Question 1194153
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For linear equations of the form
x/a + y/b = 1
we can quickly spot the x and y intercepts as 'a' and b respectively


In this case, a = -4 and b = -3 to give us
x/a + y/b = 1
x/(-4) + y/(-3) = 1
-x/4 - y/3 = 1


Notice what happens when we plug in x = 0
-x/4 - y/3 = 1
-0/4 - y/3 = 1
0 - y/3 = 1
-y/3 = 1
y = 1(-3)
y = -3
Showing that -3 is the y intercept located at (x,y) = (0,-3)
Through similar steps, you should find y = 0 leads to x = -4 to get the x intercept located at (-4,0)
This confirms we have the correct set up so far.
This example also helps show why x/a + y/b = 1 is useful.


Let's get this equation into slope-intercept form y = mx+b
-x/4 - y/3 = 1
12(-x/4 - y/3) = 12*1
-3x - 4y = 12
-4y = 12+3x
-4y = 3x+12
y = (3x+12)/(-4)
y = 3x/(-4) + 12/(-4)
y = (-3/4)x - 3
In step 2, I multiplied both sides by 12 to clear out the fractions.


The equation y = (-3/4)x - 3 is in y = mx+b form with 
m = -3/4 = slope
b = -3 = y intercept


If you need this in standard form Ax+By = C, then,
-x/4 - y/3 = 1
12(-x/4 - y/3) = 12*1
-3x - 4y = 12
-1(-3x - 4y) = -1*12
3x+4y = -12
Some math textbooks require that the x coefficient is positive. This is optional really.
Also, some textbooks may have Ax+By+C = 0 as standard form. So we'd go from 3x+4y = -12 to 3x+4y+12 = 0.
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