Solve 2*x/30-x=x+3/25-x

$$2 \cdot \frac { x } { 30 - x } = \frac { x + 3 } { 25 - x }$$

Solve for x

$$x=5$$

$$x=18$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 25,30 since division by zero is not defined. Multiply both sides of the equation by \left(x-30\right)\left(x-25\right), the least common multiple of 30-x,25-x.

$$2\left(25-x\right)x=\left(30-x\right)\left(x+3\right)$$

Use the distributive property to multiply 2 by 25-x.

$$\left(50-2x\right)x=\left(30-x\right)\left(x+3\right)$$

Use the distributive property to multiply 50-2x by x.

$$50x-2x^{2}=\left(30-x\right)\left(x+3\right)$$

Use the distributive property to multiply 30-x by x+3 and combine like terms.

$$50x-2x^{2}=27x+90-x^{2}$$

Subtract 27x from both sides.

$$50x-2x^{2}-27x=90-x^{2}$$

Combine 50x and -27x to get 23x.

$$23x-2x^{2}=90-x^{2}$$

Subtract 90 from both sides.

$$23x-2x^{2}-90=-x^{2}$$

Add x^{2} to both sides.

$$23x-2x^{2}-90+x^{2}=0$$

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

$$23x-x^{2}-90=0$$

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.

$$-x^{2}+23x-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 23 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-23±\sqrt{23^{2}-4\left(-1\right)\left(-90\right)}}{2\left(-1\right)}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\left(-1\right)\left(-90\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-23±\sqrt{529+4\left(-90\right)}}{2\left(-1\right)}$$

Multiply 4 times -90.

$$x=\frac{-23±\sqrt{529-360}}{2\left(-1\right)}$$

Add 529 to -360.

$$x=\frac{-23±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-23±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-23±13}{-2}$$

Now solve the equation x=\frac{-23±13}{-2} when ± is plus. Add -23 to 13.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

Now solve the equation x=\frac{-23±13}{-2} when ± is minus. Subtract 13 from -23.

$$x=-\frac{36}{-2}$$

Divide -36 by -2.

$$x=18$$

The equation is now solved.

$$x=5 x=18$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 25,30 since division by zero is not defined. Multiply both sides of the equation by \left(x-30\right)\left(x-25\right), the least common multiple of 30-x,25-x.

$$2\left(25-x\right)x=\left(30-x\right)\left(x+3\right)$$

Use the distributive property to multiply 2 by 25-x.

$$\left(50-2x\right)x=\left(30-x\right)\left(x+3\right)$$

Use the distributive property to multiply 50-2x by x.

$$50x-2x^{2}=\left(30-x\right)\left(x+3\right)$$

Use the distributive property to multiply 30-x by x+3 and combine like terms.

$$50x-2x^{2}=27x+90-x^{2}$$

Subtract 27x from both sides.

$$50x-2x^{2}-27x=90-x^{2}$$

Combine 50x and -27x to get 23x.

$$23x-2x^{2}=90-x^{2}$$

Add x^{2} to both sides.

$$23x-2x^{2}+x^{2}=90$$

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

$$23x-x^{2}=90$$

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.

$$-x^{2}+23x=90$$

Divide both sides by -1.

$$\frac{-x^{2}+23x}{-1}=\frac{90}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{23}{-1}x=\frac{90}{-1}$$

Divide 23 by -1.

$$x^{2}-23x=\frac{90}{-1}$$

Divide 90 by -1.

$$x^{2}-23x=-90$$

Divide -23, the coefficient of the x term, by 2 to get -\frac{23}{2}. Then add the square of -\frac{23}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-23x+\left(-\frac{23}{2}\right)^{2}=-90+\left(-\frac{23}{2}\right)^{2}$$

Square -\frac{23}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-23x+\frac{529}{4}=-90+\frac{529}{4}$$

Add -90 to \frac{529}{4}.

$$x^{2}-23x+\frac{529}{4}=\frac{169}{4}$$

Factor x^{2}-23x+\frac{529}{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}.

$$\left(x-\frac{23}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{23}{2}=\frac{13}{2} x-\frac{23}{2}=-\frac{13}{2}$$

Add \frac{23}{2} to both sides of the equation.

$$x=18 x=5$$