Question 13258: How do you solve an equation and leava it in standard form?
Ex. parallel to 3x+5y=8,when passing through (-1,7)?
Answer by bam878s(77)   (Show Source):
You can put this solution on YOUR website! Start with your original equation 3x+5y=8. FIRST, you need to solve for y. 3x+5y-8 = 8-8. We have 3x+5y-8=0. Now move the term with y variable to the other side of the equation. 3x+5y-5y-8 = 0-5y. We have, 3x-8=-5y. Now divde both sides of the equation by -5. We have (3/-5)x-(8/-5)=y or (-3/5)x+(8/5)=y.
We now have the slope of this equation -3/5 (the coefficient of the x-variable). So, we know that any line of this same slope will be parallel to this line. Your equation will look something like this, (-3/5)x + constant = y. To get a line to pass through a certain point you need to know the point-slope form of a line. It is (y-y1) = slope(x-x1). So with a certain point (x1,y1), in this case the point(x1= -1, y1= 7), we have (y-7)= (-3/5)*{x-(-1)}. Distributing the x on the right hand side of this equation yields (y-7)= (-3/5)x - (3/5). Now adding 7 to both sides we have (y-7+7)= (-3/5)x - (3/5) + 7. Note 7 = (35/5) so we have y = (-3/5)x - (3/5) + (35/5). Ultimately the parallel line passing through point (-1,7) is y = (-3/5)x + (32/5)
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