Solve k-8/3k=6/k-8

$$\frac { k - 8 } { 3 k } = \frac { 6 } { k - 8 }$$

Solve for k

$$k=2$$

$$k=32$$

Steps Using Factoring

Variable k cannot be equal to any of the values 0,8 since division by zero is not defined. Multiply both sides of the equation by 3k\left(k-8\right), the least common multiple of 3k,k-8.

$$\left(k-8\right)\left(k-8\right)=3k\times 6$$

Multiply k-8 and k-8 to get \left(k-8\right)^{2}.

$$\left(k-8\right)^{2}=3k\times 6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-8\right)^{2}.

$$k^{2}-16k+64=3k\times 6$$

Multiply 3 and 6 to get 18.

$$k^{2}-16k+64=18k$$

Subtract 18k from both sides.

$$k^{2}-16k+64-18k=0$$

Combine -16k and -18k to get -34k.

$$k^{2}-34k+64=0$$

To solve the equation, factor k^{2}-34k+64 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.

$$a+b=-34 ab=64$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 64.

$$-1,-64 -2,-32 -4,-16 -8,-8$$

Calculate the sum for each pair.

$$-1-64=-65 -2-32=-34 -4-16=-20 -8-8=-16$$

The solution is the pair that gives sum -34.

$$a=-32 b=-2$$

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

$$\left(k-32\right)\left(k-2\right)$$

To find equation solutions, solve k-32=0 and k-2=0.

$$k=32 k=2$$

Steps Using Factoring By Grouping

Variable k cannot be equal to any of the values 0,8 since division by zero is not defined. Multiply both sides of the equation by 3k\left(k-8\right), the least common multiple of 3k,k-8.

$$\left(k-8\right)\left(k-8\right)=3k\times 6$$

Multiply k-8 and k-8 to get \left(k-8\right)^{2}.

$$\left(k-8\right)^{2}=3k\times 6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-8\right)^{2}.

$$k^{2}-16k+64=3k\times 6$$

Multiply 3 and 6 to get 18.

$$k^{2}-16k+64=18k$$

Subtract 18k from both sides.

$$k^{2}-16k+64-18k=0$$

Combine -16k and -18k to get -34k.

$$k^{2}-34k+64=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+64. To find a and b, set up a system to be solved.

$$a+b=-34 ab=1\times 64=64$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 64.

$$-1,-64 -2,-32 -4,-16 -8,-8$$

Calculate the sum for each pair.

$$-1-64=-65 -2-32=-34 -4-16=-20 -8-8=-16$$

The solution is the pair that gives sum -34.

$$a=-32 b=-2$$

Rewrite k^{2}-34k+64 as \left(k^{2}-32k\right)+\left(-2k+64\right).

$$\left(k^{2}-32k\right)+\left(-2k+64\right)$$

Factor out k in the first and -2 in the second group.

$$k\left(k-32\right)-2\left(k-32\right)$$

Factor out common term k-32 by using distributive property.

$$\left(k-32\right)\left(k-2\right)$$

To find equation solutions, solve k-32=0 and k-2=0.

$$k=32 k=2$$

Steps Using the Quadratic Formula

Variable k cannot be equal to any of the values 0,8 since division by zero is not defined. Multiply both sides of the equation by 3k\left(k-8\right), the least common multiple of 3k,k-8.

$$\left(k-8\right)\left(k-8\right)=3k\times 6$$

Multiply k-8 and k-8 to get \left(k-8\right)^{2}.

$$\left(k-8\right)^{2}=3k\times 6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-8\right)^{2}.

$$k^{2}-16k+64=3k\times 6$$

Multiply 3 and 6 to get 18.

$$k^{2}-16k+64=18k$$

Subtract 18k from both sides.

$$k^{2}-16k+64-18k=0$$

Combine -16k and -18k to get -34k.

$$k^{2}-34k+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -34 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-34\right)±\sqrt{\left(-34\right)^{2}-4\times 64}}{2}$$

Square -34.

$$k=\frac{-\left(-34\right)±\sqrt{1156-4\times 64}}{2}$$

Multiply -4 times 64.

$$k=\frac{-\left(-34\right)±\sqrt{1156-256}}{2}$$

Add 1156 to -256.

$$k=\frac{-\left(-34\right)±\sqrt{900}}{2}$$

Take the square root of 900.

$$k=\frac{-\left(-34\right)±30}{2}$$

The opposite of -34 is 34.

$$k=\frac{34±30}{2}$$

Now solve the equation k=\frac{34±30}{2} when ± is plus. Add 34 to 30.

$$k=\frac{64}{2}$$

Divide 64 by 2.

$$k=32$$

Now solve the equation k=\frac{34±30}{2} when ± is minus. Subtract 30 from 34.

$$k=\frac{4}{2}$$

Divide 4 by 2.

$$k=2$$

The equation is now solved.

$$k=32 k=2$$

Steps for Completing the Square

Variable k cannot be equal to any of the values 0,8 since division by zero is not defined. Multiply both sides of the equation by 3k\left(k-8\right), the least common multiple of 3k,k-8.

$$\left(k-8\right)\left(k-8\right)=3k\times 6$$

Multiply k-8 and k-8 to get \left(k-8\right)^{2}.

$$\left(k-8\right)^{2}=3k\times 6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-8\right)^{2}.

$$k^{2}-16k+64=3k\times 6$$

Multiply 3 and 6 to get 18.

$$k^{2}-16k+64=18k$$

Subtract 18k from both sides.

$$k^{2}-16k+64-18k=0$$

Combine -16k and -18k to get -34k.

$$k^{2}-34k+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$k^{2}-34k=-64$$

Divide -34, the coefficient of the x term, by 2 to get -17. Then add the square of -17 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-34k+\left(-17\right)^{2}=-64+\left(-17\right)^{2}$$

Square -17.

$$k^{2}-34k+289=-64+289$$

Add -64 to 289.

$$k^{2}-34k+289=225$$

Factor k^{2}-34k+289. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-17\right)^{2}=225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-17\right)^{2}}=\sqrt{225}$$

Simplify.

$$k-17=15 k-17=-15$$

Add 17 to both sides of the equation.

$$k=32 k=2$$