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-2 as well, are there any more?

+/- \(i\sqrt{8}\)

Two equations    two unknowns

a+b = a/b

a*b = a/b     or   a =  a (1/b^2)

                           1 = 1/b^2            b= 1 or -1

a + 1 = a /1

a+1 = a                 no answer for when b = 1

a-1 = a /-1

a-1 = -a

2a = 1       a = 1/2               b = -1           then a- b  =   1/2 - -1 = 1 1/2

Nice Chrissy!

we find the slope of the line:

\(\begin{align*} 4x+3y&=7\\ 3y&=-4x+7\\ y&=-\frac{4}{3}x+7 \end{align*}\)

and plug in the \(x\) and \(y\) of \((2, -11)\) to find the \(y \)-intercept:

\(\begin{align*} y &= -\frac{4}{3}x + b \\ -11 &= -\frac{4}{3}(2)+b \\ -11 &= -\frac{8}{3} + b \\ b &= -\frac{25}{3} \end{align*}\)

and plug in the slope for 

\(\boxed{y=-\frac{4}{3}x-\frac{25}{3}}.\)

The domain is all of the 'x' values which are allowed in the expression....

   In this example, you cannot have a zero denominator    so  x = -8 would not be allowed

       can you see another 'x' that is not allowed? (which would result in a zero denominator?)

By "numbers" do you "integers", or did you mean "digits" ?

LOL nice!

I think.. IDK?

The product of al possible values of t is (3)(-17) = -51.

a. y = -21/2.

b. The x-intercept is (14/3,0).

a = 7/4 and b = 7.

Ok I tried it again and I got 84

Is that correct?

What do you mean wrong?

 There are \(48 + 20 = 68 \) ways

There are:  5005  5014  5023  5032  5041  5050  5104  5113  5122  5131  5140  5203  5212  5221  5230  5302  5311  5320  5401  5410  5500  Total =  21 such numbers.

Of course!!!

No problem!

Thanks!