SOLUTION: Stuck on how to solve this simultaneous equation X^2+y^2=25 X+y=7 ASAP please Thank you

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Question 967820: Stuck on how to solve this simultaneous equation
X^2+y^2=25
X+y=7
ASAP please
Thank you

Found 3 solutions by checkley77, MathLover1, Theo:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
X^2+y^2=25
X^2=25-Y^2
X+y=7
X=7-Y
(7-Y)^2=25-Y^2
49-14Y+Y^2=25-Y^2
49-25-14Y+Y^2+Y^2=0
2Y^2-14Y+24=0
2(Y^2-7Y+12)=0
Y^2-7Y+12=0
(Y-4)(Y-3)=0
Y-4=0
Y=4 ANS.
Y-3=0
Y=3 ANS
IF Y=4 THEN
X+4=7
X=7-4
X=3
IF Y=3 THEN
X+3=7
X=7-3
X=4
PROOF:
3^2+4^2=25
9+16=25
25=25

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2By%5E2=25......eq.1
x%2By=7......eq.2
_____________________
x%2By=7......eq.2......solve for y
y=7-x.....substitute in eq.1

x%5E2%2B%287-x%29%5E2=25......eq.1...solve for x
x%5E2%2B7%5E2-14x%2Bx%5E2=25
2x%5E2-14x%2B49=25
2x%5E2-14x%2B49-25=0
2x%5E2-14x%2B24=0...simplify, both sides divide by 2
x%5E2-7x%2B12=0
x%5E2-4x-3x%2B12=0
%28x%5E2-4x%29-%283x-12%29=0
x%28x-4%29-3%28x-4%29=0
%28x-3%29%28x-4%29=0
solutions:
if %28x-3%29=0=>x=3
if %28x-4%29=0=>x=4
substitute in y=7-x and find y
y=7-x if x=3
y=7-3
y=4
y=7-x if x=4
y=7-4
y=3
so, your solutions are:
x=3 and y=4
or
x=4 and y=3




Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
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:



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