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Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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15 Mar 2015, 23:04
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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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16 Mar 2015, 22:55
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Hi All, Since it looks like this Ds question has gotten 'buried', I'm going to bump it back up with a 'nudge' about how to solve it: When dealing with graphing and lines, it helps to organize your data using slopeintercept format: Y = (M)(X) + B We're given the slope of a line and one of the points that it runs through. From this, you can calculate the value of B. With the equation of the full line, you should then be able to calculate the Xintercept.... GMAT assassins aren't born, they're made, Rich
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Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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17 Mar 2015, 00:04
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slope 5/3 and point (2;7), so equation is
7=(5/3)*2+b => b=11/3
to find X intercept, rewrite the equation
0=(5/3)*x+11/3 => x=11/5=2 1/5
B



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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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17 Mar 2015, 09:31
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Here is my algebraic approach: y = (5/3)x + b 7 = (5/3)(2) + b b = 11/3
0 = (5/3)x + 11/3 x = 11/5 or 2 1/5



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Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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23 Mar 2015, 03:58
Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Think about this visually. A slope of – 5/3 means, among other things, right 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is much closer to the xintercept. Let’s think about this in the vicinity of the xintercept. Obviously, (1, 2) is at a height of 2 above the xaxis. Let b be the distance from (1, 0) to the xintercept. We know –h/b must equal the slope. h/b = 5/3 > 2/b = 5/3. 5b = 6 > b = 6/5 = 1 1/5. Now, we just have to add one to that to get the horizontal distance from the origin. xintercept = 2 1/5. Answer = (B) Attachment:
ppocg_img6.png [ 9.41 KiB  Viewed 4527 times ]
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Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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25 May 2015, 05:05
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Bunuel wrote: Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Think about this visually. A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. Let’s think about this in the vicinity of the xintercept. Attachment: ppocg_img6.png Obviously, (1, 2) is at a height of 2 above the xaxis. Let b be the distance from (1, 0) to the xintercept. We know –h/b must equal the slope. h/b = 5/3 > 2/b = 5/3. 5b = 6 > b = 6/5 = 1 1/5. Now, we just have to add one to that to get the horizontal distance from the origin. xintercept = 2 1/5. Answer = (B) Hi Bunuel, Please make me understand the below point: A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. I believe, for slope  5/3, right 3 spaces and down 5 so that the line goes through (1,2) from (2,7) downward slope. Please explain if iam wrong. Thanks
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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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25 May 2015, 08:54
balamoon wrote: Bunuel wrote: Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Think about this visually. A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. Let’s think about this in the vicinity of the xintercept. Obviously, (1, 2) is at a height of 2 above the xaxis. Let b be the distance from (1, 0) to the xintercept. We know –h/b must equal the slope. h/b = 5/3 > 2/b = 5/3. 5b = 6 > b = 6/5 = 1 1/5. Now, we just have to add one to that to get the horizontal distance from the origin. xintercept = 2 1/5. Answer = (B) Hi Bunuel, Please make me understand the below point: A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. I believe, for slope  5/3, right 3 spaces and down 5 so that the line goes through (1,2) from (2,7) downward slope. Please explain if iam wrong. Thanks Yes, that's true. Edited. Thank you.
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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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26 May 2015, 01:55
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My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.
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Re: Line A has a slope of 5/3 through the point (–2, 7). What [#permalink]
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26 May 2015, 05:26
With the equation y=mx+c we get 7=10/3 + c So c = 11/3
To find the x intercept put y=0 in the equationy=mx+c
0= 5/3 x +11/3 Multiplying by 3 throughout 11 = 5x So x=11/5 Which is 2 1/5



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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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07 Apr 2016, 23:57
Jackal wrote: My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations. HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN



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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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08 Apr 2016, 00:04
akshay4gmat wrote: HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN Hi here 11/5 is not 21/5 BUT \(2 \frac{1}{5}\)that is 2 and 1/5.. \(2 + \frac{1}{5} = \frac{(10+1)}{5} =\frac{11}{5}\)
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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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08 Apr 2016, 07:46
akshay4gmat wrote: Jackal wrote: My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations. HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN Hi Akshay Just as Chetan explained. \(= \frac{11}{5}\) \(= \frac{(10 + 1)}{5}\) \(= 2 \frac{1}{5}\) Apologies for not using Latex equivalent symbols
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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. Point intercept form a line is y=mx+b slope = m, y intercept =b (given that x=0) x intercept =b/m Here we know m=5/3 ; we need to find b and we are done lets use the equation by plugging values of x,y and m from the question stem 7=5/3*2 +b ==> 7=10/3+b b=11/3 x intercept = b/m = 11/3/5/3=11/5==> 11/5 = 2 \(\frac{1}{5}\) Answer is B
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Re: Line A has a slope of 5/3 [#permalink]
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04 Jun 2017, 00:39
slope m= 5/3 and point (2;7), so equation is
y=mx+b
We need to find b 7=(5/3)*2+b 710/3=b
b=11/3 to find x intercept, rewrite the equation
0=(5/3)*x+11/3 => x=11/5
Answer B
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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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05 Nov 2017, 03:45
Hi @bunnel, Could please explain the how you arrived at x intercept. The equation of the line is 5x3y+11 = 0 So should the y intercept be a/b = 5/3? Thanks in advance! S Jackal wrote: My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.



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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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10 Nov 2017, 03:46
Is this solution fine? y= mx+c y= 5/3*x +c Using formula to find slope: (7y)/(2x) = 5/3 3(7y)= 5(2x) 213y= 10+5x 5x+3y= 11 To find xintercept, plug y= 0 x= 11/5
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Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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14 Nov 2017, 07:10
Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5 The equation of Line A is y = (5/3)x + b. Let’s now plug in (2,7) and determine b: 7 = (5/3)(2) + b 7 = 10/3 + b Multiplying by 3, we have: 21 = 10 + 3b 11 = 3b 11/3 = b Thus, we have: y = (5/3)x + 11/3 We can substitute 0 for y to determine the xintercept, and we have: 0 = (5/3)x + 11/3 11/3 = 5x/3 11 = 5x x = 11/5 = 11/5 = 2 1/5 Answer: B
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Line A has a slope of 5/3 and passes through the point (–2, 7). What [#permalink]
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15 Nov 2017, 13:39
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Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. With a slope and a point, and asked to find the xintercept of a line, I use the slope intercept form. (It's quick). 1) Start with just the slope to construct a slopeintercept equation for the line: \(y = mx + b\), m = slope, b = yintercept Slope = \(\frac{5}{3}\),so \(y = \frac{5}{3}x + b\) 2) Next, plug in the given point (2,7) to find \(b\) \(7 = \frac{5}{3}(2) + b\) \(7 = \frac{10}{3} + b\) \((7  \frac{10}{3}) = \frac{11}{3} = b\) 3) Plug \(b\) in. The line's equation is now \(y = \frac{5}{3}x + \frac{11}{3}\) 4) Set y equal to 0 to find xintercept \(0 = \frac{5}{3}x + \frac{11}{3}\) \(\frac{5}{3}x=\frac{11}{3}\) \(x=\frac{11}{3}*\frac{3}{5}=\frac{11}{5}\) \(=2\frac{1}{5}\) Answer B Shiv2016 , yes, it's correct. I had to grin, though (with relief!) because I have done what you did . . . There is no need to find the slope. It's given. shinrai , you wrote Quote: Could please explain the how you arrived at x intercept?
The equation of the line is 5x3y+11 = 0
So should the y intercept be a/b = 5/3? Mmm, no. The yintercept is 11/3. 5/3 is the slope. You were quoting Jackal, who wrote, "So equation of line is y=(5/3)x+(11/3)" Jackal got the xintercept by setting y equal to 0: 0 = (5/3)x+(11/3) \(\frac{5}{3}x = \frac{11}{3}\) \(x=\frac{11}{3} * \frac{3}{5}=\frac{11}{5}\) The line crosses the xaxis at \(\frac{11}{5}\) (and when it crosses the xaxis, y = 0) You used this equation for the line: 5x3y+11 = 0. (A more standard form is 5x + 3y = 11.) It doesn't matter which form of equation you use; to find the yintercept, set x = 0. (x IS zero when the line crosses the yaxis.) Your way: 5x  3y + 11 = 0 (5)(0) (3)(y) + 11 = 0 3y = 11 y = \(\frac{11}{3}\) Hope it helps.
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Line A has a slope of 5/3 and passes through the point (–2, 7). What
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