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Find all solutions for the following system of equations. List the solutions as ordered pairs?

x^2+y^2-25=0, 2x-y=-5 How do you go about getting the answer? Just need some help, and preferrably the answer! Any help would be appreciated. Please dotn respond with use a homework site!

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1. the easiest way is to set the left equation equal to y. then plug in and solve. I am running out of time since im at school just IM me and im sure i can help u :)
2. x^2+y^2-25=0, 2x-y=-5 Ans {x= -4, y= -3}, {x= 0, y = 5}
3. The 1st eq is x^2 + y^2 = 25 and the 2nd eq is y = 2x +5 . Now we square the 2nd eq and put the value of y^2 from eq 2 into eq 1. By doing so we get 5x^2 + 20x = 0 => x (x + 4)=0 so x=0 or x = -4 and we get the corresponding values of y by putting the values of x in eq 2 and we get y = 5 and y = -3. The ordered pairs are (0, 5) and (-4,-3).
4. x^2 + y^2 = 25 is the equation of a circle y = 2x + 5 is the equation of a line rewrite the first as x^2 + (2x+5)^2 - 25 = 0, and solve the quadratic for two values of x put these two values into y = 2x+5 to find the corresponding y value
5. x² + y² = 25 2x - y = -5 Solve for y in the second equation (y = expression containing x) and substitute that for y in the first equation and find the value of x. y = 2x + 5 x² + (2x+5)² = 25 x² + 4x² + 20x + 25 = 25 x² + 4x² + 20x = 0 5x² + 20x = 0 5x(x + 4) = 0 x = -4, 0 2x - y = -5 2(0) - y = -5 0 - y = -5 -y = -5 y = 5 (0, 5) 2(-4) - y = -5 -8 - y = -5 y = -3 (-4, -3) --- OR --- x² + y² = 25 (rearranging the first equation) when x=0, y²=25; y=±5 (0, -5) and (0, 5) The two graphs can intersect in up to two points since one is a line, and there is another point with another value of x, so I would go with (0, +5) found using the other equation. (-4)² + y² = 25 16 + y² = 25 y² = 9 y = ±3 Again, one of the possible values has to be chosen, and using the other equation y=-3. (0, 5), (-4, -3)