How do you solve this system of non-linear equations?

x² + y² = 25

2x + 4y = -10

x² + y² = 25

2x + 4y = -10

**When solving the system of those two equations, you are finding the point(s) of intersection of a circle and a straight line.**

**x² + y² = 25**{a circle}

**2x + 4y = -10**{a straight line}

__Isolate a variable on the linear equation:__

2x + 4y = -10 {the second equation}

2x = -4y - 10 {subtracted 4y from each side}

x = -2y - 5 {divided each side by 2}

__Substitute the value of that variable into the non-linear equation:__

x² + y² = 25 {the first equation}

(-2y - 5)² + y² = 25 {substituted (-2y - 5) , in for x, into top equation}

(-2y - 5)(-2y - 5) + y² = 25 {when squaring a binomial, multiply it by itself}

4y² + 20y + 25 + y² = 25 {used foil method}

5y² + 20y + 25 = 25 {combined like terms}

5y² + 20y = 0 {subtracted 25 from each side}

5y(y + 4) = 0 {factored out the common factor, 5y}

5y = 0 or y + 4 = 0 {set each factor equal to 0}

**y = 0 or y = -4**{solved each equation}

x = -2y - 5 {the re-arranged second equation}

__If y = 0:__x = -2(0) - 5 {substituted 0 for y}

x = 0 - 5 {multiplied}

x = -5 {subtracted}

(-5,0) is a solution {a point of intersection of the line and circle}

x = -2y - 5 {the re-arranged second equation}

__If y = -4:__x = -2(-4) - 5 {substituted -4 for y}

x = 8 - 5 {multiplied}

x = 3 {subtracted}

(3,-4) is a solution {a point of intersection of the line and circle}

x = -5 when y = 0 and x = 3 when y = -4

**(-5,0)**and

**(3,-4)**are the points of intersection of the line and circle

*- Algebra House*