Solving a system of nonlinear equations: Problem type 1

Solve the following system of equations.
\[
\left\{\begin{array}{l}
y-4 x=8 \\
20 x^{2}+26 x-y=-14
\end{array}\right.
\]

If there is more than one solution, enter additional solutions with the "or" button. If there is no real solution, use the "No solution" button.
\[
(x, y)=(\mathbb{1}:[)
\]

Step 1 ：The system of equations is: \(y - 4x = 8\) and \(20x^2 + 26x - y = -14\)

Step 2 ：From equation 1), we can express y in terms of x: \(y = 4x + 8\)

Step 3 ：Substitute y into equation 2): \(20x^2 + 26x - (4x + 8) = -14\)

Step 4 ：Simplify the equation: \(20x^2 + 22x + 6 = 0\)

Step 5 ：This is a quadratic equation in the form \(ax^2 + bx + c = 0\). We can solve for x using the quadratic formula \(x = [-b ± \sqrt{b^2 - 4ac}] / (2a)\)

Step 6 ：Substitute the values into the formula: \(x = [-22 ± \sqrt{(22)^2 - 4*20*6}] / (2*20)\)

Step 7 ：Simplify the equation: \(x = [-22 ± \sqrt{4}] / 40\)

Step 8 ：So the solutions for x are: \(x1 = (-22 + 2) / 40 = -0.5\) and \(x2 = (-22 - 2) / 40 = -0.6\)

Step 9 ：Substitute \(x1\) and \(x2\) into the equation \(y = 4x + 8\) to find the corresponding y values

Step 10 ：For \(x1 = -0.5\), \(y1 = 4*(-0.5) + 8 = 6\)

Step 11 ：For \(x2 = -0.6\), \(y2 = 4*(-0.6) + 8 = 5.6\)

Step 12 ：So the solutions to the system of equations are: \(\boxed{(x, y) = (-0.5, 6)}\) or \(\boxed{(x, y) = (-0.6, 5.6)}\)