Solve 3x^2-2/3x^2+2=5x-3/5(x-1)

$$\frac { 3 x ^ { 2 } - 2 } { 3 x ^ { 2 } + 2 } = \frac { 5 x - 3 } { 5 ( x - 1 ) }$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=-4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 5\left(x-1\right)\left(3x^{2}+2\right), the least common multiple of 3x^{2}+2,5\left(x-1\right).

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

Use the distributive property to multiply 5x-5 by 3x^{2}-2.

$$15x^{3}-10x-15x^{2}+10=\left(3x^{2}+2\right)\left(5x-3\right)$$

Use the distributive property to multiply 3x^{2}+2 by 5x-3.

$$15x^{3}-10x-15x^{2}+10=15x^{3}-9x^{2}+10x-6$$

Subtract 15x^{3} from both sides.

$$15x^{3}-10x-15x^{2}+10-15x^{3}=-9x^{2}+10x-6$$

Combine 15x^{3} and -15x^{3} to get 0.

$$-10x-15x^{2}+10=-9x^{2}+10x-6$$

Add 9x^{2} to both sides.

$$-10x-15x^{2}+10+9x^{2}=10x-6$$

Combine -15x^{2} and 9x^{2} to get -6x^{2}.

$$-10x-6x^{2}+10=10x-6$$

Subtract 10x from both sides.

$$-10x-6x^{2}+10-10x=-6$$

Combine -10x and -10x to get -20x.

$$-20x-6x^{2}+10=-6$$

Add 6 to both sides.

$$-20x-6x^{2}+10+6=0$$

Add 10 and 6 to get 16.

$$-20x-6x^{2}+16=0$$

Divide both sides by 2.

$$-10x-3x^{2}+8=0$$

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

$$-3x^{2}-10x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-3\times 8=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=2 b=-12$$

Rewrite -3x^{2}-10x+8 as \left(-3x^{2}+2x\right)+\left(-12x+8\right).

$$\left(-3x^{2}+2x\right)+\left(-12x+8\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(3x-2\right)-4\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(-x-4\right)$$

To find equation solutions, solve 3x-2=0 and -x-4=0.

$$x=\frac{2}{3} x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 5\left(x-1\right)\left(3x^{2}+2\right), the least common multiple of 3x^{2}+2,5\left(x-1\right).

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

Use the distributive property to multiply 5x-5 by 3x^{2}-2.

$$15x^{3}-10x-15x^{2}+10=\left(3x^{2}+2\right)\left(5x-3\right)$$

Use the distributive property to multiply 3x^{2}+2 by 5x-3.

$$15x^{3}-10x-15x^{2}+10=15x^{3}-9x^{2}+10x-6$$

Subtract 15x^{3} from both sides.

$$15x^{3}-10x-15x^{2}+10-15x^{3}=-9x^{2}+10x-6$$

Combine 15x^{3} and -15x^{3} to get 0.

$$-10x-15x^{2}+10=-9x^{2}+10x-6$$

Add 9x^{2} to both sides.

$$-10x-15x^{2}+10+9x^{2}=10x-6$$

Combine -15x^{2} and 9x^{2} to get -6x^{2}.

$$-10x-6x^{2}+10=10x-6$$

Subtract 10x from both sides.

$$-10x-6x^{2}+10-10x=-6$$

Combine -10x and -10x to get -20x.

$$-20x-6x^{2}+10=-6$$

Add 6 to both sides.

$$-20x-6x^{2}+10+6=0$$

Add 10 and 6 to get 16.

$$-20x-6x^{2}+16=0$$

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.

$$-6x^{2}-20x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -20 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-6\right)\times 16}}{2\left(-6\right)}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-6\right)\times 16}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-20\right)±\sqrt{400+24\times 16}}{2\left(-6\right)}$$

Multiply 24 times 16.

$$x=\frac{-\left(-20\right)±\sqrt{400+384}}{2\left(-6\right)}$$

Add 400 to 384.

$$x=\frac{-\left(-20\right)±\sqrt{784}}{2\left(-6\right)}$$

Take the square root of 784.

$$x=\frac{-\left(-20\right)±28}{2\left(-6\right)}$$

The opposite of -20 is 20.

$$x=\frac{20±28}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{20±28}{-12}$$

Now solve the equation x=\frac{20±28}{-12} when ± is plus. Add 20 to 28.

$$x=\frac{48}{-12}$$

Divide 48 by -12.

$$x=-4$$

Now solve the equation x=\frac{20±28}{-12} when ± is minus. Subtract 28 from 20.

$$x=-\frac{8}{-12}$$

Reduce the fraction \frac{-8}{-12} to lowest terms by extracting and canceling out 4.

$$x=\frac{2}{3}$$

The equation is now solved.

$$x=-4 x=\frac{2}{3}$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 5\left(x-1\right)\left(3x^{2}+2\right), the least common multiple of 3x^{2}+2,5\left(x-1\right).

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

Use the distributive property to multiply 5x-5 by 3x^{2}-2.

$$15x^{3}-10x-15x^{2}+10=\left(3x^{2}+2\right)\left(5x-3\right)$$

Use the distributive property to multiply 3x^{2}+2 by 5x-3.

$$15x^{3}-10x-15x^{2}+10=15x^{3}-9x^{2}+10x-6$$

Subtract 15x^{3} from both sides.

$$15x^{3}-10x-15x^{2}+10-15x^{3}=-9x^{2}+10x-6$$

Combine 15x^{3} and -15x^{3} to get 0.

$$-10x-15x^{2}+10=-9x^{2}+10x-6$$

Add 9x^{2} to both sides.

$$-10x-15x^{2}+10+9x^{2}=10x-6$$

Combine -15x^{2} and 9x^{2} to get -6x^{2}.

$$-10x-6x^{2}+10=10x-6$$

Subtract 10x from both sides.

$$-10x-6x^{2}+10-10x=-6$$

Combine -10x and -10x to get -20x.

$$-20x-6x^{2}+10=-6$$

Subtract 10 from both sides.

$$-20x-6x^{2}=-6-10$$

Subtract 10 from -6 to get -16.

$$-20x-6x^{2}=-16$$

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.

$$-6x^{2}-20x=-16$$

Divide both sides by -6.

$$\frac{-6x^{2}-20x}{-6}=-\frac{16}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\left(-\frac{20}{-6}\right)x=-\frac{16}{-6}$$

Reduce the fraction \frac{-20}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{10}{3}x=-\frac{16}{-6}$$

Reduce the fraction \frac{-16}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{10}{3}x=\frac{8}{3}$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{8}{3}+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{8}{3}+\frac{25}{9}$$

Add \frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{49}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{7}{3} x+\frac{5}{3}=-\frac{7}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=\frac{2}{3} x=-4$$