Question

question #7 solve each system of equations. 4x² - 5x - 2y + 20 = 0 -x + y - 4 = 0 (2, 5), (-10, 5) (5, -8) no solution (-10, 5), (5, -8) question #8 solve each system of equations. -x² + 3x - 3y + 28 = 0 x - y = -4 (4, 8) no solution (-4, 0) (-4, 0), (4, 8)

Explanation:

Step1: Solve for y in the linear - equation of Question #7

From \(-x + y-4=0\), we get \(y=x + 4\).

Step2: Substitute y into the quadratic equation

Substitute \(y=x + 4\) into \(4x^{2}-5x-2y + 20=0\). Then \(4x^{2}-5x-2(x + 4)+20=0\). Expand it: \(4x^{2}-5x-2x-8 + 20=0\), which simplifies to \(4x^{2}-7x + 12=0\). The discriminant \(\Delta=b^{2}-4ac\) where \(a = 4\), \(b=-7\), \(c = 12\). \(\Delta=(-7)^{2}-4\times4\times12=49-192=-143<0\), so there is no real - solution for Question #7.

Step3: Solve for y in the linear - equation of Question #8

From \(x-y=-4\), we get \(y=x + 4\).

Step4: Substitute y into the quadratic equation

Substitute \(y=x + 4\) into \(-x^{2}+3x-3y + 28=0\). We have \(-x^{2}+3x-3(x + 4)+28=0\). Expand it: \(-x^{2}+3x-3x-12 + 28=0\), which simplifies to \(-x^{2}+16=0\). Rearrange to \(x^{2}=16\), so \(x=\pm4\).
When \(x = 4\), \(y=x + 4=8\); when \(x=-4\), \(y=x + 4=0\).

Answer:

Question #7: No solution
Question #8: (-4, 0), (4, 8)