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

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

Solve for a

$$a=2$$

Steps Using Factoring

Variable a cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by -a+3.

$$3-a=-a^{2}+4a-3$$

Add a^{2} to both sides.

$$3-a+a^{2}=4a-3$$

Subtract 4a from both sides.

$$3-a+a^{2}-4a=-3$$

Combine -a and -4a to get -5a.

$$3-5a+a^{2}=-3$$

Add 3 to both sides.

$$3-5a+a^{2}+3=0$$

Add 3 and 3 to get 6.

$$6-5a+a^{2}=0$$

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

$$a^{2}-5a+6=0$$

To solve the equation, factor a^{2}-5a+6 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=6$$

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 -5.

$$a=-3 b=-2$$

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

$$\left(a-3\right)\left(a-2\right)$$

To find equation solutions, solve a-3=0 and a-2=0.

$$a=3 a=2$$

Variable a cannot be equal to 3.

$$a=2$$

Steps Using Factoring By Grouping

Variable a cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by -a+3.

$$3-a=-a^{2}+4a-3$$

Add a^{2} to both sides.

$$3-a+a^{2}=4a-3$$

Subtract 4a from both sides.

$$3-a+a^{2}-4a=-3$$

Combine -a and -4a to get -5a.

$$3-5a+a^{2}=-3$$

Add 3 to both sides.

$$3-5a+a^{2}+3=0$$

Add 3 and 3 to get 6.

$$6-5a+a^{2}=0$$

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

$$a^{2}-5a+6=0$$

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.

$$a+b=-5 ab=1\times 6=6$$

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 -5.

$$a=-3 b=-2$$

Rewrite a^{2}-5a+6 as \left(a^{2}-3a\right)+\left(-2a+6\right).

$$\left(a^{2}-3a\right)+\left(-2a+6\right)$$

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

$$a\left(a-3\right)-2\left(a-3\right)$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(a-2\right)$$

To find equation solutions, solve a-3=0 and a-2=0.

$$a=3 a=2$$

Variable a cannot be equal to 3.

$$a=2$$

Steps Using the Quadratic Formula

Variable a cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by -a+3.

$$3-a=-a^{2}+4a-3$$

Add a^{2} to both sides.

$$3-a+a^{2}=4a-3$$

Subtract 4a from both sides.

$$3-a+a^{2}-4a=-3$$

Combine -a and -4a to get -5a.

$$3-5a+a^{2}=-3$$

Add 3 to both sides.

$$3-5a+a^{2}+3=0$$

Add 3 and 3 to get 6.

$$6-5a+a^{2}=0$$

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.

$$a^{2}-5a+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Square -5.

$$a=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$a=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Add 25 to -24.

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

Take the square root of 1.

$$a=\frac{-\left(-5\right)±1}{2}$$

The opposite of -5 is 5.

$$a=\frac{5±1}{2}$$

Now solve the equation a=\frac{5±1}{2} when ± is plus. Add 5 to 1.

$$a=\frac{6}{2}$$

Divide 6 by 2.

$$a=3$$

Now solve the equation a=\frac{5±1}{2} when ± is minus. Subtract 1 from 5.

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

Divide 4 by 2.

$$a=2$$

The equation is now solved.

$$a=3 a=2$$

Variable a cannot be equal to 3.

$$a=2$$

Steps for Completing the Square

Variable a cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by -a+3.

$$3-a=-a^{2}+4a-3$$

Add a^{2} to both sides.

$$3-a+a^{2}=4a-3$$

Subtract 4a from both sides.

$$3-a+a^{2}-4a=-3$$

Combine -a and -4a to get -5a.

$$3-5a+a^{2}=-3$$

Subtract 3 from both sides.

$$-5a+a^{2}=-3-3$$

Subtract 3 from -3 to get -6.

$$-5a+a^{2}=-6$$

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.

$$a^{2}-5a=-6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-5a+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-5a+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$a^{2}-5a+\frac{25}{4}=\frac{1}{4}$$

Factor a^{2}-5a+\frac{25}{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}.

$$\left(a-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$a-\frac{5}{2}=\frac{1}{2} a-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$a=3 a=2$$

Variable a cannot be equal to 3.

$$a=2$$