Y = x2 – 14x + 23
y = - 3x + 5

let first equation be  y=x^2-14x+23

second equation be y=-3x+5

now substitute the vale of y in equation second

y=-3x+5

x^2-14x+23=-3x+5

x^2-14x+3x=5-23

x^2-11x=-18

x^2-11x+18=0

x^2-(9+2)x+18=0

x^2-9x-2x+18=0

x(x-9)-2(x-9)=0

(x-2)(x-9)=0

either x-2=0                                        Or, x-9=0

x-2=0

x=0+2

x=2

x-9=0

x=0+9

x=9

therefore x=2,9

Step-by-step explanation:

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(2, - 1 ) and (9, - 22 )

Step-by-step explanation:

Given the 2 equations

y = x² - 14x + 23 → (1)

y = - 3x + 5 → (2)

Substitute y = x² - 14x + 23 into (2)

x² - 14x + 23 = - 3x + 5 ( subtract - 3x + 5 from both sides )

x² - 11x + 18 = 0 ← in standard form

(x - 2)(x - 9) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 2 = 0 ⇒ x = 2

x - 9 = 0 ⇒ x = 9

Substitute these values into (2) for corresponding values of y

x = 2 : y = - 3(2) + 5 = - 6 + 5 = - 1 ⇒ (2, - 1 )

x = 9 : y = - 3(9) + 5 = - 27 + 5 = - 22 ⇒ (9, - 22 )

i used to know how to do this in 7th but i forget easily

step-by-step explanation: