Solve by Substitution x+y+1=0 , x^2+y^2+2y-3x=-1

Problem

{[x+y+1=0],[x2+y2+2*y−3*x=−1])

Solution

1. Isolate a variable in the linear equation. We can solve the first equation for y in terms of x

y=−x−1

2. Substitute the expression for y into the second equation.

x2+(−x−1)2+2*(−x−1)−3*x=−1

3. Expand the squared term and distribute the constants.

x2+(x2+2*x+1)−2*x−2−3*x=−1

4. Combine like terms to simplify the quadratic equation.

2*x2−3*x−1=−1

5. Solve for x by adding 1 to both sides and factoring.

2*x2−3*x=0

x*(2*x−3)=0

x=0* or *x=3/2

6. Find the corresponding y values by substituting the x values back into the linear equation y=−x−1

If *x=0⇒y=−(0)−1=−1

If *x=3/2⇒y=−(3/2)−1=−5/2

Final Answer

(0,−1),(3/2,−5/2)

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