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## GMAT - Problem Solving - Algebra

### If x ? 0 and x?1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to (more)

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

If x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

$A.\quad { \left( \frac { x+1 }{ x-1 } \right) }^{ 2 }\\ B.\quad { \left( \frac { x-1 }{ x+1 } \right) }^{ 2 }\\ C.\quad \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \\ D.\quad \frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \\ E.-{ \left( \frac { x-1 }{ x+1 } \right) }^{ 2 }\\ \\ \\$

${ \left( \frac { x+1 }{ x-1 } \right) }^{ 2 }\\ replacing\quad x\quad =\quad 1/x\\ =\quad { \left( \frac { \frac { 1 }{ x } +1 }{ \frac { 1 }{ x } -1 } \right) }^{ 2 }\\ ={ \left( \frac { \frac { 1+x }{ x } }{ \frac { 1-x }{ x } } \right) }^{ 2 }\\ since\quad x\quad is\quad not\quad equal\quad to\quad zero\\ ={ \left( \frac { x+1 }{ 1-x } \right) }^{ 2 }\\ since\quad { (1-x })^{ 2 }={ (1+x) }^{ 2 }\\ ={ \left( \frac { x+1 }{ x-1 } \right) }^{ 2 }$

### Which of the following equations is NOT equivalent to 10y^2=(x+2)(x?2) ? (more)

Which of the following equations is NOT equivalent to 10y2=(x 2)(x2) ?

(A) 30y2=3x212

(B) 20y2=(2x4)(x 2)

(C) 10y2 4=x2

(D) 5y2=x22

(E) y2=(x24)/10

Analysing Options:

A. 30y2=3x212

dividing both sides by 3

10y2 = x2 - 4 = (x + 2) (x - 2)

Equivalent

(B) 20y2=(2x4)(x+2)

20y2 = 2(x - 2) (x + 2)

dividing both sides by 2

10y2 = (x - 2) (x + 2)

Equivalent

(C) 10y2+4=x2

10y2 = x2 - 4 = (x + 2) (x - 2)

Equivalent

(D) 5y2=x22

Not equivalent

(E) y2=(x24)/10

10y2 = x2 - 4 = (x + 2) (x - 2)

### If 1 + 1/x = 2 - 2/x, then x = (more)

If 1 1/x = 2 - 2/x, then x = ?

1 + 1/x = 2 - 2/x

adding 2/x - 1 to both sides

1/x + 2/x = 2 - 1

3/x = 1

x = 3

### If xy not equal 0 and x^2*y^2 -xy = 6, which of the following could be y in terms of x? (more)

If xy not equal 0 and x2*y2 -xy = 6, which of the following could be y in terms of x?

I. 1/2x
II. -2/x
III. 3/x

Given:

x2y2 - xy = 6

assume xy = p

we can write the equation as

p2 - p = 6

p2 - p - 6 = 0

p2 - 3p + 2p - 6 = 0

p(p - 3) + 2 (p - 3) = 0

(p - 3) (p + 2) = 0

p = 3 or p = -2

xy = 3 or xy = -2

y = 3/x or y = -2/x

II and III

### If x(2x+1)=0 and (x+1/2)(2x-3)=0, then x= (more)

If x(2x 1)=0 and (x 1/2)(2x-3)=0, then x=?

Equation 1

x(2x+1)=0

2x2 + x = 0

x (2x + 1) = 0

x = 0 or x = -1/2

Equation 2

(x+1/2)(2x-3) = 0

2x2 - 3x + x - 3/2 = 0

2x2 - 2x - 3/2 = 0

4x2 - 4x - 3 = 0

4x2 - 6x + 2x - 3 = 0

2x (2x - 3) + (2x - 3) = 0

(2x - 3) (2x + 1) = 0

x = 3/2 or x = -1/2

x = -1/2 satisfies both the equation

hence answer is x = -1/2