We have been given a system of nonlinear equations.
[tex]\\y=40x^{2},y=19x+3[/tex]
In order to solve this system we can first equate the two equations in order to get a quadratic in x.
[tex]\\40x^{2}=19x+3[/tex]
Our next step is to bring all the terms on one side.
[tex]\\40x^{2}-19x-3=0[/tex]
Now we have to solve this equation. We can solve it by factoring using the splitting middle term method.
[tex]\\40x^{2}-24x+5x-3=0[/tex]
[tex]\\8x(5x-3)+(5x-3)=0[/tex]
[tex]\\(8x+1)(5x-3)=0[/tex]
Upon setting each of these factors equal to zero using zero product property we get
[tex]\\8x+1=0 \text{ or } 5x-3 = 0[/tex]
Upon solving both these equations we get
[tex]\\x=-\frac{1}{8} \text{ or } x=\frac{3}{5}[/tex]
Now we can substitute these values of x in the equation
[tex]y=19x+3[/tex]
We get
[tex]\text{For }x=- \frac{1}{8} \Rightarrow y=19(- \frac{1}{8})+3= \frac{5}{8}\\[/tex]
[tex]\text{For }x= \frac{3}{5}\Rightarrow y=19( \frac{3}{5})+3= \frac{72}{5}\\[/tex]
Therefore, our final set of solutions are
[tex](x,y)=(- \frac{1}{8},\frac{5}{8}) \text{ and } ( \frac{3}{5}, \frac{72}{5})[/tex]
