Solve -f/10f+42=1/f+3

$$\frac { - f } { 10 f + 42 } = \frac { 1 } { f + 3 }$$

Solve for f

$$f=-7$$

$$f=-6$$

Steps Using the Quadratic Formula

Variable f cannot be equal to any of the values -\frac{21}{5},-3 since division by zero is not defined. Multiply both sides of the equation by 2\left(f+3\right)\left(5f+21\right), the least common multiple of 10f+42,f+3.

$$\left(f+3\right)\left(-f\right)=10f+42$$

Use the distributive property to multiply f+3 by -f.

$$f\left(-f\right)+3\left(-f\right)=10f+42$$

Subtract 10f from both sides.

$$f\left(-f\right)+3\left(-f\right)-10f=42$$

Subtract 42 from both sides.

$$f\left(-f\right)+3\left(-f\right)-10f-42=0$$

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

$$f^{2}\left(-1\right)+3\left(-1\right)f-10f-42=0$$

Multiply 3 and -1 to get -3.

$$f^{2}\left(-1\right)-3f-10f-42=0$$

Combine -3f and -10f to get -13f.

$$f^{2}\left(-1\right)-13f-42=0$$

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.

$$-f^{2}-13f-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -13 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$f=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\left(-1\right)\left(-42\right)}}{2\left(-1\right)}$$

Square -13.

$$f=\frac{-\left(-13\right)±\sqrt{169-4\left(-1\right)\left(-42\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$f=\frac{-\left(-13\right)±\sqrt{169+4\left(-42\right)}}{2\left(-1\right)}$$

Multiply 4 times -42.

$$f=\frac{-\left(-13\right)±\sqrt{169-168}}{2\left(-1\right)}$$

Add 169 to -168.

$$f=\frac{-\left(-13\right)±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$f=\frac{-\left(-13\right)±1}{2\left(-1\right)}$$

The opposite of -13 is 13.

$$f=\frac{13±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$f=\frac{13±1}{-2}$$

Now solve the equation f=\frac{13±1}{-2} when ± is plus. Add 13 to 1.

$$f=\frac{14}{-2}$$

Divide 14 by -2.

$$f=-7$$

Now solve the equation f=\frac{13±1}{-2} when ± is minus. Subtract 1 from 13.

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

Divide 12 by -2.

$$f=-6$$

The equation is now solved.

$$f=-7 f=-6$$

Steps for Completing the Square

Variable f cannot be equal to any of the values -\frac{21}{5},-3 since division by zero is not defined. Multiply both sides of the equation by 2\left(f+3\right)\left(5f+21\right), the least common multiple of 10f+42,f+3.

$$\left(f+3\right)\left(-f\right)=10f+42$$

Use the distributive property to multiply f+3 by -f.

$$f\left(-f\right)+3\left(-f\right)=10f+42$$

Subtract 10f from both sides.

$$f\left(-f\right)+3\left(-f\right)-10f=42$$

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

$$f^{2}\left(-1\right)+3\left(-1\right)f-10f=42$$

Multiply 3 and -1 to get -3.

$$f^{2}\left(-1\right)-3f-10f=42$$

Combine -3f and -10f to get -13f.

$$f^{2}\left(-1\right)-13f=42$$

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.

$$-f^{2}-13f=42$$

Divide both sides by -1.

$$\frac{-f^{2}-13f}{-1}=\frac{42}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$f^{2}+\left(-\frac{13}{-1}\right)f=\frac{42}{-1}$$

Divide -13 by -1.

$$f^{2}+13f=\frac{42}{-1}$$

Divide 42 by -1.

$$f^{2}+13f=-42$$

Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$f^{2}+13f+\left(\frac{13}{2}\right)^{2}=-42+\left(\frac{13}{2}\right)^{2}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$f^{2}+13f+\frac{169}{4}=-42+\frac{169}{4}$$

Add -42 to \frac{169}{4}.

$$f^{2}+13f+\frac{169}{4}=\frac{1}{4}$$

Factor f^{2}+13f+\frac{169}{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}.

$$\left(f+\frac{13}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(f+\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$f+\frac{13}{2}=\frac{1}{2} f+\frac{13}{2}=-\frac{1}{2}$$

Subtract \frac{13}{2} from both sides of the equation.

$$f=-6 f=-7$$