# Simultaneous Equations.

• Oct 12th 2011, 03:18 PM
JibblyJabbly
Simultaneous Equations.
I was wondering if you could doublecheck this question please:

g) $\displaystyle y=2x$ and$\displaystyle y = 2x^2$
y = 4 and x = 2

How would I do this?

h)$\displaystyle y= 6 - x^2$ and $\displaystyle y = 4x +1$
• Oct 12th 2011, 03:37 PM
skeeter
Re: Simultaneous Equations.
Quote:

Originally Posted by JibblyJabbly
I was wondering if you could doublecheck this question please:

g) $\displaystyle y=2x$ and$\displaystyle y = 2x^2$
y = 4 and x = 2

no

How would I do this?

h)$\displaystyle y= 6 - x^2$ and $\displaystyle y = 4x +1$

g)

$\displaystyle 2x = 2x^2$

$\displaystyle 2x - 2x^2 = 0$

$\displaystyle 2x(1 - x) = 0$

$\displaystyle x = 0$ and $\displaystyle x = 1$

h)

$\displaystyle 6-x^2 = 4x+1$

$\displaystyle 0 = x^2 + 4x - 5$

factor and use the zero product property to solve for x
• Oct 13th 2011, 01:48 AM
sbhatnagar
Re: Simultaneous Equations.
In case a graph helps...

g)$\displaystyle y=2x,y=2x^2$. The red line is $\displaystyle y=2x^2$ and the blue line is $\displaystyle y=2x$.

Clearly solutions are $\displaystyle x=0,1$

h)$\displaystyle y=6-x^2,y=4x+1$. The red line is $\displaystyle y=4x+1$ and the blue line is $\displaystyle y=6-x^2$.

Clearly solutions are $\displaystyle x=1,-5$
Yet another way to do (g): $\displaystyle y=2x$ and $\displaystyle y=2x^2$:
If x and y are not 0, $\displaystyle \frac{y}{y}= /frac{2x^2}{2x}$, [tex]1= x[/itex]. And if x= 1, what is y?