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A diffcult math prob

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Source: — Problem Solving |

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by Anurag@Gurome » Sat Sep 03, 2011 4:44 am
tracyyahoo wrote:In the xy-plane, what is the x-intercept of the line whose equation is 3y-4x=15?
Method 1:
If we write (3y - 4x = 15) in intercept form we'll get y/(15/3) + x/(-15/4) = 1
Hence, x-intercept = -15/4

Method 2:
The x-intercept is nothing but the x-coordinate of the point where the line cuts the x-axis. Now the y-coordinate of the point must be zero as that point will lie on the x-axis. Thus if we make y = 0 int he given equation, we'll get the value of x-intercept.

3*0 - 4x = 15 ----> x = -15/4

The correct answer is A.
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by GmatKiss » Sat Sep 03, 2011 10:12 am
Anurag@Gurome wrote:
tracyyahoo wrote:In the xy-plane, what is the x-intercept of the line whose equation is 3y-4x=15?
Method 1:
If we write (3y - 4x = 15) in intercept form we'll get y/(15/3) + x/(-15/4) = 1
Hence, x-intercept = -15/4

Method 2:
The x-intercept is nothing but the x-coordinate of the point where the line cuts the x-axis. Now the y-coordinate of the point must be zero as that point will lie on the x-axis. Thus if we make y = 0 int he given equation, we'll get the value of x-intercept.

3*0 - 4x = 15 ----> x = -15/4

The correct answer is A.
Thanks for method(2)
Does this hold true for finding the y intercept(making x=0)
Also please let us know what else can be found with the given equation?

TIA,
GK
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by thestartupguy » Sat Sep 03, 2011 12:18 pm
Yes GmatKiss, Method (2) works for the y-intercept when x=0 :)
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