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December 19, 2014

December 19, 2014

Posted by **John** on Monday, October 15, 2007 at 9:18pm.

Drawing a blank on this one, not sure where to start.

Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 12 and passing through (9, -5).

I have half a solution to this one:

Solve the system of equations using the substitution method.

If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”

4x + y = 4

2x + 8y = 0

First I did

4x - 4x + y = 4 - 4x

y = 4 - 4x

then

2x + 8(4-4x) = 0

2x + 32 - 32x = 0

2x + 32 - 32x = 0-32

2x - 32x = -32

-30x = -32

-30x/30 = -32/30

x = 32/30

x = 16/15

then

y = 4 - 4x

y = 4 - 4(16/15)

This is where I am stuck...

So i got x = 16/15 an y = ?

- Algebra -
**Anonymous**, Monday, October 15, 2007 at 9:28pm(I did not check you "x" if it is correct or not)

First, you need to simplify 4{16/15)

Then, find a common denominator

New Equation:

y = 60/15 - 64/15

Answer:

y = 4/15

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