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| Question 224549:  Write the slope intercept form of the equation of the line described.
 through: (-3,-3), parallel to y=7/3x+3
 Answer by drj(1380)
      (Show Source): 
You can put this solution on YOUR website! Write the slope intercept form of the equation of the line described. through: (-3,-3), parallel to y=7x/3+3
 
 Step 1.  We can find the slope by recognizing that parallel lines have the same slope.  Since
  is in slope-intercept form given as y=mx+b where the slope m=7/3 and the y-intercept b=3 when x=0 or at point (0,b) or (0,3). 
 Step 2.  Now we have to find the line with slope m=7/3 going through point (-3,-3).
 
 Step 3.  Given two points (x1,y1) and (x2,y2), then the slope m is given as
 
 
   
 Step 4.  Let (x1,y1)=(-3,-3) or x1=-3 and y1=-3.  Let other point be ((x2,y2)=(x,y) or x2=x and y2=y.
 
 Step 5.  Now we're given
  .  Substituting above values and variables in the slope equation m yields the following steps: 
 
   
 
   
 Step 6.  Multiply x+3 to both sides to get rid of denominator on right side of equation.
 
 
   
 
   
 
   
 Step 7.  Now subtract 3 form both sides of equation to solve for y.
 
 
   
 
   
 Step 8.  ANSWER:  The equation in slope-intercept form is
   
 
 Note:  the above equation can be rewritten as
 
 
   
 And the graph is shown below which is consistent with the above steps.
 
 
 
 | Solved by pluggable solver: DESCRIBE a linear EQUATION: slope, intercepts, etc |  | Equation  describes a sloping line. For any equation ax+by+c = 0, slope is
  . X intercept is found by setting y to 0: ax+by=c becomes ax=c. that means that x =  c/a. 12/-7 = -1.71428571428571.Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is 12/3 = 4.Slope is --7/3 = 2.33333333333333. Equation in slope-intercept form: y=2.33333333333333*x+4.
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 I hope the above steps and explanation were helpful.
 
 For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
 
 And good luck in your studies!
 
 Respectfully,
 Dr J
 http://www.FreedomUniversity.TV
 
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