# Thread: IM stuck on algebra!

1. ## IM stuck on algebra!

I have homework due tomorrow which I as usual forgot about and left till the last minute...its algebra which im hopeless at, any help would be much appreciated.

1) -x < 4

2) 12x > 2x^2

3) x - 3 2x - 1
____ = _____

x - 1 4x + 5

4) 3(2x + 5)
_________ = 6x + 1
4

5) find the x coordinates of the points of intersection of the two parabolas
y = 3x^2 + 5x - 6 and y = -2x^2 - 3x + 18

Any help wud be much appreciated

2. Originally Posted by Brownhash
I have homework due tomorrow which I as usual forgot about and left till the last minute...its algebra which im hopeless at, any help would be much appreciated.

1) -x < 4

2) 12x > 2x^2

3) x - 3 2x - 1
____ = _____

x - 1 4x + 5

4) 3(2x + 5)
_________ = 6x + 1
4

5) find the x coordinates of the points of intersection of the two parabolas
y = 3x^2 + 5x - 6 and y = -2x^2 - 3x + 18

Any help wud be much appreciated
I'll post the solutions. If you can't follow them, say which steps don't make sense and I'll explain them...

1. $\displaystyle -x<4$ gives $\displaystyle 0<x+4$ which means $\displaystyle -4<x$ or $\displaystyle x>-4$

2. $\displaystyle 12x > 2x^2$
$\displaystyle 6x > x^2$
$\displaystyle 6>x$ if $\displaystyle x\geq 0$ else $\displaystyle 6<x$ if $\displaystyle x<0$

3. $\displaystyle \frac{x-3}{x-1}=\frac{2x-1}{4x+5}$
$\displaystyle (x-3)(4x+5)=(2x-1)(x-1)$
$\displaystyle 4x^2+5x-12x-15=2x^2-2x-x+1$
$\displaystyle 4x^2-7x-15=2x^2-3x+1$
$\displaystyle 2x^2-4x-16=0$
$\displaystyle 2(x^2-2x-8)=0$
$\displaystyle x^2-2x-8=0$
$\displaystyle (x-4)(x+2)=0$
$\displaystyle x-4 = 0$ or $\displaystyle x+2=0$
$\displaystyle x=4$ or $\displaystyle x=-2$

4. $\displaystyle \frac{3(2x+5)}{4}=6x+1$
$\displaystyle 3(2x+5)=4(6x+1)$
$\displaystyle 6x+30=24x+4$
$\displaystyle 30-4=24x-6x$
$\displaystyle 26=18x$
$\displaystyle \frac{26}{18}=x$
$\displaystyle \frac{13}{9}=x$

5. Intersection points between functions occur when the functions are equal. Since they are both equal to $\displaystyle y$, they are equal to each other...

$\displaystyle 3x^2 + 5x - 6 = -2x^2 - 3x + 18$
$\displaystyle 5x^2+8x-24=0$

$\displaystyle x=\frac{-8\pm\sqrt{8^2-4\times5\times(-24)}}{2\times5}$ using the Quadratic Formula
$\displaystyle =\frac{-8\pm\sqrt{64+480}}{10}=\frac{-8\pm\sqrt{544}}{10}=\frac{-8\pm4\sqrt{34}}{10}=\frac{-4\pm2\sqrt{34}}{5}$

So $\displaystyle x=\frac{-4+2\sqrt{34}}{5}$ or $\displaystyle x=\frac{-4-2\sqrt{34}}{5}$

3. ## Thankyou

I understand them now thanks for the help.