Question

This system has one solution. What is the y-coordinate of the solution?

y = –2x + 4
y = x2 + 4x + 13

Answer 1

`\left \{ {{y=-2x+4} \atop {y= x^(2) +4x+13}} \right. \\ x^(2) +4x+13=-2x+4 \\ x^(2) +6x+9=0 \\ з=6^2-4(9)=36-36=0 \\ x= (-6)/(2) =-3 \\ y=-2x+4 \\ y=-2(-3)+4=6+4=10`

Answer 2

y = -2x + 4
y = x² + 4x + 13

-2x + 4 = x² + 4x + 13
+ 2x + 2x
4 = x² + 6x + 13
- 4 - 4
0 = x² + 6x + 9
x = -(6) ± √((6)² - 4(1)(9))
2(1)
x = -6 ± √(36 - 36)
2
x = -6 ± √(0)
2
x = -6 ± 0
2
x = -6
2
x = -3
y = -2x + 4
y = -2(-3) + 4
y = 6 + 4
y = 10
(x, y) = (-3, 10)

The y - coordinate of the solution is 10.