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 $
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 $\displaystyle x=1,-5$
The value of y can be calculated just by substituting the value of x or by the graph.
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.