Solve the system.
{x^2 + y^2 = 85
{3x + y = 15
If you re-arrange 3x + y = 15 into y = -3x + 15 then sub it into the first equation you can solve for y : )
(-3x+15)^2 + y^2 = 85
(-3x +15)(-3x + 15) + x^2 = 85
(9x)^2 - 90x + 225 + x^2 = 85
(10x)^2 - 90x +225 - 85 = 0
either factor or use the quadratic formula to solve for x, then sub it back into 3x +y =15 and solve for y : )
To solve the system of equations {x^2 + y^2 = 85} and {3x + y = 15}, we can start by solving one equation for one variable and then substitute that value into the other equation. Let's solve the second equation for y:
3x + y = 15
Subtract 3x from both sides:
y = 15 - 3x
Now we can substitute this value of y into the first equation:
x^2 + (15 - 3x)^2 = 85
Expand and simplify:
x^2 + (225 - 90x + 9x^2) = 85
Combine like terms:
10x^2 - 90x + 140 = 0
Divide by 10 to simplify:
x^2 - 9x + 14 = 0
Now we can factor this quadratic equation:
(x - 7)(x - 2) = 0
Setting each factor equal to zero, we get:
x - 7 = 0 or x - 2 = 0
Solving for x:
x = 7 or x = 2
Now substitute these values of x back into the equation y = 15 - 3x to find the corresponding values of y:
For x = 7:
y = 15 - 3(7) = 15 - 21 = -6
For x = 2:
y = 15 - 3(2) = 15 - 6 = 9
So the solutions to the system of equations are:
(x, y) = (7, -6) and (2, 9)