Solve Equations

do you mean

[MATH]3x + 8x^2 = 20\\ 45x + 30x^2 = 180[/MATH]
 
These are poly equations...........have you been taught to solve them?
 
I haven’t done this in over 20 years and I’m trying to get a sense of how to solve.
 
You can graph them and see where the x-intercepts land. That's the easiest way to get an answer.

There are other ways if you wish me to elaborate.
 
Romsek said:
do you mean

[MATH]3x + 8x^2 = 20\\ 45x + 30x^2 = 180[/MATH]
Click to expand...

Well, the system is possibly overdetermined. We can solve each one and see if any solutions overlap.

[MATH]8x^2 + 3x - 20 = 0\\ \text{I was never very good at factoring these when the coefficient of $x^2$ isn't one so let's use the quadratic formula}\\ a x^2 + bx + c = 0 \Rightarrow x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\~\\ x = \dfrac{-3 \pm \sqrt{9 + (4)(8)(20)}}{16} = \\ \dfrac{-3 \pm \sqrt{649}}{16} [/MATH]
[MATH]30x^2 + 45x - 180 = 0\\ 2x^2 + 3x - 12 = 0\\ x = \dfrac{-3 \pm \sqrt{9+(4)(2)(12)}}{4} = \\ \dfrac{-3 \pm \sqrt{105}}{4}[/MATH]
Neither of these two quadratic equations have any roots in common so there is no solution to the system of the two equations.
 
vocampo0811 said:
How do I solve for:
3x1 + 8x2 = 20
45x1 + 30x2 = 180
Click to expand...
Is it possible that your problem is actually
[math]3x_1 + 8x_2 - 20[/math][math]45x_1 + 30x_2 = 180[/math]where [math]x_1[/math] and [math]x_2[/math] are two different numbers?

Maybe do this as
[math]3x + 8y = 20[/math][math]45x + 30y = 180[/math]
I'll show you how to start.

Solve, say, the top equation for y:
[math]3x +8y = 20[/math]
[math]y = \dfrac{20 - 3x}{8}[/math]
Now put that into the other equation:
[math]45x + 30 \left ( \dfrac{20 - 3x}{8} \right ) = 180[/math]
[math]45x \cdot 8 + 30 \cdot 20 - 30 \cdot 3x = 180 \cdot 8[/math]
etc. Once you get x then you can put it into either of the original equations to get y.

Try it and see where it takes you.

-Dan
 
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