Solve a+6/a=35/a-6

$$\frac { a + 6 } { a } = \frac { 35 } { a - 6 }$$

Solve for a

$$a=-1$$

$$a=36$$

Steps Using Factoring

Variable a cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by a\left(a-6\right), the least common multiple of a,a-6.

$$\left(a-6\right)\left(a+6\right)=a\times 35$$

Consider \left(a-6\right)\left(a+6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$a^{2}-36=a\times 35$$

Subtract a\times 35 from both sides.

$$a^{2}-36-a\times 35=0$$

Multiply -1 and 35 to get -35.

$$a^{2}-36-35a=0$$

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

$$a^{2}-35a-36=0$$

To solve the equation, factor a^{2}-35a-36 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.

$$a+b=-35 ab=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -35.

$$a=-36 b=1$$

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

$$\left(a-36\right)\left(a+1\right)$$

To find equation solutions, solve a-36=0 and a+1=0.

$$a=36 a=-1$$

Steps Using Factoring By Grouping

Variable a cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by a\left(a-6\right), the least common multiple of a,a-6.

$$\left(a-6\right)\left(a+6\right)=a\times 35$$

Consider \left(a-6\right)\left(a+6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$a^{2}-36=a\times 35$$

Subtract a\times 35 from both sides.

$$a^{2}-36-a\times 35=0$$

Multiply -1 and 35 to get -35.

$$a^{2}-36-35a=0$$

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

$$a^{2}-35a-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-36. To find a and b, set up a system to be solved.

$$a+b=-35 ab=1\left(-36\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -35.

$$a=-36 b=1$$

Rewrite a^{2}-35a-36 as \left(a^{2}-36a\right)+\left(a-36\right).

$$\left(a^{2}-36a\right)+\left(a-36\right)$$

Factor out a in a^{2}-36a.

$$a\left(a-36\right)+a-36$$

Factor out common term a-36 by using distributive property.

$$\left(a-36\right)\left(a+1\right)$$

To find equation solutions, solve a-36=0 and a+1=0.

$$a=36 a=-1$$

Steps Using the Quadratic Formula

Variable a cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by a\left(a-6\right), the least common multiple of a,a-6.

$$\left(a-6\right)\left(a+6\right)=a\times 35$$

Consider \left(a-6\right)\left(a+6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$a^{2}-36=a\times 35$$

Subtract a\times 35 from both sides.

$$a^{2}-36-a\times 35=0$$

Multiply -1 and 35 to get -35.

$$a^{2}-36-35a=0$$

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.

$$a^{2}-35a-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -35 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\left(-36\right)}}{2}$$

Square -35.

$$a=\frac{-\left(-35\right)±\sqrt{1225-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$a=\frac{-\left(-35\right)±\sqrt{1225+144}}{2}$$

Add 1225 to 144.

$$a=\frac{-\left(-35\right)±\sqrt{1369}}{2}$$

Take the square root of 1369.

$$a=\frac{-\left(-35\right)±37}{2}$$

The opposite of -35 is 35.

$$a=\frac{35±37}{2}$$

Now solve the equation a=\frac{35±37}{2} when ± is plus. Add 35 to 37.

$$a=\frac{72}{2}$$

Divide 72 by 2.

$$a=36$$

Now solve the equation a=\frac{35±37}{2} when ± is minus. Subtract 37 from 35.

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

Divide -2 by 2.

$$a=-1$$

The equation is now solved.

$$a=36 a=-1$$

Steps for Completing the Square

Variable a cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by a\left(a-6\right), the least common multiple of a,a-6.

$$\left(a-6\right)\left(a+6\right)=a\times 35$$

Consider \left(a-6\right)\left(a+6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$a^{2}-36=a\times 35$$

Subtract a\times 35 from both sides.

$$a^{2}-36-a\times 35=0$$

Multiply -1 and 35 to get -35.

$$a^{2}-36-35a=0$$

Add 36 to both sides. Anything plus zero gives itself.

$$a^{2}-35a=36$$

Divide -35, the coefficient of the x term, by 2 to get -\frac{35}{2}. Then add the square of -\frac{35}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-35a+\left(-\frac{35}{2}\right)^{2}=36+\left(-\frac{35}{2}\right)^{2}$$

Square -\frac{35}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-35a+\frac{1225}{4}=36+\frac{1225}{4}$$

Add 36 to \frac{1225}{4}.

$$a^{2}-35a+\frac{1225}{4}=\frac{1369}{4}$$

Factor a^{2}-35a+\frac{1225}{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}.

$$\left(a-\frac{35}{2}\right)^{2}=\frac{1369}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{35}{2}\right)^{2}}=\sqrt{\frac{1369}{4}}$$

Simplify.

$$a-\frac{35}{2}=\frac{37}{2} a-\frac{35}{2}=-\frac{37}{2}$$

Add \frac{35}{2} to both sides of the equation.

$$a=36 a=-1$$