Variables

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ISEE Upper Level Quantitative Reasoning › Variables

Questions 1 - 10

1

Multiply:

$15 x ^{2}+31xy -24 y^{2}$

$15 x ^{2}-49xy +24 y^{2}$

$15x^{2} +49xy - 24y^{2}$

$15x^{2} - 24y^{2}$

$15x^{2} -31xy+ 24y^{2}$

Explanation

$= 5x \cdot 3x+5x \cdot 8y - 3y \cdot 3x - 3y \cdot 8y$

$= 5\cdot3 \cdot x \cdot x+5\cdot 8 \cdot x \cdot y - 3\cdot 3\cdot y\cdot x - 3\cdot 8 \cdot y \cdot y$

$= 15 x ^{2}+40xy - 9xy -24 y^{2}$

$= 15 x ^{2}+31xy -24 y^{2}$

2

Multiply:

$15 x ^{2}+31xy -24 y^{2}$

$15 x ^{2}-49xy +24 y^{2}$

$15x^{2} +49xy - 24y^{2}$

$15x^{2} - 24y^{2}$

$15x^{2} -31xy+ 24y^{2}$

Explanation

$= 5x \cdot 3x+5x \cdot 8y - 3y \cdot 3x - 3y \cdot 8y$

$= 5\cdot3 \cdot x \cdot x+5\cdot 8 \cdot x \cdot y - 3\cdot 3\cdot y\cdot x - 3\cdot 8 \cdot y \cdot y$

$= 15 x ^{2}+40xy - 9xy -24 y^{2}$

$= 15 x ^{2}+31xy -24 y^{2}$

3

Multiply:

$15 x ^{2}+31xy -24 y^{2}$

$15 x ^{2}-49xy +24 y^{2}$

$15x^{2} +49xy - 24y^{2}$

$15x^{2} - 24y^{2}$

$15x^{2} -31xy+ 24y^{2}$

Explanation

$= 5x \cdot 3x+5x \cdot 8y - 3y \cdot 3x - 3y \cdot 8y$

$= 5\cdot3 \cdot x \cdot x+5\cdot 8 \cdot x \cdot y - 3\cdot 3\cdot y\cdot x - 3\cdot 8 \cdot y \cdot y$

$= 15 x ^{2}+40xy - 9xy -24 y^{2}$

$= 15 x ^{2}+31xy -24 y^{2}$

4

Simplify if $xy\neq \pm 2$ .

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

$x^{-1}y^2$

Explanation

Start by factoring the numerator. can be removed from each term.

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

Next, expand the denominator.

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

Simplify by canceling terms.

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

5

Simplify if $xy\neq \pm 2$ .

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

$x^{-1}y^2$

Explanation

Start by factoring the numerator. can be removed from each term.

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

Next, expand the denominator.

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

Simplify by canceling terms.

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

6

Simplify if $xy\neq \pm 2$ .

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

$x^{-1}y^2$

Explanation

Start by factoring the numerator. can be removed from each term.

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

Next, expand the denominator.

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

Simplify by canceling terms.

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

7

Multiply:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$x^{2} - 6xy +9y ^{2} - 1$

$x^{2} +9y ^{2} - 1$

$x^{2} + 6xy -9y ^{2} - 1$

$x^{2} + 3xy -9y ^{2} - 1$

$x^{2} - 3xy +9y ^{2} - 1$

Explanation

This can be achieved by using the pattern of difference of squares:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$=\left [ \left ( x -3y \right ) + 1\right ] \left \left[( x - 3y \right ) - 1\right ]$

$= ( x -3y ) ^{2} - 1 ^{2}$

$= (x -3y) ^{2} - 1$

Applying the binomial square pattern:

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

$= x^{2} - 6xy +9y ^{2} - 1$

8

Add:

Explanation

In order to simplify this expression, we will need to add like terms.

There is a lone positive three.

Combine all the terms.

The answer is:

9

Add:

Explanation

In order to simplify this expression, we will need to add like terms.

There is a lone positive three.

Combine all the terms.

The answer is:

10

Multiply:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$x^{2} - 6xy +9y ^{2} - 1$

$x^{2} +9y ^{2} - 1$

$x^{2} + 6xy -9y ^{2} - 1$

$x^{2} + 3xy -9y ^{2} - 1$

$x^{2} - 3xy +9y ^{2} - 1$

Explanation

This can be achieved by using the pattern of difference of squares:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$=\left [ \left ( x -3y \right ) + 1\right ] \left \left[( x - 3y \right ) - 1\right ]$

$= ( x -3y ) ^{2} - 1 ^{2}$

$= (x -3y) ^{2} - 1$

Applying the binomial square pattern:

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

$= x^{2} - 6xy +9y ^{2} - 1$

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