simultaneous equations?

x^2 + (2x-5)^2 = 25
x^2 + 4x^2 - 20x + 25 = 25
5x^2 -20x = 0
x^2 - 4x = 0
x(x - 4) = 0
Therefore x = 0 or x = 4
when x = 0, y = -5
when x = 4, y = 3.
Thus solution is (x,y) = (0,-5) or (4,3).

x ² + (2x - 5) ² = 25

x ² + 4x ² - 20 x + 25 = 25
5 x ² - 20 x = 0
5 x (x - 4) = 0
x = 0 , x = 4
y = - 5 , y = 3

(0,- 5) , (4,3)

x^2 + y^2 = 25 (solve by using substitution)
y = 2x - 5

x^2 + (2x - 5)^2 = 25
x^2 + (2x - 5)(2x - 5) = 25
x^2 + 4x^2 - 20x + 25 = 25
5x^2 - 20x + 25 - 25 = 0
5x^2 - 20x = 0
x^2 - 4x = 0
x(x - 4) = 0

x = 0

x - 4 = 0
x = 4

y = 2x - 5
y = 2(0) - 5
y = -5

y = 2x - 5
y = 2(4) - 5
y = 3

∴ (x = 0 , y = -5) , (x = 4 , y = 3)

x^2+y^2=25

y=2x-5

x^2 +(2x-5)^2=25
x^2+4x^2-20x+25=25
5x^2-20x=0

5x(x-4)=0
x=0 or 4

I needed help with one of these questions the other week

What you do is not put the equations together, but substitue them

You know what y is

so

x^2 + (2x-5)^2 = 25

x^2 + (2x-5)(2x-5) = 25

x^2 +4x^2 - 20x + 25

5x^2 -20x + 25 = 25

Then factorise

(5x - 25)(x - 1) = 25

This then gives two answers for x

x = 1 or x = 5 (25/5)

Then substitute these into one of the first equations

y = 2x - 5

y = (2 X 1) - 5

y = 2 - 5

y = -3


OR

y = (2 X 5) - 5

y = 10 - 5

y = 5


So your final answers are:

x = 1 y = -3

Or

x = 5 y = 5


Hope this helps

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x^2 + y^2 = 25__________(1)
y = 2x - 5______________(2)
substitute equation (2) into equation (1);
x^2 + (2x - 5)^2 = 25
x^2 + 4x^2 - 20x + 25 - 25 = 0
5x^2 - 20x = 0
x^2 - 4x = 0
x(x - 4) = 0
x = 0 ...OR... x = 4

when x = 0,
y = 2(0) - 5
y = 5

when x = 4'
y = 2(4) - 5
y = 8 - 5
y = 3

you change x^2 + y^2 = 25 into a y= equation...


x^2 + y^2 = 25

y^2 = 25 - x^2

y= SQRT(25 - x^2)

y=5 - x <-----This is your new first equation



Then go from there!!