Solve 4/t-3=-t/4t+9

$$\frac { 4 } { t - 3 } = \frac { - t } { 4 t + 9 }$$

Solve for t

$$t=-9$$

$$t=-4$$

Steps Using Factoring

Variable t cannot be equal to any of the values -\frac{9}{4},3 since division by zero is not defined. Multiply both sides of the equation by \left(t-3\right)\left(4t+9\right), the least common multiple of t-3,4t+9.

$$\left(4t+9\right)\times 4=\left(t-3\right)\left(-t\right)$$

Use the distributive property to multiply 4t+9 by 4.

$$16t+36=\left(t-3\right)\left(-t\right)$$

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

$$16t+36=t\left(-t\right)-3\left(-t\right)$$

Multiply -3 and -1 to get 3.

$$16t+36=t\left(-t\right)+3t$$

Subtract t\left(-t\right) from both sides.

$$16t+36-t\left(-t\right)=3t$$

Subtract 3t from both sides.

$$16t+36-t\left(-t\right)-3t=0$$

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

$$16t+36-t^{2}\left(-1\right)-3t=0$$

Multiply -1 and -1 to get 1.

$$16t+36+t^{2}-3t=0$$

Combine 16t and -3t to get 13t.

$$13t+36+t^{2}=0$$

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

$$t^{2}+13t+36=0$$

To solve the equation, factor t^{2}+13t+36 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.

$$a+b=13 ab=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=4 b=9$$

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

$$\left(t+4\right)\left(t+9\right)$$

To find equation solutions, solve t+4=0 and t+9=0.

$$t=-4 t=-9$$

Steps Using Factoring By Grouping

Variable t cannot be equal to any of the values -\frac{9}{4},3 since division by zero is not defined. Multiply both sides of the equation by \left(t-3\right)\left(4t+9\right), the least common multiple of t-3,4t+9.

$$\left(4t+9\right)\times 4=\left(t-3\right)\left(-t\right)$$

Use the distributive property to multiply 4t+9 by 4.

$$16t+36=\left(t-3\right)\left(-t\right)$$

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

$$16t+36=t\left(-t\right)-3\left(-t\right)$$

Multiply -3 and -1 to get 3.

$$16t+36=t\left(-t\right)+3t$$

Subtract t\left(-t\right) from both sides.

$$16t+36-t\left(-t\right)=3t$$

Subtract 3t from both sides.

$$16t+36-t\left(-t\right)-3t=0$$

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

$$16t+36-t^{2}\left(-1\right)-3t=0$$

Multiply -1 and -1 to get 1.

$$16t+36+t^{2}-3t=0$$

Combine 16t and -3t to get 13t.

$$13t+36+t^{2}=0$$

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

$$t^{2}+13t+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt+36. To find a and b, set up a system to be solved.

$$a+b=13 ab=1\times 36=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=4 b=9$$

Rewrite t^{2}+13t+36 as \left(t^{2}+4t\right)+\left(9t+36\right).

$$\left(t^{2}+4t\right)+\left(9t+36\right)$$

Factor out t in the first and 9 in the second group.

$$t\left(t+4\right)+9\left(t+4\right)$$

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

$$\left(t+4\right)\left(t+9\right)$$

To find equation solutions, solve t+4=0 and t+9=0.

$$t=-4 t=-9$$

Steps Using the Quadratic Formula

Variable t cannot be equal to any of the values -\frac{9}{4},3 since division by zero is not defined. Multiply both sides of the equation by \left(t-3\right)\left(4t+9\right), the least common multiple of t-3,4t+9.

$$\left(4t+9\right)\times 4=\left(t-3\right)\left(-t\right)$$

Use the distributive property to multiply 4t+9 by 4.

$$16t+36=\left(t-3\right)\left(-t\right)$$

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

$$16t+36=t\left(-t\right)-3\left(-t\right)$$

Multiply -3 and -1 to get 3.

$$16t+36=t\left(-t\right)+3t$$

Subtract t\left(-t\right) from both sides.

$$16t+36-t\left(-t\right)=3t$$

Subtract 3t from both sides.

$$16t+36-t\left(-t\right)-3t=0$$

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

$$16t+36-t^{2}\left(-1\right)-3t=0$$

Multiply -1 and -1 to get 1.

$$16t+36+t^{2}-3t=0$$

Combine 16t and -3t to get 13t.

$$13t+36+t^{2}=0$$

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.

$$t^{2}+13t+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 13 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-13±\sqrt{13^{2}-4\times 36}}{2}$$

Square 13.

$$t=\frac{-13±\sqrt{169-4\times 36}}{2}$$

Multiply -4 times 36.

$$t=\frac{-13±\sqrt{169-144}}{2}$$

Add 169 to -144.

$$t=\frac{-13±\sqrt{25}}{2}$$

Take the square root of 25.

$$t=\frac{-13±5}{2}$$

Now solve the equation t=\frac{-13±5}{2} when ± is plus. Add -13 to 5.

$$t=-\frac{8}{2}$$

Divide -8 by 2.

$$t=-4$$

Now solve the equation t=\frac{-13±5}{2} when ± is minus. Subtract 5 from -13.

$$t=-\frac{18}{2}$$

Divide -18 by 2.

$$t=-9$$

The equation is now solved.

$$t=-4 t=-9$$

Steps for Completing the Square

Variable t cannot be equal to any of the values -\frac{9}{4},3 since division by zero is not defined. Multiply both sides of the equation by \left(t-3\right)\left(4t+9\right), the least common multiple of t-3,4t+9.

$$\left(4t+9\right)\times 4=\left(t-3\right)\left(-t\right)$$

Use the distributive property to multiply 4t+9 by 4.

$$16t+36=\left(t-3\right)\left(-t\right)$$

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

$$16t+36=t\left(-t\right)-3\left(-t\right)$$

Multiply -3 and -1 to get 3.

$$16t+36=t\left(-t\right)+3t$$

Subtract t\left(-t\right) from both sides.

$$16t+36-t\left(-t\right)=3t$$

Subtract 3t from both sides.

$$16t+36-t\left(-t\right)-3t=0$$

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

$$16t+36-t^{2}\left(-1\right)-3t=0$$

Multiply -1 and -1 to get 1.

$$16t+36+t^{2}-3t=0$$

Combine 16t and -3t to get 13t.

$$13t+36+t^{2}=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$13t+t^{2}=-36$$

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.

$$t^{2}+13t=-36$$

Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+13t+\left(\frac{13}{2}\right)^{2}=-36+\left(\frac{13}{2}\right)^{2}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+13t+\frac{169}{4}=-36+\frac{169}{4}$$

Add -36 to \frac{169}{4}.

$$t^{2}+13t+\frac{169}{4}=\frac{25}{4}$$

Factor t^{2}+13t+\frac{169}{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}.

$$\left(t+\frac{13}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{13}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$t+\frac{13}{2}=\frac{5}{2} t+\frac{13}{2}=-\frac{5}{2}$$

Subtract \frac{13}{2} from both sides of the equation.

$$t=-4 t=-9$$