Hey! I'm from connections academy too. Which was the answer??
Im honestly not sure but I’m thinking answer A?
The solution is
and ![(7,-3)](https://tex.z-dn.net/?f=%287%2C-3%29)
Step-by-step explanation:
The expression is
and ![x+y=4](https://tex.z-dn.net/?f=x%2By%3D4)
Using substitution method we can solve the expression.
Let us substitute
in ![x^{2} +y^{2} -x+3y-42=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2By%5E%7B2%7D%20-x%2B3y-42%3D0)
![(4-y)^{2} +y^{2} -(4-y)+3y-42=0](https://tex.z-dn.net/?f=%284-y%29%5E%7B2%7D%20%2By%5E%7B2%7D%20-%284-y%29%2B3y-42%3D0)
Expanding and simplifying the expression, we get,
![\begin{array}{r}{16-8 y+y^{2}+y^{2}-4+y+3 y-42=0} \\{2 y^{2}-4 y-30=0}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Br%7D%7B16-8%20y%2By%5E%7B2%7D%2By%5E%7B2%7D-4%2By%2B3%20y-42%3D0%7D%20%5C%5C%7B2%20y%5E%7B2%7D-4%20y-30%3D0%7D%5Cend%7Barray%7D)
Let us use the quadratic equation formula to solve this equation,
![\begin{aligned}y &=\frac{4 \pm \sqrt{16-4(2)(-30)}}{2(2)} \\&=\frac{4 \pm \sqrt{16+240}}{4} \\&=\frac{4 \pm 16}{4} \\y &=1 \pm 4\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dy%20%26%3D%5Cfrac%7B4%20%5Cpm%20%5Csqrt%7B16-4%282%29%28-30%29%7D%7D%7B2%282%29%7D%20%5C%5C%26%3D%5Cfrac%7B4%20%5Cpm%20%5Csqrt%7B16%2B240%7D%7D%7B4%7D%20%5C%5C%26%3D%5Cfrac%7B4%20%5Cpm%2016%7D%7B4%7D%20%5C%5Cy%20%26%3D1%20%5Cpm%204%5Cend%7Baligned%7D)
Thus,
and ![y=-3](https://tex.z-dn.net/?f=y%3D-3)
Substituting y-values in the equation
, we get the value of x.
For
⇒ ![x=4-5=-1](https://tex.z-dn.net/?f=x%3D4-5%3D-1)
For
⇒ ![x=4+3=7](https://tex.z-dn.net/?f=x%3D4%2B3%3D7)
Thus, the solution set is
and ![(7,-3)](https://tex.z-dn.net/?f=%287%2C-3%29)