Question 6420: can you solve these problems in step?
Set2
1. Solve the equation 5x+3y=-15, for x if y=0.
2. find the x- intercept for the equation 6x-y= -12
3. determine the equation of the line with slope -3 and containing (-7,2)
4. given the following , write an equation in standard form. The line has y-intercept 5 and slope 2
5. write an equation of the line in slope- intercept form if contains (-1,2) and ( 5, -4)
6. write an equation slope intercept form of the line that is parallel to the graph of 3y-4x=1 and passes through (0,6)
7. write the equation in standard form for the line that is perpendicular to the graph of y=5x+1 and has a y-intercept of 4.
8. write the equation of the vertical line that contains (-5,-4).
9. find the slope for the equation x-2y=6
10. for the equation x-2y=6, is the point (4,-1) on the line?
Answer by sabanasir(22)   (Show Source):
You can put this solution on YOUR website! Solutions:
1) 5x + 3y =15
when y = 0
5x + 3(0) =15
5x = 15
x = 3
3) slope = -3 and (x,y) = (-7,2)
we use the equation formula:
m= (y-y1) / (x - x1)
m is slope and (x1,y1) = (-7,2)
so,
-3 = ( y - 2 ) / ( x + 7 )
-3 ( x + 7 ) = y - 2
-3x - 21 = y - 2
3x + y - 2 + 21 = 0
3x + y + 20 = 0
5) consider (x1,y1) = (-1,2) and (x2,y2) = ( 5,-4)
slope is m
m = (y2-y1)/(x2-x1)
m = (-4 -2) / ( 5 + 1)
m = -6/ 6
m = -1
considering any point as (x1,y1)
here we take (x1,y1) = (-1,2)
using
m = (y - y1) / ( x - x1)
putting m = -1
-1 = ( y - 2 )/ ( x + 1)
-x - 1 = y - 2
x + y -2 + 1 = 0
x + y -1 = 0
6) for parallel line slope is the same. so is we take out 3y - 4x =1 's slope it would be useed for the point ( 0 , 6) also.
FINDING SLOPE:
3y - 4x = 1
y = ( 4x +1 ) / 3
y = 4x/3 + 1/3
comparing with y=mx+c
we get m = 4/3
now putting it in m = (y - y1) / ( x - x1)
where m = 4/3, (x1,y1) = ( 0 , 6 )
we get:
4/3 = ( y - 6 ) / ( x - 0 )
4 = (3y - 18) / x
4x = 3y - 18
4x - 3y =18 ANSWER.
9) y = mx + c
here c is constant
x-2y = 6
y = -(- x + 6) / 2
y = x/2 - 6/2
y = x/2 - 3
comparing with y = mx+c and m is slope
here,
m = 1/2.
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