How do you solve the system of equations {(y =7 - x), (x^2 + y^2 = 65):}?

Mar 15, 2017

There are two points that are solutions, $\left(8 , - 1\right) \mathmr{and} \left(- 1 , 8\right)$

Explanation:

Given:

${y}^{2} + {x}^{2} = 65 \mathmr{and} y + x = 7$

Write the line as y as a function of x

${y}^{2} + {x}^{2} = 65 \mathmr{and} y = 7 - x$

Substitute $\left(7 - x\right)$ for y into the circle:

${\left(7 - x\right)}^{2} + {x}^{2} = 65$

Square first term:

$49 - 14 x + {x}^{2} + {x}^{2} = 65$

$2 {x}^{2} - 14 x - 16 = 0$

${x}^{2} - 7 x - 8 = 0$

$\left(x - 8\right) \left(x + 1\right) = 0$

$x = 8 \mathmr{and} x = - 1$

Find the corresponding y values, using the equation, $y = 7 - x$

y = -1 and y = 8

Mar 15, 2017

(8, -1);(-1,8)

Explanation:

From the second equation, we have $y = 7 - x$. Substituting:

${\left(7 - x\right)}^{2} + {x}^{2} = 65$

$49 - 14 x + {x}^{2} + {x}^{2} = 65$

$2 {x}^{2} - 14 x - 16 = 0$

$2 \left({x}^{2} - 7 x - 8\right) = 0$

${x}^{2} - 7 x - 8 = 0$

$\left(x - 8\right) \left(x + 1\right) = 0$

$x = 8 \mathmr{and} - 1$

Since this system involves the equation of a circle (e.g ${x}^{2} + {y}^{2} = 65$), the solution set will be of the form (a, b); (b, a).

The solution set therefore is (8, -1);(-1, 8).

Hopefully this helps!