Answer:
The solution is (3,13) and (-1,-3). So none of the mentioned options is correct.
Step-by-step explanation:
Given that
[tex]y = 4x + 1 (i)[/tex]
[tex]y = x^2 + 2x -2 (ii)[/tex]
Now, by susbstituting the value of 'y' from equation i to equation ii, we get
[tex]4x+1=x^2 + 2x -2[/tex]
[tex]0=x^2 + 2x -2-4x-1[/tex]
[tex]0=x^2 + (2x -4x) +(-2-1)[/tex]
[tex]0=x^2 + (-2x) +(-3)[/tex]
[tex]0=x^2 -2x -3[/tex]
[tex]x^2 -2x -3 = 0 (iii)[/tex]
Now by factorization, equation iii can be written as
[tex]x^2 -3x +x -3 = 0[/tex]
[tex]x(x -3) + 1(x -3) = 0[/tex]
[tex](x -3)(x +1) = 0[/tex]
x = 3 and x = -1
By putting the values of x in equation i, we get
y = 4(3) + 1
y = 12 +1
y = 13
and
y = 4(-1) + 1
y = -4 +1
y = -3
Therefore, the solution is (3,13) and (-1,-3). So none of the mentioned options is correct.
