Question 967820
you have 2 equations.
they are:


x^2 + y^2 = 25
x + y = 7


solve for y in the second equation to get y = 7 - x
replace y in the first equation to get x^2 + (7-x)^2 = 25


now that your first equation is in terms of x only, you can solve for x.
once you solve for x, then use that value to solve for y.


start with x^2 + (7-x)^2 = 25
simplify to get x^2 + 49 - 14x + x^2 = 25
combine like terms to get 2x^2 - 14x + 49 = 25
subttract 25 from both sides of the equation to get 2x^2 - 14x + 24 = 0
divide both sides of the equation by 2 to get x^2 - 7x + 12 = 0
factor that quadratic equation to get (x - 4) * (x - 3) = 0
solve for x to get x = 4 or x = 3.


go back to the second original equation and solve for y using those values of x.
when x = 3, y = 4
when x = 4, y = 3


you have two possible pairs of solutions:


(x,y) = (4,3)
(x,y) = (3,4)


go back to both original equations and see if these solutions make those equation true.


both original equations are:


x^2 + y^2 = 25
x + y = 7


when x = 3 and y = 4, the equations become:


3^2 + 4^2 = 25 which becomes 9 + 16 = 24 which becomes 25 = 25 which is true.
3 + 4 = 7 which becomes 7 = 7 which is true.


when x = 4 and y = 3, the equations becomes:


4^2 + 3^2 = 25 which becomes 25 = 25 which is true.
4 + 3 = 7 which becomes 7 = 7 which is true.


looks like those are your solutions.


the graph of both equations is shown below:


{{{graph(500,500,-1,6,-1,6,7-x,sqrt(25-x^2),-sqrt(25-x^2),90(x-3),90(x-4))}}}


the 2 vertical lines at x = 3 and x = 4 show where the intersection of the graph of the two equations is.