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From the second equation, you express y = 4-x and substitute it into the first equation, replacing "y" there
4 - x = -x^2 + 2x + 4.
Now you just have one single quadratic equation.
Reduce it to the standard form quadratic equation, collecting all the terms in one side (on the left, in this case)
x^2 - 3x = 0.
Solve by factoring
x*(x-3) = 0.
The roots are x= 0 and x= 3.
The solutions to the system are (x,y) = (0,4) and (x,y) = (3,1). ANSWER
You can check the answer by substituting the found solution values of x and y into the original system.
Solved and explained in all details.
Is everything clear to you ?
If you still have questions, do not hesitate to ask me.
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If you want to see many other similar solved problems, look into the lesson
- Solving systems of algebraic equations of degree 2 and degree 1
in this site.