Answer:
(4,0) and (1,-3) are the solutions sets.
Step-by-step explanation:
Given that:
y = x²-4x Eqn 1
y = x - 4 Eqn 2
In substitution method, the equation is solved for one variable and then put in the other.
Putting value of y from Eqn 2 in Eqn 1
x - 4 = x²- 4x
Adding -x+4 on both sides
[tex]x-4-x+4=x^2-4x-x+4\\0=x^2-5x+4\\x^2-5x+4=0[/tex]
Factorizing the equation;
[tex]x^2-4x-x+4=0\\x(x-4)-1(x-4)=0\\(x-4)(x-1)=0[/tex]
Either,
x-4 = 0, => x=4
Or,
x-1=0, => x=1
Putting x=4 in Eqn 2
y = 4-4
y = 0
Putting x=1 in Eqn 2
y = 1-4
y = -3
Hence,
(4,0) and (1,-3) are the solutions sets.
