=k^{2}$$

Use the distributive property to multiply k-6 by 2k+5 and combine like terms.

$$2k^{2}-7k-30=k^{2}$$

Subtract k^{2} from both sides.

$$2k^{2}-7k-30-k^{2}=0$$

Combine 2k^{2} and -k^{2} to get k^{2}.

$$k^{2}-7k-30=0$$

Add 30 to both sides. Anything plus zero gives itself.

$$k^{2}-7k=30$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-7k+\left(-\frac{7}{2}\right)^{2}=30+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-7k+\frac{49}{4}=30+\frac{49}{4}$$

Add 30 to \frac{49}{4}.

$$k^{2}-7k+\frac{49}{4}=\frac{169}{4}$$

Factor k^{2}-7k+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{7}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{7}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$k-\frac{7}{2}=\frac{13}{2} k-\frac{7}{2}=-\frac{13}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$k=10 k=-3$$