How do you write an equation of a line through the the point (-5, -3) and is parallel to the line 7x+2y=5?

Since they're parallel, this means that they have the same gradient. So start by finding the gradient of the line by using the line equation provided,

To do this, it as to be in the #y= mx+c# format.

So:

2y= 5- #7x#

y= #5/2#- #7/2##x#

The gradient is the co-efficient of #x#, which is #-7/2#.

So:

m= -#7/2#

Next step is finding the y-intercept (c). To do this we make use of the gradient and the point provided.

So you get:

-3 = (-#7/2#) x (-5) +c

Here, notice that I'm using #y-mx+c# to find my answer. Where y= -3, #x#= -5 and m= -#7/2#.

Now, let's find c:

-3 = #35/2# +c

c= -#41/2#

Now, SUBSTITUTION TIME!

y= -#7/2##x#- #41/2#

Take 2 to the other side by multiplication:

2y= #-7x#- 41

Then:

2y+ #7x#= -41

#• " Parallel lines have equal slopes"#

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"rearrange "7x+2y=5" into this form"#

#"subtract "7x" from both sides and divide by 2"#

#2y=-7x+5#

#y=-7/2x+5/2larrcolor(blue)"in slope-intercept form"#

#"with slope m "=-7/2#

#y=-7/2x+blarrcolor(blue)"is the partial equation"#

#"to find b substitute "(-5,-3)" into the partial equation"#

#-3=35/2+brArrb=-6/2-35/2=-41/2#

#y=-7/2x-41/2larrcolor(red)"in slope-intercept form"#

#"multiply through by 2"#

#2y=-7x-41#

#7x+2y=-41larrcolor(red)"in standard form"#