solve this simultaneous question: y = x - 5 and y = x^2 - 5x?

y = x - 5 ------- 1
y = x² - 5x--- 2

Sub eqn 1 into eqn 2:
x - 5 = x² - 5x
x² - 5x - x + 5 = 0
x² - 6x + 5 = 0
(x - 1)(x - 5) = 0
x = 1, 5 ------ 3

Sub eqn 3 into eqn 1:
x = 1 --> y = (1) - 5 = -4
x = 5 --> y = (5) - 5 = 0

Therefore x = 1, y = -4 or x = 5, y = 0

x - 5 = x² - 5x
x² - 6x = - 5
x² - 3x = - 5 + (- 3)²
x² - 3x = - 5 + 9
(x - 3)² = 4
x - 3 = 2

Values of x:
x = 2 + 3, x = 5
x = - 2 + 3, x = 1

Values of y:
Where x = 5:
= 5 - 5
= 0

Where x = 1:
= 1 - 5
= - 4

Answer: x = 5, y = 0; x = 1, y = - 4

Proof of the 1st set of values in the 2nd equation:
0 = 5² - 5(5)
0 = 25 - 25
0 = 0

Proof of the 2nd set of values in the 2nd equation:
- 4 = 1² - 5(1)
- 4 = 1 - 5
- 4 = - 4

y = x - 5
y = x² - 5x
x - 5 = x² - 5x
0 = x² - 4x + 5
x² - 4x + 5 = 0
(x + 1)(x - 5) = 0

(x + 1) = 0
x = -1
y = -1 -5 = -6
Try in 2nd equation
y = (-1)² - 5(-1) = 1 + 5
y = 6
Doesn't check. So x = -1 does not check

(x - 5) = 0
x = 5
y = 5 - 5 = 0
In 2nd equation
y = 0 - 5(0) = 0
OK
(x, y) = (5, 0) is a solution.

you replace y=x-5 into the other equation : x-5=x^2-5x then you move all to one side : x^2-5x-x+5=0, you do the calculations : x^2-6x+5=0 after that you find the (i don't know how to translate!! in greek is symbolized Δ) D=(-6)^2-4(1)(5)=36-20=16>0 so the equation has two solutions which are calculated by this type : x=[-(-6)+sqrt(16)]/2 and x=[-(-6)-sqrt(16)]/2 and finally you find that the two solutions are x=1 and x=5. then you replace the two solutions : y=1-5=-4 and y=5-5=0 . so the points that solve these two equations are (1,-4) and (5,0). i hope that i helped you

Notice!: the ''D'' is used only when the equation has x^2 . if there is x^3 etc. we use another way to solve the equations.

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

y = x - 5
x^2 - 5x = x - 5
x^2 - 5x - x + 5 = 0
x^2 - 6x + 5 = 0
x^2 - x - 5x + 5 = 0
(x^2 - x) - (5x - 5) = 0
x(x - 1) - 5(x - 1) = 0
(x - 1)(x - 5) = 0

x - 1 = 0
x = 1

x - 5 = 0
x = 5

y = x - 5
y = 1 - 5
y = -4

y = 5 - 5
y = 0

∴ (x = 1 , y = -4) , (x = 5 , y = 0)

x² - 5x = x - 5
x² - 6x + 5 = 0
(x - 5) (x - 1) = 0
x = 1 , x = 5
y = - 4, y = 0

(1,-4) , (5,0)

y= x-5
y= x^2-5x
y=y
x-5=x^2-5x
x+5x-x^2=5
6x-x^2=5
x(6-x) = 5
Can't continue sorry!

y = x - 5
y = x^2 - 5x

It's best to use substitution in cases where one is linear and one is not.
Since y is equal to both of them, just equate them to each other.

x - 5 = x^2 - 5x

Factor the right hand side.

x - 5 = x(x - 5)

Move to the left hand side.

x - 5 - x(x - 5) = 0

Factor (x - 5) from the entire expression.

(x - 5)(1 - x) = 0

Showing that
x = { 5, 1 }

When x = 5, y = 5 - 5 = 0.
When x = 1, y = 1 - 5 = -4

Therefore, we have two sets of solutions:
(5, 0)
(1, -4)