Solve 4/t-3+2t/t-3=12/t^2-6t+9

$$\frac { 4 } { t - 3 } + \frac { 2 t } { t - 3 } = \frac { 12 } { t ^ { 2 } - 6 t + 9 }$$

Solve for t

$$t=-3$$

$$t=4$$

Steps Using Factoring By Grouping

Variable t cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(t-3\right)^{2}, the least common multiple of t-3,t^{2}-6t+9.

$$\left(t-3\right)\times 4+\left(t-3\right)\times 2t=12$$

Use the distributive property to multiply t-3 by 4.

$$4t-12+\left(t-3\right)\times 2t=12$$

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

$$4t-12+\left(2t-6\right)t=12$$

Use the distributive property to multiply 2t-6 by t.

$$4t-12+2t^{2}-6t=12$$

Combine 4t and -6t to get -2t.

$$-2t-12+2t^{2}=12$$

Subtract 12 from both sides.

$$-2t-12+2t^{2}-12=0$$

Subtract 12 from -12 to get -24.

$$-2t-24+2t^{2}=0$$

Divide both sides by 2.

$$-t-12+t^{2}=0$$

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

$$t^{2}-t-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-12. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-12\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite t^{2}-t-12 as \left(t^{2}-4t\right)+\left(3t-12\right).

$$\left(t^{2}-4t\right)+\left(3t-12\right)$$

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

$$t\left(t-4\right)+3\left(t-4\right)$$

Factor out common term t-4 by using distributive property.

$$\left(t-4\right)\left(t+3\right)$$

To find equation solutions, solve t-4=0 and t+3=0.

$$t=4 t=-3$$

Steps Using the Quadratic Formula

Variable t cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(t-3\right)^{2}, the least common multiple of t-3,t^{2}-6t+9.

$$\left(t-3\right)\times 4+\left(t-3\right)\times 2t=12$$

Use the distributive property to multiply t-3 by 4.

$$4t-12+\left(t-3\right)\times 2t=12$$

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

$$4t-12+\left(2t-6\right)t=12$$

Use the distributive property to multiply 2t-6 by t.

$$4t-12+2t^{2}-6t=12$$

Combine 4t and -6t to get -2t.

$$-2t-12+2t^{2}=12$$

Subtract 12 from both sides.

$$-2t-12+2t^{2}-12=0$$

Subtract 12 from -12 to get -24.

$$-2t-24+2t^{2}=0$$

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.

$$2t^{2}-2t-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

Square -2.

$$t=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-24\right)}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-\left(-2\right)±\sqrt{4-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$t=\frac{-\left(-2\right)±\sqrt{4+192}}{2\times 2}$$

Add 4 to 192.

$$t=\frac{-\left(-2\right)±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$t=\frac{-\left(-2\right)±14}{2\times 2}$$

The opposite of -2 is 2.

$$t=\frac{2±14}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{2±14}{4}$$

Now solve the equation t=\frac{2±14}{4} when ± is plus. Add 2 to 14.

$$t=\frac{16}{4}$$

Divide 16 by 4.

$$t=4$$

Now solve the equation t=\frac{2±14}{4} when ± is minus. Subtract 14 from 2.

$$t=-\frac{12}{4}$$

Divide -12 by 4.

$$t=-3$$

The equation is now solved.

$$t=4 t=-3$$

Steps for Completing the Square

Variable t cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(t-3\right)^{2}, the least common multiple of t-3,t^{2}-6t+9.

$$\left(t-3\right)\times 4+\left(t-3\right)\times 2t=12$$

Use the distributive property to multiply t-3 by 4.

$$4t-12+\left(t-3\right)\times 2t=12$$

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

$$4t-12+\left(2t-6\right)t=12$$

Use the distributive property to multiply 2t-6 by t.

$$4t-12+2t^{2}-6t=12$$

Combine 4t and -6t to get -2t.

$$-2t-12+2t^{2}=12$$

Add 12 to both sides.

$$-2t+2t^{2}=12+12$$

Add 12 and 12 to get 24.

$$-2t+2t^{2}=24$$

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.

$$2t^{2}-2t=24$$

Divide both sides by 2.

$$\frac{2t^{2}-2t}{2}=\frac{24}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$t^{2}+\left(-\frac{2}{2}\right)t=\frac{24}{2}$$

Divide -2 by 2.

$$t^{2}-t=\frac{24}{2}$$

Divide 24 by 2.

$$t^{2}-t=12$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-t+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-t+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$t^{2}-t+\frac{1}{4}=\frac{49}{4}$$

Factor t^{2}-t+\frac{1}{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}.

$$\left(t-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$t-\frac{1}{2}=\frac{7}{2} t-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$t=4 t=-3$$