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## Comment on

Value of x+2y## Can you show the

## Sure thing.

Sure thing.

We want to factor x^2 + 4xy - 5y^2

First IGNORE the y's and factor x^2 + 4x - 5

We get: x^2 + 4x - 5 = (x + 5)(x - 1)

So, what do we need to do when we put the y's back in?

We need only add a y to each binomial.

When we do so, we get: x^2 + 4xy - 5y^2 = (x + 5y)(x - 1y)

## x^2+4xy-5y^2

we should find factors of -5y^2 so that if we add/sub those factors should yield 4xy

therefore when 5y and -y multiplied will give -5y^2 and if we add 5y and -y will give

4xy, then

x^2-xy+5xy-5y^2

x(x-y)+5y(x-y)

(x-y)(x+5y) we are left with factors of x^2+4xy-5y^2

## I am very slow in solving

## Sometimes it's a matter of

Sometimes it's a matter of considering all of your options first, and then choosing the fastest approach. The great thing about GRE quant questions is that they can typically be solved using more than one approach. So, the key is to become familiar with all the ways to approach each question type.

So, while you're doing practice questions, be sure to review ALL of the different solutions.

## Hello Brent,

I have difficulty in solving OG 2nd edition 118(3) question. Could you please explain that question for me ?

Thanks :-)

## That's a VERY tricky question

That's a VERY tricky question.

The fastest approach is to recognize that the right side of the graph of y = f(x) has a slope of 2 (we can get that by finding two points on the graph and finding the slope between them).

Answer choices A and B are equations of lines that have slope 1. Since those two lines are LESS STEEP than the right side of the graph of y = f(x), neither of those two lines will intersect the given graph.

Answer choices C and D are equations of lines that have slope 2. Since those two lines have the SAME SLOPE as the right side of the graph of y = f(x), those two lines will both be PARALLEL with the given graph.

Answer choice E is the equation of a line with slope 3. Since this line is STEEPER than the right side of the graph of y = f(x), it will eventually intersect the given graph.

Answer: E

## ALTERNATIVE (much longer)

ALTERNATIVE (much longer) solution:

If a line intersects another line, then those two lines share a common point. That means, they must share a solution.

So, we can take each answer choice and set it EQUAL to the equation of the given graph (y = |2x| + 4) and see which answer choice shares a common solution.

Here's what I mean.

A: g(x) = x - 2

Create the equation x - 2 = |2x| + 4

When we apply the equation solving technique for absolute values (here https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...), we get the "solutions" x = -6 and x = 2.

However, when we plug those solutions back into the original equation (x - 2 = |2x| + 4), we find that they are extraneous roots.

This means there is NO SOLUTION to the equation x - 2 = |2x| + 4, which also means that the graph of y = x - 2 does NOT intersect the graph of y = |2x| + 4

If we continue this strategy for all 5 answer choices, we find that only answer choice E yields a solution.

## Is X^2-5y^2 not a difference

## We CAN express x² - 5y² as a

We CAN express x² - 5y² as a difference of squares, since 5 = (√5)²

So, x² - 5y² = (x + √5y)(x - √5y)

That said, writing x² - 5y² as a difference of squares will not help us solve this question.

Cheers,

Brent

## I have a question regarding

## I'm happy to help.

I'm happy to help.

In the future, please post questions under the related video lesson (this question concerns parallelograms, so it would best placed under the Quadrilateral lesson: https://www.greenlighttestprep.com/module/gre-geometry/video/874

Okay, onto the question!

When we create a RIGHT TRIANGLE (as you have done), the resulting triangle is neither a 30-60-90 triangle nor a 45-45-90 triangle.

Here's my detailed solution: https://gre.myprepclub.com/forum/parallelogram-opqr-lies-in-the-xy-plane...

Cheers,

Brent

## I have a question from

As P(-2,6) and R(5,0) then the slope is going to be -6/7. Then I use the R values (5,0) to find out the value of c (y = mx +c) and get 5 answer. My equation becomes y = -6/7x + 5. This is of course the wrong answer. Can you please highlight my mistake

## Aside: On the Official Guide

Aside: On the Official Guide (3rd edition), this is question 17e on page 262

Once we know that the slope is -6/7, the equation of the line becomes y = (-6/7)x + b

Since the point (5, 0), lies on the line, then x = 5 and y = 0 is a SOLUTION to the equation

Plug those values into the equation y = (-6/7)x + b

We get: 0 = (-6/7)(5) + b

Simplify: 0 = -30/7 + b

Add 30/7 to both sides to get: 30/7 = b

So, the equation of the line is y = (-6/7)x + 30/7

Does that help?

Cheers,

Brent