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 $

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- Oct 12th 2011, 03:18 PMJibblyJabblySimultaneous 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 PMskeeterRe: Simultaneous Equations.
- Oct 13th 2011, 01:48 AMsbhatnagarRe: 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 $\displaystyle y=2x$.

https://lh3.googleusercontent.com/-X...s627/graph.png

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 $\displaystyle y=6-x^2$.

https://lh5.googleusercontent.com/-p...ph%2525202.png

Clearly solutions are $\displaystyle x=1,-5$

The value of y can be calculated just by substituting the value of x or by the graph. - Oct 13th 2011, 12:57 PMHallsofIvyRe: Simultaneous Equations.
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?

That was "if x and y are not 0". It is easy to check that y is 0 if and only if x is and x= y= 0 satifies both equations.