Solve by Substitution y^2+49y+x^2-5x-50=0 , y+49+(5x-50)/y=0

Problem

{[y^2+49*y+x^2−5*x−50=0],[y+49+(5*x−50)/y=0])

Solution

1. Rearrange the second equation to eliminate the fraction by multiplying the entire equation by y

y*(y+49+(5*x−50)/y)=0*(y)

y^2+49*y+5*x−50=0

2. Isolate the term y^2+49*y in the modified second equation to prepare for substitution.

y^2+49*y=50−5*x

3. Substitute the expression for y^2+49*y into the first equation.

(50−5*x)+x^2−5*x−50=0

4. Simplify the resulting equation by combining like terms.

x^2−10*x=0

5. Factor the quadratic equation to find the possible values for x

x*(x−10)=0

x=0* or *x=10

6. Substitute x=0 back into the isolated equation y^2+49*y=50−5*x to find y

y^2+49*y=50−5*(0)

y^2+49*y−50=0

(y+50)*(y−1)=0

y=−50,y=1

7. Substitute x=10 back into the isolated equation y^2+49*y=50−5*x to find y

y^2+49*y=50−5*(10)

y^2+49*y=0

y*(y+49)=0

y=0,y=−49

8. Check for extraneous solutions by ensuring y≠0 (since y is in the denominator of the original second equation).

y≠0⇒(10,0)* is excluded.

Final Answer

(x,y)∈{(0,1),(0,−50),(10,−49)}

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