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 April 16th, 2013, 09:55 PM #1 Newbie   Joined: Apr 2013 Posts: 2 Thanks: 0 Linear Equations find the equation of a line parallel to 3x + 4y = 8 with the same y-intercept as 5x - 3y = 10 Find the equation of a line parallel to 2x +7y = 10 with the same x-intercept as 3x - 4y = 5
 April 17th, 2013, 04:31 AM #2 Member   Joined: Jan 2012 Posts: 80 Thanks: 0 Re: Linear Equations Hi Blaze, You need to find the gradient of the line you want the new line to be parallel to. So you need to rearrange to the form $y= mx + c$ (m is the gradient): $3x + 4y= 8$ $4y= -3x + 8$ $y= -\frac{3}{4}x + 2$ The gradient of this line is $-\frac{3}{4}$. Now you need to find the y intercept of the other line. Again you need to rearrange to the form $y= mx + c$ (c is the y intercept): $5x - 3y= 10$ $3y= 5x - 10$ $y= \frac{5}{3}x - \frac{10}{3}$ The y intercept of this line is $-\frac{10}{3}$. Now you can use that gradient and y intercept to give the equation for the new line: $y= -\frac{3}{4}x -\frac{10}{3}$ You should be able to apply the same steps to work out the second question.
 April 17th, 2013, 05:37 AM #3 Global Moderator   Joined: Dec 2006 Posts: 16,791 Thanks: 1238 To find the y-intercept for 5x - 3y = 10, one puts x = 0 and solves for y, getting y = -10/3. Substituting (x, y) = (0, -10/3) into 3x + 4y gives -40/3. Hence the line parallel to 3x + 4y =8 with y-intercept (0, -10/3) is 3x + 4y = -40/3, or 9x + 12y = -40.
 April 17th, 2013, 01:21 PM #4 Newbie   Joined: Apr 2013 Posts: 2 Thanks: 0 Re: Linear Equations How would you find the x-intercept though? I figured out how to do the first question
 April 17th, 2013, 10:07 PM #5 Global Moderator   Joined: Dec 2006 Posts: 16,791 Thanks: 1238 To find the y-intercept for 3x - 4y = 5, one puts y = 0 and solves for x, getting x = 5/3.

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