Thread: Simultaneous equation

By eliminating y, find the solutions to the simultaneous equations.

x^2+y^2=25
y=x-7

I've tried many different approaches, come up with 4+ different answers so I have no idea what to do.

Originally Posted by Mukilab

By eliminating y, find the solutions to the simultaneous equations.

x^2+y^2=25
y=x-7

I've tried many different approaches, come up with 4+ different answers so I have no idea what to do.

Substitute the second equation into the first:

x^2 + (x - 7)^2 = 25.

Expand, simplify, re-arrange to get a quadratic = 0 and solve for x in the usual way.

If you need more help, please show all your work and say where you get stuck.

Originally Posted by Mukilab

By eliminating y, find the solutions to the simultaneous equations.

x^2+y^2=25
y=x-7

I've tried many different approaches, come up with 4+ different answers so I have no idea what to do.

You should get two pairs of solutions. It tells you to eliminate y so put x-7 in for y in the first equation

When it's expanded and put into standard form we get ......

Solve using your favourite method although this one factorises to .....

From this find your value of x and for each one put the value into either equation to find y.

Got 2x^2-14x+24=0 after expanding and simplifying.

divided by 2

x^2-7x+12=0

put into the quadratic equation got
or

correct?

Sorry how is this calculas.
Actually. What is calculas? All I know that it has some meagre connection to engineering and Newton, I think he created it?

Time for research.

Originally Posted by Mukilab

Got 2x^2-14x+24=0 after expanding and simplifying.

divided by 2

x^2-7x+12=0

put into the quadratic equation got
or

correct?

That is correct but since you can simplify your answers

This is pre-calc rather than calculus and although I don't know how the US system works I'd imagine pre-calc sets you up for calculus. By and large calculus is largely about differentiation (rate of change) and integration (area under a graph). Of course it's far more complicated than that.

Thanks, got x=3,4

Didn't know about the squareroute of 1 being 1, thanks. I guess I just didn't bother to think about that. How silly of me.

Originally Posted by Mukilab

Thanks, got x=3,4

Didn't know about the squareroute of 1 being 1, thanks. I guess I just didn't bother to think about that. How silly of me.

I got those answers too. Now you can use those values to find y using either equation (equation 2 is simplest). You should get two pairs of values