PSAT Math - How to simplify a fraction

Simplify the expression,

$\small \frac{a^2+b^2}{a+b}$ .

Cannot be simplified

$\small a- b$

$\small \frac{a}{b}+\frac{b}{a}$

$\small a+b$

Explanation

The numerator of the expression cannot be factored. Therefore, the denominator cannot divide into the numerator, and the expression is in its simplest form.

Simplify:

$\frac{8x^3y^2z^2}{4xy^4z^3}$

$\frac{2x^2}{y^2z}$

$\frac{y^2z}{2x^2}$

Explanation

$\frac{8x^3y^2z^2}{4xy^4z^3}$

To simply a fraction with variables, subtract exponents of like bases:

This leaves us with the expression:

$2x^2y^{-2}z^{-1}$

Next, we know that to change negative exponents in the numerator into positive exponents, we place them in the denominator. Thus, our expression simplifies to:

$\frac{2x^2}{y^2z}$

Simplify the expression, $\left(\frac{x^2}{y^{-3}}\right)\left(\frac{x^{-1}}{y^3}\right)$ .

Can't be simplified

$\frac{x}{y^6}$

Explanation

The and variables with negative exponents can be rewritten with positive exponents by moving them from the denominator to the numerator, and vice versa. Therefore, the expression can be rewritten as

$\left(x^2y^3 \right) \left(\frac{1}{xy^3}\right) = \frac{x^2y^3}{xy^3}$ .

The exponents on the denominator can then be subtracted from the exponent in the numerator to give

$x^{2-1}y^{3-3} = x^1y^0 = x \cdot1 = x$

Simplify the expression

$\left(\frac{A^{\frac{3}{2}}\sqrt{B}}{C^{-\frac{1}{2}}}\right)\left(\frac{B^{-\frac{1}{2}}}{2\sqrt{AC}}\right)$ .

$\frac{A}{2}$

$\frac{A^2}{2}$

Explanation

The expression can be rewritten as

$\left(\frac{A^{\frac{3}{2}}B^\frac{1}{2}}{C^{-\frac{1}{2}}}\right)\left(\frac{B^{-\frac{1}{2}}}{2A^\frac{1}{2}C^\frac{1}{2}}\right)$

We can now move the variables to the numerator and combine alike variables

$(A^\frac{3}{2}B^\frac{1}{2}C^\frac{1}{2})\left(\frac{A^{-\frac{1}{2}}B^{-\frac{1}{2}}C^{-\frac{1}{2}}}{2}\right) = \frac{A^{(\frac{3}{2}-\frac{1}{2})}B^{(\frac{1}{2}-\frac{1}{2})}C^{(\frac{1}{2}-\frac{1}{2})}}{2}$

This becomes

$\frac{A^{\frac{2}{2}}B^0C^0}{2} = \frac{A^1}{2} = \frac{A}{2}$

Simplify x/2 – x/5

2x/7

3x/10

3x/7

7x/10

5x/3

Explanation

Simplifying this expression is similar to 1/2 – 1/5. The denominators are relatively prime (have no common factors) so the least common denominator (LCD) is 2 * 5 = 10. So the problem becomes 1/2 – 1/5 = 5/10 – 2/10 = 3/10.

Simplify the expression $\left(\frac{2x^2}{ab^4}\right)(a^5b^2x^{-1})$ .

$\frac{2a^4x}{b^2}$

$\frac{xa^4}{b^2}$

$\frac{2x^3a^4}{b^2}$

$\frac{x^2a^3}{b^2}$

Explanation

The expression can be rewritten as

$\left(\frac{2x^2}{ab^4}\right)\left(\frac{a^5b^2}{x}\right)$

Now the expression can be combined by adding and subtracting exponents

$\left(\frac{2x^2a^5b^2}{ab^4x}\right) = 2x^{(2-1)}a^{(5-1)}b^{(2-4)} = 2xa^4b^{-2} = \frac{2xa^4}{b^2} = \frac{2a^4x}{b^2}$

Simplify the expression

$(12X^2Y^{-3}Z^2)\left(\frac{X^3Y^4Z^{-3}}{4}\right)$ .

$\frac{3X^5Y}{Z}$

$\frac{3Y}{XZ}$

$\frac{3}{XYZ}$

Explanation

We can combine the expression into

$\left(\frac{12X^2X^3Y^{-3}Y^4Z^2Z^{-3}}{4}\right) = \frac{12}{4} X^{(2+3)}Y^{(-3+4)}Z^{(2-3)}$

The combined expression can then be simplified to

$3X^5Y^1Z^{-1} = \frac{3X^5Y}{Z}$

Simplify: $\frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}$

$\frac{x^{2}}{3y^{3}z}$

$\frac{3x^{2}y^{3}}{z}$

$\frac{1}{3x^{2}y^{3}z}$

$\frac{x^{2}}{8y^{3}z}$

Explanation

$\frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}=\frac{x^{2}}{3y^{3}z}$

First, let's simplify $\frac{4}{12}$ . The greatest common factor of 4 and 12 is 4. 4 divided by 4 is 1 and 12 divided by 4 is 3. Therefore $\frac{4}{12}=\frac{1}{3}$ .

To simply fractions with exponents, subtract the exponent in the numerator from the exponent in the denominator. That leaves us with $\frac{1}{3}x^{2}y^{-3}z^{-1}$ or $\frac{x^{2}}{3y^{3}z}$

Simplify the expression

$\frac{a^2-b^2}{a-b}$ .

$\frac{a}{b}-\frac{b}{a}$

Cannot be simplified

Explanation

The numerator of this expression can be factored, giving the expression

$\frac{(a+b)(a-b)}{(a-b)}$

The will cancel itself out leaving,

$\frac{a+b\cdot1}{1} = a+b$

The expression $(\frac{a^{2}}{b^{3}})(\frac{a^{-2}}{b^{-3}}) = ?$

$1$

$0$

$\frac{a^{-4}}{b^{-9}}$

$\frac{b^{9}}{a^{4}}$

$b^{-9}$

Explanation

A negative exponent in the numerator of a fraction can be rewritten with a positive exponent in the denominator. The same is true for a negative exponent in the denominator. Thus, $\frac{a^{-2}}{b^{-3}} =\frac{b^{3}}{a^{2}}$ .

When $\frac{a^{2}}{b^{3}}$ is multiplied by $\frac{b^{3}}{a^{2}}$ , the numerators and denominators cancel out, and you are left with 1.

How to simplify a fraction

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