Algebra II - Solving Rational Expressions

$\frac{y+1}{8y+2} + \frac{3}{y} = \frac{5}{3}$

Solve for .

, $y=-\frac{9}{37}$

, $y=-\frac{7}{15}$

$y=-\frac{5}{2}$

Explanation

The two fractions on the left side of the equation need a common denominator. We can easily do find one by multiplying both the top and bottom of each fraction by the denominator of the other.

$\frac{y+1}{8y+2}\cdot \frac{y}{y}$ becomes $\frac{y^{2}+y}{(y)(8y+2)}$ .

$\frac{3}{y}\cdot \frac{8y+2}{8y+2}$ becomes $\frac{24y+6}{(y)(8y+2)}$ .

Now add the two fractions: $\frac{y^{2}+25y+6}{(y)(8y+2)}$

To solve, multiply both sides of the equation by , yielding

$y^{2}+25y+6 = \frac{(40y^{2}+10y)}{3}$ .

Multiply both sides by 3:

$3y^{2}+75y+18 = 40y^{2}+ 10y$

Move all terms to the same side:

$37y^{2}-65y-18 = 0$

This looks like a complicated equation to factor, but luckily, the only factors of 37 are 37 and 1, so we are left with

Our solutions are therefore

and

$y=-\frac{9}{37}$ .

Solve for :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$x = 1 \frac{2}{5} \textrm{ or } x = 4$

$x = 1 \frac{2}{5}$

$x = 4\textrm{ or }x = 7$

Explanation

Multiply both sides by :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$\frac{3}{x-1} \cdot (x-1) (x-3)+ \frac{4}{x- 3 } \cdot (x-1) (x-3) = 5\cdot (x-1) (x-3)$

$3 \cdot (x-3)+4\cdot (x-1) = 5\cdot (x-1) (x-3)$

$3x-9 + 4x-4 = 5 (x ^{2 } -4x+3)$

$7x-13= 5 x ^{2 } -20x+15$

$7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13$

$5 x ^{2 } -27x + 28 = 0$

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is $5 \cdot28 = 140$ . These integers are :

$5 x ^{2 } -20x -7x + 28 = 0$

$\left (5 x ^{2 } -20x \right ) - \left (7x - 2 8\right ) = 0$

$5x \left (x-4 \right ) - 7 \left (x-4 \right ) = 0$

$\left (5x - 7 \right )\left (x-4 \right ) = 0$

Set each factor equal to 0 and solve separately:

$5x \div 5 = 7 \div 5$

$x = \frac{7}{5} = 1 \frac{2}{5}$

Solve for , given the equation below.

$\frac{x + 2}{x - 1} = \frac{2x + 4}{x}$

No solutions

Explanation

$\frac{x + 2}{x - 1} = \frac{2x + 4}{x}$

Begin by cross-multiplying.

Distribute the on the left side and expand the polynomial on the right.

Combine like terms and rearrange to set the equation equal to zero.

Now we can isolate and solve for by adding to both sides.

Solve for .

$\frac{x}{4}=\frac{3}{4}$

Explanation

To solve for the variable , isolate the variable on one side of the equation with all other constants on the other side. To accomplish this perform the opposite operation to manipulate the equation.

First cross multiply.

$\frac{x}{4}=\frac{3}{4}$

Next, divide by four on both sides.

$\frac{4x}{4}=\frac{12}{4}=\frac{4\cdot 3}{4}=3$

Solve for .

$\frac{x}{5}=\frac{7}{10}$

Explanation

To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

$\frac{x}{5}=\frac{7}{10}$

Cross multiply.

Divide on both sides.

Simplify: $\frac{2}{4+x}-\frac{3}{5x}$

$\frac{7x-12}{5x^2+20}$

$\frac{12x-12}{5x^2+20}$

$\frac{13x-12}{5x^2+20}$

$-\frac{1}{6x+4}$

$-\frac{2x-3}{5x^2+20}$

Explanation

To simplify the expressions, we will need a least common denominator.

Multiply the two denominators together to obtain the least common denominator.

Convert the fractions.

$\frac{2}{4+x}-\frac{3}{5x} = \frac{2(5x)}{20x+5x^2}-\frac{3(4+x)}{20x+5x^2}$

Combine the fractions as one fraction.

$\frac{2(5x)-3(4+x)}{20x+5x^2}$

Simplify the numerator and combine like-terms.

$\frac{10x-12-3x}{20x+5x^2}= \frac{7x-12}{5x^2+20}$

The answer is: $\frac{7x-12}{5x^2+20}$

Solve for .

$\frac{2x}{5}=\frac{18}{5}$

Explanation

To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

$\frac{2x}{5}=\frac{18}{5}$

Cross multiply.

Divide on both sides.

Solve: $\frac{2}{5x}+\frac{1}{6x}+\frac{x}{30}$

$\frac{x^2+17}{30x}$

$\frac{x^3+17}{30x}$

$\frac{x^3+17x}{30}$

$\frac{x+17}{30}$

$\frac{3}{5x}$

Explanation

Convert the fractions to a common denominator.

$\frac{2}{5x}+\frac{1}{6x}+\frac{x}{30} = \frac{2(6)}{5x(6)}+\frac{1(5)}{6x(5)}+\frac{x(x)}{30(x)}$

Simplify the top and bottom and combine like terms on the numerator.

$\frac{12}{30x}+\frac{5}{30x}+\frac{x^2}{30x} = \frac{x^2+17}{30x}$

The answer is: $\frac{x^2+17}{30x}$

Simplify the following expression:

$\large \frac{x^2-5x+6}{x-2}$

$\large x-3$

This expression is already simplified.

$\large x-5$

$\large (x-3)^2$

$\large (x-2)^2$

Explanation

The first step of problems like this is to try to factor the quadratic and see if it shares a factor with the linear polynomial in the denominator. And as it turns out,

$\large x^2-5x+6 = (x-2)(x-3)$

So our rational function is equal to

$\large \frac{(x-2)(x-3)}{x-2} = (x-3)$

which is as simplified as it can get.

Solve for x

$\frac{12}{(x+4)}=\frac{9}{(5-x)}$

Explanation

The correct answer is . Cross multiplying the equation in the question will give . This is simplified to . Combining like terms gives . Finally, isolating gives or .

Solving Rational Expressions

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation