Solve 6x+15/x=19

$$6 x + \frac { 15 } { x } = 19$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$6xx+15=19x$$

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

$$6x^{2}+15=19x$$

Subtract 19x from both sides.

$$6x^{2}+15-19x=0$$

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

$$6x^{2}-19x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-19 ab=6\times 15=90$$

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 90.

$$-1,-90 -2,-45 -3,-30 -5,-18 -6,-15 -9,-10$$

Calculate the sum for each pair.

$$-1-90=-91 -2-45=-47 -3-30=-33 -5-18=-23 -6-15=-21 -9-10=-19$$

The solution is the pair that gives sum -19.

$$a=-10 b=-9$$

Rewrite 6x^{2}-19x+15 as \left(6x^{2}-10x\right)+\left(-9x+15\right).

$$\left(6x^{2}-10x\right)+\left(-9x+15\right)$$

Factor out 2x in the first and -3 in the second group.

$$2x\left(3x-5\right)-3\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(2x-3\right)$$

To find equation solutions, solve 3x-5=0 and 2x-3=0.

$$x=\frac{5}{3} x=\frac{3}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$6xx+15=19x$$

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

$$6x^{2}+15=19x$$

Subtract 19x from both sides.

$$6x^{2}+15-19x=0$$

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.

$$6x^{2}-19x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -19 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 6\times 15}}{2\times 6}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 6\times 15}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-19\right)±\sqrt{361-24\times 15}}{2\times 6}$$

Multiply -24 times 15.

$$x=\frac{-\left(-19\right)±\sqrt{361-360}}{2\times 6}$$

Add 361 to -360.

$$x=\frac{-\left(-19\right)±\sqrt{1}}{2\times 6}$$

Take the square root of 1.

$$x=\frac{-\left(-19\right)±1}{2\times 6}$$

The opposite of -19 is 19.

$$x=\frac{19±1}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{19±1}{12}$$

Now solve the equation x=\frac{19±1}{12} when ± is plus. Add 19 to 1.

$$x=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{19±1}{12} when ± is minus. Subtract 1 from 19.

$$x=\frac{18}{12}$$

Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=\frac{5}{3} x=\frac{3}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$6xx+15=19x$$

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

$$6x^{2}+15=19x$$

Subtract 19x from both sides.

$$6x^{2}+15-19x=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$6x^{2}-19x=-15$$

Divide both sides by 6.

$$\frac{6x^{2}-19x}{6}=-\frac{15}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{19}{6}x=-\frac{15}{6}$$

Reduce the fraction \frac{-15}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{19}{6}x=-\frac{5}{2}$$

Divide -\frac{19}{6}, the coefficient of the x term, by 2 to get -\frac{19}{12}. Then add the square of -\frac{19}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{19}{6}x+\left(-\frac{19}{12}\right)^{2}=-\frac{5}{2}+\left(-\frac{19}{12}\right)^{2}$$

Square -\frac{19}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{19}{6}x+\frac{361}{144}=-\frac{5}{2}+\frac{361}{144}$$

Add -\frac{5}{2} to \frac{361}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{19}{6}x+\frac{361}{144}=\frac{1}{144}$$

Factor x^{2}-\frac{19}{6}x+\frac{361}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{19}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$x-\frac{19}{12}=\frac{1}{12} x-\frac{19}{12}=-\frac{1}{12}$$

Add \frac{19}{12} to both sides of the equation.

$$x=\frac{5}{3} x=\frac{3}{2}$$