Solve 5-3a/a^2+4a+3-2a+2/a+3=3-a/a+1

$$\frac { 5 - 3 a } { a ^ { 2 } + 4 a + 3 } - \frac { 2 a + 2 } { a + 3 } = \frac { 3 - a } { a + 1 }$$

Solve for a

$$a=-6$$

Steps Using Factoring By Grouping

Variable a cannot be equal to any of the values -3,-1 since division by zero is not defined. Multiply both sides of the equation by \left(a+1\right)\left(a+3\right), the least common multiple of a^{2}+4a+3,a+3,a+1.

$$5-3a-\left(a+1\right)\left(2a+2\right)=\left(a+3\right)\left(3-a\right)$$

Use the distributive property to multiply a+1 by 2a+2 and combine like terms.

$$5-3a-\left(2a^{2}+4a+2\right)=\left(a+3\right)\left(3-a\right)$$

To find the opposite of 2a^{2}+4a+2, find the opposite of each term.

$$5-3a-2a^{2}-4a-2=\left(a+3\right)\left(3-a\right)$$

Combine -3a and -4a to get -7a.

$$5-7a-2a^{2}-2=\left(a+3\right)\left(3-a\right)$$

Subtract 2 from 5 to get 3.

$$3-7a-2a^{2}=\left(a+3\right)\left(3-a\right)$$

Consider \left(a+3\right)\left(3-a\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3-7a-2a^{2}=9-a^{2}$$

Subtract 9 from both sides.

$$3-7a-2a^{2}-9=-a^{2}$$

Subtract 9 from 3 to get -6.

$$-6-7a-2a^{2}=-a^{2}$$

Add a^{2} to both sides.

$$-6-7a-2a^{2}+a^{2}=0$$

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

$$-6-7a-a^{2}=0$$

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

$$-a^{2}-7a-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-\left(-6\right)=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-1 b=-6$$

Rewrite -a^{2}-7a-6 as \left(-a^{2}-a\right)+\left(-6a-6\right).

$$\left(-a^{2}-a\right)+\left(-6a-6\right)$$

Factor out a in the first and 6 in the second group.

$$a\left(-a-1\right)+6\left(-a-1\right)$$

Factor out common term -a-1 by using distributive property.

$$\left(-a-1\right)\left(a+6\right)$$

To find equation solutions, solve -a-1=0 and a+6=0.

$$a=-1 a=-6$$

Variable a cannot be equal to -1.

$$a=-6$$

Steps Using the Quadratic Formula

Variable a cannot be equal to any of the values -3,-1 since division by zero is not defined. Multiply both sides of the equation by \left(a+1\right)\left(a+3\right), the least common multiple of a^{2}+4a+3,a+3,a+1.

$$5-3a-\left(a+1\right)\left(2a+2\right)=\left(a+3\right)\left(3-a\right)$$

Use the distributive property to multiply a+1 by 2a+2 and combine like terms.

$$5-3a-\left(2a^{2}+4a+2\right)=\left(a+3\right)\left(3-a\right)$$

To find the opposite of 2a^{2}+4a+2, find the opposite of each term.

$$5-3a-2a^{2}-4a-2=\left(a+3\right)\left(3-a\right)$$

Combine -3a and -4a to get -7a.

$$5-7a-2a^{2}-2=\left(a+3\right)\left(3-a\right)$$

Subtract 2 from 5 to get 3.

$$3-7a-2a^{2}=\left(a+3\right)\left(3-a\right)$$

Consider \left(a+3\right)\left(3-a\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3-7a-2a^{2}=9-a^{2}$$

Subtract 9 from both sides.

$$3-7a-2a^{2}-9=-a^{2}$$

Subtract 9 from 3 to get -6.

$$-6-7a-2a^{2}=-a^{2}$$

Add a^{2} to both sides.

$$-6-7a-2a^{2}+a^{2}=0$$

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

$$-6-7a-a^{2}=0$$

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.

$$-a^{2}-7a-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square -7.

$$a=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-\left(-7\right)±\sqrt{49+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$a=\frac{-\left(-7\right)±\sqrt{49-24}}{2\left(-1\right)}$$

Add 49 to -24.

$$a=\frac{-\left(-7\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$a=\frac{-\left(-7\right)±5}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$a=\frac{7±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{7±5}{-2}$$

Now solve the equation a=\frac{7±5}{-2} when ± is plus. Add 7 to 5.

$$a=\frac{12}{-2}$$

Divide 12 by -2.

$$a=-6$$

Now solve the equation a=\frac{7±5}{-2} when ± is minus. Subtract 5 from 7.

$$a=\frac{2}{-2}$$

Divide 2 by -2.

$$a=-1$$

The equation is now solved.

$$a=-6 a=-1$$

Variable a cannot be equal to -1.

$$a=-6$$

Steps for Completing the Square

Variable a cannot be equal to any of the values -3,-1 since division by zero is not defined. Multiply both sides of the equation by \left(a+1\right)\left(a+3\right), the least common multiple of a^{2}+4a+3,a+3,a+1.

$$5-3a-\left(a+1\right)\left(2a+2\right)=\left(a+3\right)\left(3-a\right)$$

Use the distributive property to multiply a+1 by 2a+2 and combine like terms.

$$5-3a-\left(2a^{2}+4a+2\right)=\left(a+3\right)\left(3-a\right)$$

To find the opposite of 2a^{2}+4a+2, find the opposite of each term.

$$5-3a-2a^{2}-4a-2=\left(a+3\right)\left(3-a\right)$$

Combine -3a and -4a to get -7a.

$$5-7a-2a^{2}-2=\left(a+3\right)\left(3-a\right)$$

Subtract 2 from 5 to get 3.

$$3-7a-2a^{2}=\left(a+3\right)\left(3-a\right)$$

Consider \left(a+3\right)\left(3-a\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3-7a-2a^{2}=9-a^{2}$$

Add a^{2} to both sides.

$$3-7a-2a^{2}+a^{2}=9$$

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

$$3-7a-a^{2}=9$$

Subtract 3 from both sides.

$$-7a-a^{2}=9-3$$

Subtract 3 from 9 to get 6.

$$-7a-a^{2}=6$$

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.

$$-a^{2}-7a=6$$

Divide both sides by -1.

$$\frac{-a^{2}-7a}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a^{2}+\left(-\frac{7}{-1}\right)a=\frac{6}{-1}$$

Divide -7 by -1.

$$a^{2}+7a=\frac{6}{-1}$$

Divide 6 by -1.

$$a^{2}+7a=-6$$

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.

$$a^{2}+7a+\left(\frac{7}{2}\right)^{2}=-6+\left(\frac{7}{2}\right)^{2}$$

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

$$a^{2}+7a+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$a^{2}+7a+\frac{49}{4}=\frac{25}{4}$$

Factor a^{2}+7a+\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}.

$$\left(a+\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$a+\frac{7}{2}=\frac{5}{2} a+\frac{7}{2}=-\frac{5}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$a=-1 a=-6$$

Variable a cannot be equal to -1.

$$a=-6$$