Answer:
[tex]-\frac{(2q-p)^{2}}{16} +\frac{2q-p}{4}=q[/tex]
Step-by-step explanation:
The given equations are:
[tex]-2y^{2}-2y=p[/tex] Equation 1
[tex]-y^{2}+y=q[/tex] Equation 2
Multiplying the Equation 2 with -2, we get:
[tex]-2(-y^{2}+y)=-2(q)\\[/tex]
[tex]2y^{2}-2y=-2q[/tex] Equation 3
Adding Equation 1 and Equation 3, we get:
[tex]-2y^{2}-2y + (2y^{2}-2y)=p+(-2q)\\\\ -2y^{2}-2y + 2y^{2}-2y=p-2q\\\\ -4y=p-2q\\\\ y= \frac{p-2q}{-4}\\\\ y=\frac{2q-p}{4}[/tex]
Using this value of y, in either of the equations will give the relation independent of y.
Using the value of y in Equation 2, we get:
[tex]-(\frac{2q-p}{4})^{2}+(\frac{2q-p}{4} )=q\\\\ -\frac{(2q-p)^{2}}{16} +\frac{2q-p}{4}=q[/tex]
