How to find the solution of a rational equation with a binomial denominator - Algebra

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Question

Which of the following fractions is NOT equivalent to $- \frac{x-5}{2x + 3}$ ?

Answer

We know that $-\frac{a}{b}$ is equivalent to $\frac{-a}{b}$ or $\frac{a}{-b}$ .

By this property, there is no way to get $\frac{x+5}{2x+3}$ from $-\frac{x-5}{2x+3}$ .

Therefore the correct answer is $\frac{x+5}{2x+3}$ .

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Question

Find the values of which will make the given rational expression undefined:

$\frac{x+5}{\left ( x-1 \right )\left ( x+2 \right )}$

Answer

If $\left ( x-1 \right ) = 0$ or $\left ( x+2 \right ) =0$ , the denominator is 0, which makes the expression undefined.

This happens when x = 1 or when x = -2.

Therefore the correct answer is .

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Question

Simplify:

$\frac{2x-3}{3-2x}$

Answer

Factor out from the numerator which gives us

$\left -(3-2x \right )$

Hence we get the following

$\frac{\left -(3-2x \right )}{3-2x}$

which is equal to

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Question

Simplify:

$\frac{x+3}{x^{2}+6x+9}$

Answer

If we factors the denominator we get

$\left ( x+3 \right )^{2}$

Hence the rational expression becomes equal to

$\frac{\left ( x+3 \right )}{\left ( x+3 \right )^{2}}$

which is equal to $\frac{1}{\left ( x+3 \right )}$

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Question

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

Solve for .

Answer

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

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

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Question

Solve for .

$2=\frac{10}{3y+2x^{2}}$

Answer

Multiply each side by $3y+2x^{2}$

$2(3y+2x^{2})=10$

Distribute 2 to each term of the polynomial.

$6y+4x^{2}=10$

Divide the polynomial by 6.

$6(y+\frac{2}{3}x^{2})=10$

Divide each side by 6.

$y+\frac{2}{3}x^{2}=\frac{5}{3}$

Subtract the term from each side.

$y=\frac{5}{3}-\frac{2}{3}x^{2}$

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Question

Solve for .

$3=\frac{10}{k+2}$

Answer

Multiply each side by

Distribute 3 to the terms in parentheses.

Subtract 6 from each side of the equation.

Divide each side by 3.

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

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Question

Solve for .

$5=\frac{105}{n^{2}+5}$

Answer

$5=\frac{105}{n^{2}+5}$

Multiply each side of the equation by $n^{2}+5$

$5(n^{2}+5)=105$

Distribute 5 to each term in parentheses.

$5n^{2}+25=105$

Subtract 25 from each side of equation.

$5n^{2}=80$

Divide each side of equation by 5.

$n^{2}=16$

Square root of each side of equation.

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Question

$\small \frac{x^2-5x-6}{x+1}$

For all values , which of the following is equivalent to the expression above?

Answer

First, factor the numerator. We need factors that multiply to $\small -6$ and add to $\small -5$ .

$\small 1*-6=-6\ \text{and}\ 1+(-6)=-5$

$\small x^2-5x-6=(x-6)(x+1)$

We can plug the factored terms into the original expression.

$\small \frac{x^2-5x-6}{x+1}=\frac{(x-6)(x+1)}{(x+1)}$

Note that $\small (x+1)$ appears in both the numerator and the denominator. This allows us to cancel the terms.

$\frac{(x-6)(x+1)}{(x+1)}=(x-6)$

This is our final answer.

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Question

Simplify the expression:

$\frac{x^{3}-5x^{2}+6x}{x-3}$

Answer

First, factor out x from the numerator:

$\frac{x^{3}-5x^{2}+6x}{x-3}$ $=\frac{x(x^{2}-5x+6)}{x-3}$

Notice that the resultant expression in the parentheses is quadratic. This expression can be further factored:

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

We can then cancel the (x-3) which appears in both the numerator and denominator:

$=\frac{x(x-2)}{1}$

Finally, distribute the x outside of the parentheses to reach our answer:

$=x^{2}-2x$

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Question

Which of the following fractions is NOT equivalent to $- \frac{x-5}{2x + 3}$ ?

Answer

We know that $-\frac{a}{b}$ is equivalent to $\frac{-a}{b}$ or $\frac{a}{-b}$ .

By this property, there is no way to get $\frac{x+5}{2x+3}$ from $-\frac{x-5}{2x+3}$ .

Therefore the correct answer is $\frac{x+5}{2x+3}$ .

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Question

Find the values of which will make the given rational expression undefined:

$\frac{x+5}{\left ( x-1 \right )\left ( x+2 \right )}$

Answer

If $\left ( x-1 \right ) = 0$ or $\left ( x+2 \right ) =0$ , the denominator is 0, which makes the expression undefined.

This happens when x = 1 or when x = -2.

Therefore the correct answer is .

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Question

Simplify:

$\frac{2x-3}{3-2x}$

Answer

Factor out from the numerator which gives us

$\left -(3-2x \right )$

Hence we get the following

$\frac{\left -(3-2x \right )}{3-2x}$

which is equal to

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Question

Simplify:

$\frac{x+3}{x^{2}+6x+9}$

Answer

If we factors the denominator we get

$\left ( x+3 \right )^{2}$

Hence the rational expression becomes equal to

$\frac{\left ( x+3 \right )}{\left ( x+3 \right )^{2}}$

which is equal to $\frac{1}{\left ( x+3 \right )}$

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Question

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

Solve for .

Answer

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

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

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Question

Solve for .

$2=\frac{10}{3y+2x^{2}}$

Answer

Multiply each side by $3y+2x^{2}$

$2(3y+2x^{2})=10$

Distribute 2 to each term of the polynomial.

$6y+4x^{2}=10$

Divide the polynomial by 6.

$6(y+\frac{2}{3}x^{2})=10$

Divide each side by 6.

$y+\frac{2}{3}x^{2}=\frac{5}{3}$

Subtract the term from each side.

$y=\frac{5}{3}-\frac{2}{3}x^{2}$

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Question

Solve for .

$3=\frac{10}{k+2}$

Answer

Multiply each side by

Distribute 3 to the terms in parentheses.

Subtract 6 from each side of the equation.

Divide each side by 3.

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

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Question

Solve for .

$5=\frac{105}{n^{2}+5}$

Answer

$5=\frac{105}{n^{2}+5}$

Multiply each side of the equation by $n^{2}+5$

$5(n^{2}+5)=105$

Distribute 5 to each term in parentheses.

$5n^{2}+25=105$

Subtract 25 from each side of equation.

$5n^{2}=80$

Divide each side of equation by 5.

$n^{2}=16$

Square root of each side of equation.

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Question

$\small \frac{x^2-5x-6}{x+1}$

For all values , which of the following is equivalent to the expression above?

Answer

First, factor the numerator. We need factors that multiply to $\small -6$ and add to $\small -5$ .

$\small 1*-6=-6\ \text{and}\ 1+(-6)=-5$

$\small x^2-5x-6=(x-6)(x+1)$

We can plug the factored terms into the original expression.

$\small \frac{x^2-5x-6}{x+1}=\frac{(x-6)(x+1)}{(x+1)}$

Note that $\small (x+1)$ appears in both the numerator and the denominator. This allows us to cancel the terms.

$\frac{(x-6)(x+1)}{(x+1)}=(x-6)$

This is our final answer.

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Question

Simplify the expression:

$\frac{x^{3}-5x^{2}+6x}{x-3}$

Answer

First, factor out x from the numerator:

$\frac{x^{3}-5x^{2}+6x}{x-3}$ $=\frac{x(x^{2}-5x+6)}{x-3}$

Notice that the resultant expression in the parentheses is quadratic. This expression can be further factored:

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

We can then cancel the (x-3) which appears in both the numerator and denominator:

$=\frac{x(x-2)}{1}$

Finally, distribute the x outside of the parentheses to reach our answer:

$=x^{2}-2x$

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