How to subtract exponential variables

ISEE Upper Level Quantitative Reasoning · Learn by Concept

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ISEE Upper Level Quantitative Reasoning › How to subtract exponential variables

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1

is a positive number. is a negative number.

Which is the greater quantity?

(a) $x^{2} - y ^{2}$

(b) $(x-y) ^{2}$

(b) is greater.

CORRECT

(a) is greater.

0

It is impossible to tell from the information given.

0

(a) and (b) are equal.

0

Explanation

$(x-y) ^{2} = x ^{2} - 2xy + y ^{2}$

is a positive number and is a negative number, so and $y^{2} > 0$ . Therefore,

$(x-y) ^{2} = x ^{2} - 2xy + y ^{2} > x ^{2} + y ^{2} > x ^{2} - y ^{2}$ .

(b) is always greater.

2

Simplify the following expression:

CORRECT

0

0

0

Explanation

Simplify the following expression:

We can treat this basically as a regular subtraction problem.

We want to subtract our coefficients, while leaving the exponents the same.

3

Assume that $x \neq 0$ . Which of the following expressions is equivalent to:

$\frac{\left (x+1 \right ) ^{3} - \left ( x^{3 } + 1\right ) }{x^{2}}$ ?

$3 + \frac{ 3 }{x}$

CORRECT

$2 + \frac{ 2 }{x}$

0

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

0

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

0

0

Explanation

$\frac{\left (x+1 \right ) ^{3} - \left ( x^{3 } + 1\right ) }{x^{2}}$

$= \frac{\left (x^{3} +3x^{2} + 3x + 1 ) - \left ( x^{3 } + 1\right ) }{x^{2}}$

$= \frac{x^{3} -x^{3} +3x^{2} + 3x + 1-1}{x^{2}}$

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

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

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

4

Simplify the expression:

$10x - 4x^{2} - 5x$

$-4x^{2} +5x$

CORRECT

$-4x^{2} +15x$

0

$4x^{2} -5x$

0

$x^ {3}$

0

The expression cannot be simplified further.

0

Explanation

Group, then collect like terms.

$10x - 4x^{2} - 5x$

$= - 4x^{2} + 10x - 5x$

$= - 4x^{2} + \left ( 10 - 5 \right ) x$

$= - 4x^{2} + 5 x$

5

Simplify:

$\frac{(x+2y)^{2} - (x - 2y)^{2}}{xy}$

CORRECT

0

0

0

$\frac{2x^{2}+8y^{2}}{xy}$

0

Explanation

$\frac{(x+2y)^{2} - (x - 2y)^{2}}{xy}$

$=\frac{(x^{2}+4xy+4y^{2})-(x^{2}-4xy+4y^{2})}{xy}$

$=\frac{x^{2}-x^{2}+4xy+4xy+4y^{2}-4y^{2}}{xy}$

$=\frac{8xy}{xy} = 8$

6

If $x\neq 0$ , simplify:

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

CORRECT

0

0

0

0

Explanation

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

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

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

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

$=\frac{2}{-1}=-2$

7

If $x,y\neq 0$ , simplify:

$\frac{(x+y)^2-(x-y)^2}{2xy}$

CORRECT

0

0

0

0

Explanation

$\frac{(x+y)^2-(x-y)^2}{2xy}=\frac{(x^2+2xy+y^2)-(x^2-2xy+y^2)}{2xy}$

$=\frac{x^{2}-x^{2}+y^{2}-y^{2}+2xy+2xy}{2xy}$

$=\frac{4xy}{xy}=4$

8

If $x\neq 0$ , simplify:

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

CORRECT

0

0

$\frac{4}{x}$

0

$\frac{16}{x}$

0

Explanation

$(x+\frac{4}{x})^2-(x-\frac{4}{x})^2=(x^{2}+2\times x\times \frac{4}{x}+\frac{16}{x^2})-(x^{2}-2\times x\times \frac{4}{x}+\frac{16}{x^2})$

$=(x^{2}+8+\frac{16}{x^2})-(x^{2}-8+\frac{16}{x^2})$

$=x^2-x^2+8+8+\frac{16}{x^2}-\frac{16}{x^2}$

9

If $x\neq 0$ , simplify:

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

CORRECT

0

0

0

$\frac{2}{x}$

0

Explanation

$\frac{(x^{2}+3)^2-(x^2-3)^2}{6x^2}=\frac{(x^4+6x^2+9)-(x^4-6x^2+9)}{6x^2}$

$=\frac{x^4-x^4+6x^2+6x^2+9-9}{6x^2}$

$=\frac{12x^2}{6x^2}=2$

10

If $x,y\neq 0$ , simplify:

$\frac{(x^2+y^2)-(x-y)^{2}}{xy}$

CORRECT

0

0

0

0

Explanation

$\frac{(x^2+y^2)-(x-y)^{2}}{xy}=\frac{x^2+y^2-(x^2-2xy+y^2)}{xy}$

$=\frac{x^2-x^2+y^2-y^2+2xy}{xy}$

$=\frac{2xy}{xy}=2$

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