# Math Help - Point of Intersection (2)

1. ## Point of Intersection (2)

Sorry for starting another topic, but I've tried for hours to try and get an answer.

I need to find the point of intersection (which I know is (4,7)), but I don't know how I'm supposed to get there?

$\displaystyle y=- \frac {4x}{3}+12.33$
$\displaystyle y= \frac {3x}{4}+4$

$\displaystyle y= \frac {3x}{4}+4 \Rightarrow 4y=3x+16 \Rightarrow 3x=4y-16 \Rightarrow x= \frac {4y-16}{3}$

$\displaystyle y=- \frac {4x}{3}+12.33$

$\displaystyle y=- \frac {4(\frac {4y-16}{3} )}{3}+12.33$

$\displaystyle y=- \frac {\frac {16y-64}{3} }{3} +12.33$

If I move the 4y to the other side? Then do $\frac{-16}{3} + 12.33$, I get 6.9997, but that's for 4y.

What am I doing wrong?

x
$\displaystyle x = \frac {4y-16}{3}$

2. Originally Posted by watp
Sorry for starting another topic, but I've tried for hours to try and get an answer.

I need to find the point of intersection (which I know is (4,7)), but I don't know how I'm supposed to get there?

$\displaystyle y=- \frac {4x}{3}+12.33$
$\displaystyle y= \frac {3x}{4}+4$

$\displaystyle y= \frac {3x}{4}+4 \Rightarrow 4y=3x+16 \Rightarrow 3x=4y-16 \Rightarrow x= \frac {4y-16}{3}$

$\displaystyle y=- \frac {4x}{3}+12.33$

$\displaystyle y=- \frac {4(\frac {4y-16}{3} )}{3}+12.33$

$\displaystyle y=- \frac {\frac {16y-64}{3} }{3} +12.33$

If I move the 4y to the other side? Then do $\frac{-16}{3} + 12.33$, I get 6.9997, but that's for 4y.

What am I doing wrong?

x
$\displaystyle x = \frac {4y-16}{3}$
12.33 should be 37/3 if you want (4,7) to work.

$- \frac {4x}{3}+\frac{37}{3} = \frac {3x}{4}+4$

Multiply through by 12 to remove denominators.

$-16x + 148 = 9x + 48$

$25x = 100$

$x = 4$

Then go back and find y, easier to use this equation

$\displaystyle y= \frac {3x}{4}+4$

than the other since the 4's cancel.

3. Thanks undefined