The first equation is [tex] y-2x+1=0 [/tex]
[tex] y=2x-1 [/tex] (Equation 1)
The second equation is [tex] 4x^{2}+3y^{2}-2xy=7 [/tex] (Equation 2)
Putting the value of x from equation 1 in equation 2.
we get,
[tex] 4x^{2}+3(2x-1)^{2}-2x(2x-1)=7 [/tex]
[tex] 4x^{2}+3(4x^{2}+1-4x)-4x^{2}+2x=7 [/tex]
by simplifying the given equation,
[tex] 12x^{2}-10x-4=0 [/tex]
[tex] 6x^{2}-5x-2=0 [/tex]
Using discriminant formula,
[tex] D=b^{2}-4ac [/tex]
[tex] D=25-4 \times 6 \times -2 = 73 [/tex]
Now the formula for solution 'x' of quadratic equation is given by:
[tex] x=\frac{-b+\sqrt{D}}{2a} and x=\frac{-b-\sqrt{D}}{2a} [/tex]
[tex] x=\frac{5+\sqrt{73}}{12} and x=\frac{5-\sqrt{73}}{12} [/tex]
Hence, these are the required solutions.
