Solve for x 7/(x+1)=(2x-1)/36

$\frac{7}{x + 1} = \frac{2 x - 1}{36}$

Step 1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$7 \cdot 36 = (x + 1) (2 x - 1)$

Step 2

Solve the equation for $x$ .

Rewrite the equation as $(x + 1) (2 x - 1) = 7 \cdot 36$ .

$(x + 1) (2 x - 1) = 7 \cdot 36$

Simplify $(x + 1) (2 x - 1)$ .

Expand $(x + 1) (2 x - 1)$ using the FOIL Method.

Apply the distributive property.

$x (2 x - 1) + 1 (2 x - 1) = 7 \cdot 36$

Apply the distributive property.

$x (2 x) + x \cdot - 1 + 1 (2 x - 1) = 7 \cdot 36$

Apply the distributive property.

$x (2 x) + x \cdot - 1 + 1 (2 x) + 1 \cdot - 1 = 7 \cdot 36$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot - 1 + 1 (2 x) + 1 \cdot - 1 = 7 \cdot 36$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x) + x \cdot - 1 + 1 (2 x) + 1 \cdot - 1 = 7 \cdot 36$

Multiply $x$ by $x$ .

$2 x^{2} + x \cdot - 1 + 1 (2 x) + 1 \cdot - 1 = 7 \cdot 36$

Move $- 1$ to the left of $x$ .

$2 x^{2} - 1 \cdot x + 1 (2 x) + 1 \cdot - 1 = 7 \cdot 36$

Rewrite $- 1 x$ as $- x$ .

$2 x^{2} - x + 1 (2 x) + 1 \cdot - 1 = 7 \cdot 36$

Multiply $2 x$ by $1$ .

$2 x^{2} - x + 2 x + 1 \cdot - 1 = 7 \cdot 36$

Multiply $- 1$ by $1$ .

$2 x^{2} - x + 2 x - 1 = 7 \cdot 36$

Add $- x$ and $2 x$ .

$2 x^{2} + x - 1 = 7 \cdot 36$

Multiply $7$ by $36$ .

$2 x^{2} + x - 1 = 252$

Subtract $252$ from both sides of the equation.

$2 x^{2} + x - 1 - 252 = 0$

Subtract $252$ from $- 1$ .

$2 x^{2} + x - 253 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 253 = - 506$ and whose sum is $b = 1$ .

Multiply by $1$ .

$2 x^{2} + 1 x - 253 = 0$

Rewrite $1$ as $- 22$ plus $23$

$2 x^{2} + (- 22 + 23) x - 253 = 0$

Apply the distributive property.

$2 x^{2} - 22 x + 23 x - 253 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 22 x) + 23 x - 253 = 0$

Factor out the greatest common factor (GCF) from each group.

$2 x (x - 11) + 23 (x - 11) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 11$ .

$(x - 11) (2 x + 23) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 11 = 0$

$2 x + 23 = 0$

Set $x - 11$ equal to $0$ and solve for $x$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

Set $2 x + 23$ equal to $0$ and solve for $x$ .

Set $2 x + 23$ equal to $0$ .

$2 x + 23 = 0$

Solve $2 x + 23 = 0$ for $x$ .

Subtract $23$ from both sides of the equation.

$2 x = - 23$

Divide each term in $2 x = - 23$ by $2$ and simplify.

Divide each term in $2 x = - 23$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

The final solution is all the values that make $(x - 11) (2 x + 23) = 0$ true.

$x = 11, - \frac{23}{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 11, - \frac{23}{2}$

Decimal Form:

$x = 11, - 11.5$

Mixed Number Form:

$x = 11, - 11 \frac{1}{2}$

Solve for x 7/(x+1)=(2x-1)/36

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