tertre

Well, I'll solve one of them.

1. $6x+4x-6=24+9x, 10x-6=24+9x, x=30$, so one solution.

Remember infinite solutions, 0=0, etc.

Thanks, CPhill!

Divide 9n^4+49n^3-7n^2-102n-41 by -4/9, 9  49  -7 -102 -41 divided by -4/9,

Solving synthetic, 9x^3+45x^2-27x-90-1/x-4/9?

Is it wrong?

Wow! Great Solution, CPhill! 

Let x represent the cost of one of Tanya's items, and y represent the cost of one of Tony's items. So, we have y+1.25=x.

So, 4(y+1.25)=5(y), 4y+5=5y, y=5. And, x=5+1.25=$6.25. Tanya-6.25*4=25, Tony-5*5=$25.

Yeah, I didn't use synthetic division. 

1.  25-4x=15-3x+10-x, true for all x, Infinite Solutions... In the end, after rearranging, you get 0=0.

2. 4x+8=2x+7+2x-20, no solutions, because at the end, after rearranging. you attain 0=-21.

$\frac{(80x^3-50x^2+7)}{(8x-5)}$. First, divide the leading coefficients of $80x^3-50x^2+7\mathrm{\:and\:the\:divisor\:}8x-5$, so this means that: 

$\frac{80x^3}{8x}$, which is $10x^2.$ That is our quotient, and now we have to find the remainder. So, multiply 8x-5 by 10x^2, to attain $80x^3-50x^2.$ Now, subtract this result from $80x^3-50x^2+7$ , to get $7$ as our new remainder. Therefore, $\frac{(80x^3-50x^2+7)}{(8x-5)}=\boxed{10x^2+\frac{7}{8x-5}}$.

Wait, is this good? IDK.

12t, 2400 ft, 45m

1t, 2400ft, 540m

9ft, 2400ft, 60m

9ft, 1ft, 1/40m

9ft, 3600ft, 1/40*3600=360/4=$\boxed{90}$ minutes!

-5+7g=3g+9, 7g-3g=14, 4g=14, g=3.5.