Question

Solve the system of equations y = 40x2 and y = 19x + 3 algebraically.

Answer 1

We have been given a system of nonlinear equations.

$\\y=40x^{2},y=19x+3$

In order to solve this system we can first equate the two equations in order to get a quadratic in x.

$\\40x^{2}=19x+3$

Our next step is to bring all the terms on one side.

$\\40x^{2}-19x-3=0$

Now we have to solve this equation. We can solve it by factoring using the splitting middle term method.

$\\40x^{2}-24x+5x-3=0$

$\\8x(5x-3)+(5x-3)=0$

$\\(8x+1)(5x-3)=0$

Upon setting each of these factors equal to zero using zero product property we get

$\\8x+1=0 \text{ or } 5x-3 = 0$

Upon solving both these equations we get

$\\x=-\frac{1}{8} \text{ or } x=\frac{3}{5}$

Now we can substitute these values of x in the equation

$y=19x+3$

We get

$\text{For }x=- \frac{1}{8} \Rightarrow y=19(- \frac{1}{8})+3= \frac{5}{8}$

$\text{For }x= \frac{3}{5}\Rightarrow y=19( \frac{3}{5})+3= \frac{72}{5}$

Therefore, our final set of solutions are

$(x,y)=(- \frac{1}{8},\frac{5}{8}) \text{ and } ( \frac{3}{5}, \frac{72}{5})$

Answer 2

Plug it in So you should get 19x+3=40x2 then solve for x