Rearrange in the form ax+by+c=0
1) y+4 = 3/5(x-1)
Thank you
y+4 = (3/5) (x-1)
or
y+4 = 3/ [ 5(x-1) ]
????
The first one you wrote
Sorry i should've made it clear
If
y+4 = (3/5) (x-1)
then
5(y+4) = 3 (x-1)
5 y + 20 = 3 x - 3
5 y = 3 x - 23
y = (3/5) x - 23/5
y + 4 = 3 / 5 ( x - 1 ) Subtract 3 / 5 ( x - 1 )
y + 4 - 3 / 5 ( x - 1 ) = 3 / 5 ( x - 1 ) - 3 / 5 ( x - 1 )
y + 4 - 3 / 5 ( x - 1 ) = 0
- 3 / 5 ( x - 1 ) + y + 4 = 0
a = - 3 / 5
b = 1
c = 47
or
5 y - 3 x + 23 = 0
Sorry
c = 4
Thankyou so much both ♡
- 3 / 5 ( x - 1 ) + y + 4 = 0 Multiply both sides by 5
- 3 x + 5 y + 20 = 0
To rearrange the equation y + 4 = (3/5)(x - 1) in the form ax + by + c = 0, we need to eliminate the fractions.
First, let's distribute the fraction (3/5) to both terms within the parentheses:
y + 4 = (3/5)x - 3/5
y + 4 = 3x/5 - 3/5
Next, we need to move the term with y to the other side of the equation. To do this, we can subtract y from both sides:
y - y + 4 = 3x/5 - 3/5 - y
4 = 3x/5 - y - 3/5
Now, let's combine like terms:
4 = (3x - 5y + 3) / 5
To eliminate the fraction, we can multiply both sides of the equation by 5:
5 * 4 = 5 * (3x - 5y + 3) / 5
20 = 3x - 5y + 3
Finally, we can rearrange the equation to the form ax + by + c = 0:
3x - 5y + 3 - 20 = 0
3x - 5y - 17 = 0
So, the rearranged form of the equation y + 4 = (3/5)(x - 1) is 3x - 5y - 17 = 0.