Find the equation of a line parallel to the line 2x-y-9=0 and passing through the point of intersection of the lines 5x+y+4=0 and 2x+3y=1

Respuesta :

Answer:

[tex]y=2x-3[/tex]

Step-by-step explanation:

1. Using the point of intersections, we use the substitution method to find the coordinates of the line parallel to [tex]2x-y-9=0[/tex]

[tex]5x+y+4=0[/tex]

[tex]y= -4-5x[/tex]

substituting the value of y in [tex]2x+3y=1[/tex]:

[tex]2x +3(-4-5x)=1\\\\2x-12-15x-1=0\\-13x=13x\\x= \frac{13}{-13}= -1[/tex]

substituting x=-1 in y= -4-5x:

[tex]y= 1[/tex] (upon solving, you should get this)

[tex](x,y)= (-1,1)[/tex]

2. Using y=mx+c and making y the subject of the formula 2x-y-9=0 and using the coordinate we found earlier, we will find the equation of the parallel line. (We make y the subject of the formula to find the gradient)

[tex]2x-y-9=0\\2x-9=y\\y= 2x-9[/tex]

[tex]y= mx+c\\[/tex]

-1= 2 x 1 +c

-3=c

  • y= 2x-3

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