I have many problems.First I have to solve an equation algebraically using a number line.I know how to do it but one problem is tricky. 0 is less than or equal to -x squared -3x-70 how can I answer this?

First, it's easier if you switch it around: -x^2 - 3x - 70 is greater than or equal to 0. Then put the -x^2 - 3x - 70 in factored form. You'll have to multiply everything through by -1 in order to eliminate the leading negative which'll switch the inequality around to less than or equal to. Once you've got it factored, find the roots and do a sign analysis by using test values (you'll need three) from the intervals formed by the roots. Since the product of the factors is less than or equal to 0, you'll want the factors to have opposite signs. Your solution consists of those sets of values of x that cause the factors to have opposite signs.

I have many problems.First I have to solve an equation algebraically using a number line.I know how to do it but one problem is tricky. 0 is less than or equal to -x squared -3x-70 how can I answer this?
What

What

0 <= -x**2 -3x - 70, so x**2 + 3x + 70 >+ 0. Now the discriminate of this quadratic is negative, so it has no real root. Hence it is either always >+ 0 or always <= 0. Since it it >= 0 when x = 0, it is always >= 0

0 is less than or equal to -x squared -3x-70



Elisha Gray is correct, but I thought I'd lay it out in LaTeX to try to make it clearer.

0\,\le\,-x^2\,-3x\,-70

if you multiply both sides by a negative number (-1), the sign will change:

0\,\gt\,x^2\,+3x\,+70

I'm not sure if you know what the descriminate is but the roots of the equation are given as follows:

For the equation

0 = ax^2 + bx + c

\large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The descriminate is the part beneath the square root sign. Since a is 1, and c is 70, b^2-4ac is always negative so only imaginary roots are possible for the equation.

find the ordered triple of these equations.
x-2y+2z=-2
2x+3y-z=0
3x+2y+3z=-15

Hello latevia. Do you want to solve these equations?

And note that you should have made a new thread for your question.

Post back.