Solve p-1/p+3-2/p-3=7-3p/p^2-9

$$\frac { p - 1 } { p + 3 } - \frac { 2 } { p - 3 } = \frac { 7 - 3 p } { p ^ { 2 } - 9 }$$

Solve for p

$$p=-2$$

$$p=5$$

Steps Using Factoring

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

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p-3 by p-1 and combine like terms.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p+3 by 2.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

To find the opposite of 2p+6, find the opposite of each term.

$$p^{2}-4p+3-2p-6=7-3p$$

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

$$p^{2}-6p+3-6=7-3p$$

Subtract 6 from 3 to get -3.

$$p^{2}-6p-3=7-3p$$

Subtract 7 from both sides.

$$p^{2}-6p-3-7=-3p$$

Subtract 7 from -3 to get -10.

$$p^{2}-6p-10=-3p$$

Add 3p to both sides.

$$p^{2}-6p-10+3p=0$$

Combine -6p and 3p to get -3p.

$$p^{2}-3p-10=0$$

To solve the equation, factor p^{2}-3p-10 using formula p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right). To find a and b, set up a system to be solved.

$$a+b=-3 ab=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

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

$$\left(p-5\right)\left(p+2\right)$$

To find equation solutions, solve p-5=0 and p+2=0.

$$p=5 p=-2$$

Steps Using Factoring By Grouping

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

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p-3 by p-1 and combine like terms.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p+3 by 2.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

To find the opposite of 2p+6, find the opposite of each term.

$$p^{2}-4p+3-2p-6=7-3p$$

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

$$p^{2}-6p+3-6=7-3p$$

Subtract 6 from 3 to get -3.

$$p^{2}-6p-3=7-3p$$

Subtract 7 from both sides.

$$p^{2}-6p-3-7=-3p$$

Subtract 7 from -3 to get -10.

$$p^{2}-6p-10=-3p$$

Add 3p to both sides.

$$p^{2}-6p-10+3p=0$$

Combine -6p and 3p to get -3p.

$$p^{2}-3p-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-10\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite p^{2}-3p-10 as \left(p^{2}-5p\right)+\left(2p-10\right).

$$\left(p^{2}-5p\right)+\left(2p-10\right)$$

Factor out p in the first and 2 in the second group.

$$p\left(p-5\right)+2\left(p-5\right)$$

Factor out common term p-5 by using distributive property.

$$\left(p-5\right)\left(p+2\right)$$

To find equation solutions, solve p-5=0 and p+2=0.

$$p=5 p=-2$$

Steps Using the Quadratic Formula

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

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p-3 by p-1 and combine like terms.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p+3 by 2.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

To find the opposite of 2p+6, find the opposite of each term.

$$p^{2}-4p+3-2p-6=7-3p$$

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

$$p^{2}-6p+3-6=7-3p$$

Subtract 6 from 3 to get -3.

$$p^{2}-6p-3=7-3p$$

Subtract 7 from both sides.

$$p^{2}-6p-3-7=-3p$$

Subtract 7 from -3 to get -10.

$$p^{2}-6p-10=-3p$$

Add 3p to both sides.

$$p^{2}-6p-10+3p=0$$

Combine -6p and 3p to get -3p.

$$p^{2}-3p-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)}}{2}$$

Square -3.

$$p=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$p=\frac{-\left(-3\right)±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$p=\frac{-\left(-3\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$p=\frac{-\left(-3\right)±7}{2}$$

The opposite of -3 is 3.

$$p=\frac{3±7}{2}$$

Now solve the equation p=\frac{3±7}{2} when ± is plus. Add 3 to 7.

$$p=\frac{10}{2}$$

Divide 10 by 2.

$$p=5$$

Now solve the equation p=\frac{3±7}{2} when ± is minus. Subtract 7 from 3.

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

Divide -4 by 2.

$$p=-2$$

The equation is now solved.

$$p=5 p=-2$$

Steps for Completing the Square

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

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p-3 by p-1 and combine like terms.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

Use the distributive property to multiply p+3 by 2.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

To find the opposite of 2p+6, find the opposite of each term.

$$p^{2}-4p+3-2p-6=7-3p$$

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

$$p^{2}-6p+3-6=7-3p$$

Subtract 6 from 3 to get -3.

$$p^{2}-6p-3=7-3p$$

Add 3p to both sides.

$$p^{2}-6p-3+3p=7$$

Combine -6p and 3p to get -3p.

$$p^{2}-3p-3=7$$

Add 3 to both sides.

$$p^{2}-3p=7+3$$

Add 7 and 3 to get 10.

$$p^{2}-3p=10$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$p^{2}-3p+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$p^{2}-3p+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$p^{2}-3p+\frac{9}{4}=\frac{49}{4}$$

Factor p^{2}-3p+\frac{9}{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}.

$$\left(p-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$p-\frac{3}{2}=\frac{7}{2} p-\frac{3}{2}=-\frac{7}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$p=5 p=-2$$