Simultaneous Equation?

x^2 +y^2=25 (where ^2 means squared)
y=x-7

x^2 + x^2 +49 - 14x = 25

2x^2 -14 x +24=0

x^2 -7x + 12
x^2 -3x-4x +12 =0

x( x-3) -4 (x-3)=0

x=3 , x=4
y=-4 ,y=-3

AS

1. subsitute y = x - 7 into the first equation -i.e. replace y:
x^2 + (x-7)^2 = 25
x^2 + x^2 - 14x + 49 = 25
2x^2 - 14x + 24 = 0
2. Factorise the new function:
(2x - 6)(x - 4) = 0
3. As this equals 0, and anything multiplied by 0 = 0, each bracket must equal 0 at some point:
2x - 6 = 0 -> x = 3
x - 4 = 0 -> x = 4
4. Subsitute each value of x into y = x -7, to the values for y:
y = 3 - 7 = -4
y = 4 - 7 = -3

x² + y² = 25
y = x-7

Substitute (x-7) in place of y in the first equation

x² + (x-7)² = 25
(x-7)² = (x-7)(x-7) = x² -14x + 49

Solve for x

x² + x² -14x + 49 = 25
2x² - 14x + 24 = 0
2(x² -7x + 12) = 0

Now factor the quadratic

2(x - 3)(x-4) = 0

x = 3, x = 4
y = -4, y = -3

remember y = x-7, so when x = 3, y = 3 - 7, which is -4

x² + (x - 7)² = 25
x² + x² - 14x + 49 = 25
2x² - 14x + 24 = 0
x² - 7x + 12 = 0
(x - 4)(x - 3) = 0
x = 4 , x = 3
y = - 3 , y = - 4

(4 , - 3) , (3 , - 4)

The second equation y=x-7 tells you what y equals, so what you do is substitute this value into the first equation:

x^2 + (x-7)^2 = 25

Multiply this out to give you x^2 + x^2 - 14x + 49 = 25 which simplifyies to give 2x^2 -14x + 24 = 0

This is a quadratic, you can solve it by factorising it to
(2x - 8)(x - 3) = 0

This means that one of the factors has to equal 0 (as their product is 0)

Therefore either
(2x - 8) = 0 which means that x = 4
or (x - 3) = 0 which means that x = 3

Then, one at a time, sub both these values into the simplist inital equation i.e. y = x - 7:

y = 4 - 7 which means y = -3
and
y = 3 - 7 which means y = -4

Hope this helped. If you need anymore help try bbc bitesize.

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

x^2 + y^2 = 25
x^2 = 25 - y^2
x = √(25 - y^2)

y = x - 7
y = √(25 - y^2) - 7
y + 7 = √(25 - y^2)
y^2 + 14y + 49 = 25 - y^2
y^2 + y^2 + 14y + 49 - 25 = 0
2y^2 + 14y + 24 = 0
y^2 + 7y + 12 = 0
(y + 4)(y + 3) = 0

y + 4 = 0
y = -4

y + 3 = 0
y = -3

∴ y = -4 , -3

x^2 + y^2 = 25
x^2 + (-4)^2 = 25
x^2 + 16 = 25
x^2 = 25 - 16
x^2 = 9
x = ±√9
x = ±3

x^2 + y^2 = 25
x^2 + (-3)^2 = 25
x^2 + 9 = 25
x^2 = 25 - 9
x^2 = 16
x = ±√16
x = ±4

∴ (x = ±3 , y = -4) , (x = ±4 , y = -3)

1. substitute (x-7) for y in equation #1
2. solve equation #1 for x (its a quadratic so has 2 solutions)
3. substitute the value for x found in step #2 into equation #2
4. solve equation #2 for y (twice, once for each possible x)

step #1:
x^2 + (x-7)^2 = 25

step #2:
x^2 + x^2 - 14x +49 = 25
2x^2 -14x + 49 = 25
2x^2 - 14x + 24 = 0
x^2 - 7x + 12 = 0
(x-4)(x-3) = 0

Therefore x = 4 or x = 3

step #3:
y = x - 7
a) y = 4 - 7
b) y = 3 - 7

step #4
a) y = -3
b) y = -4

Therefore:
x = 4 and y = -3 OR x = 3 and y = -4

x² + y² = (5)². Since the value of y = (x-7), you can substitute in the equation. By doing this you get:
x² + (x-7)² = (5)².
x² + [x² - 14x + 59] = 25.
2x² - 14x = 25 - 49.
2x² - 14x = -24. Since the Left Hand Side and the Right Hand Side have a common multiple of 2, just divide the whole equation by 2. By doing this, you get:
x² - 7x = -12, which equals:
x(x - 7) = -12. So, x - 7 = (-12/x).
From here onwards, I don't have any ideas on how to continue solving this problem. You've 2 sets of option, right? Just substitute their values in my final answer. By doing this you'll get, x=3, y=-4. (First Option). I tried my best. Hope this helps for your daughter.

This is called Equation and inequasion. This is 2 unit maths.

Here is a little help, The opposite of plus is minus, the opposite of multiplication is division.

So, if I had x+5=10 then you have to MINUS (the opposite of the power) 10 (the answer) by 5 (and you get 5)

But with multiplication, it is much different.