The solution set is x = 3, y = 4 (or) x = 3, y = –4.
Solution:
Given system of algebraic equations are
[tex]y^{2}+(x-8)^{2}=41[/tex] – – – – – (1)
[tex]y^{2}-25=-x^{2}[/tex] – – – – – (2)
Expand equation (1) using algebraic identity: [tex](a-b)^2=a^2-2ab+b^2[/tex]
[tex]y^{2}+x^2-16x+64=41[/tex]
subtract 64 from both sides of the equation
[tex]y^{2}+x^2-16x+64-64=41-64[/tex]
[tex]y^{2}+x^2-16x=-23[/tex] – – – – – (3)
Now, to arrange equation (2) in order, add [tex]x^2[/tex] on both sides.
[tex]y^{2}-25+x^2=-x^{2}+x^2[/tex]
[tex]y^{2}-25+x^2=0[/tex]
Add 25 on both sides of the equation,
[tex]y^{2}+x^2=25[/tex] – – – – – (4)
To solve this subtract equation (4) from equation (3)
[tex]\Rightarrow y^{2}+x^2-16x-(y^{2}+x^2)=-23-25[/tex]
[tex]\Rightarrow y^{2}+x^2-16x-y^{2}-x^2=-23-25[/tex]
[tex]\Rightarrow -16x=-48[/tex]
Divide both sides of the equation by –16,
⇒ x = 3
Substitute x = 3 in equation (4), we get
[tex]\Rightarrow y^{2}+3^2=25[/tex]
[tex]\Rightarrow y^{2}=25-9[/tex]
[tex]\Rightarrow y^{2}=16[/tex]
[tex]\Rightarrow y=\pm 4[/tex]
i. e. y = 4 (or) y = –4
The solution set is x = 3, y = 4 (or) x = 3, y = –4.
