# 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. $-x<4$ gives $0 which means $-4 or $x>-4$

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

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

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

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

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

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

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

3. ## Thankyou

I understand them now thanks for the help.