Answer:
Given the equations:
[tex]x^2+y^2 =36[/tex] .....[1]
[tex]x =2y +6[/tex] ....[2]
Substitute the value of x in [1] we get;'
[tex](2y+6)^2+y^2 =36[/tex]
Use identity: [tex](a+b)^2= a^2+2ab+b^2[/tex]
[tex]4y^2+36+24y + y^2 =36[/tex]
Combine like terms;
[tex]5y^2+36+24y=36[/tex]
Subtract 36 from both sides we get;
[tex]5y^2+24y=0[/tex]
[tex]y (5y+24) = 0[/tex]
By zero product property, we get;
y = 0 and [tex]y =- \frac{24}{5} = -4.8[/tex]
Substitute these y values in [2] to get x values;
For y = 0 we have;
x = 2(0) +6 = 0+6 = 6
For x = -4.8
x = 2(-4.8)+6 = -9.6 + 6 = -3.6
Therefore, the solution for the given equations are; (6, 0) and (-3.6, -4.8)
