How do you solve the system by using substitution: x^2+y^2=5, y= -2x+5?

Answer 1

x² + y² = 5 and y = -2x + 5

So, x² + (-2x+5)² = 5

x² + 4x² - 20x + 25 = 5

5x² - 20x + 20 = 0

x² - 4x + 4 = 0

(x - 2)² = 0 x = 2 and y = -2(2) + 5 = 1

The graphs intersect at the point (2,1).

Answer 2

To solve a system by substitution, we isolate one variable (here we already have y isolated) and insert it into another equation to solve for one variable.

So we have two equations:

(1) x² + y² = 5

(2) y = -2x + 5

If we substitute (2) into (1), we get:

x² + (-2x + 5)² = 5

Now we expand the parenthesis:

x² + (-2x + 5)(-2x + 5) = 5

x² + -4x² -10x - 10x + 25 = 5

-3x² - 20x + 25 = 5

Simplified, this is:

3x² + 20x - 20 = 0

From here, we can use the quadratic equation

x = (-b ± √(b² - 4ac))/2a

x = (-20 ± √(20² - 4(2)(-20)))/2(3)

And simplify from there.