Answer:
The y-coordinate of the solution is, 10
Step-by-step explanation:
Given the system of equation:
[tex]y = -2x+4[/tex] ....[1]
[tex]y=x^2+4x+13[/tex] ....[2]
Equate the equation [1] and [2] we have;
[tex]-2x+4 = x^2+4x+13[/tex]
Add 2x to both sides of an equation:
[tex]4 = x^2+6x+13[/tex]
Subtract 4 from both sides we have;
[tex]0= x^2+6x+9[/tex]
or
[tex]x^2+6x+9=0[/tex]
Using perfect square:
[tex](x+a)^2 = x^2+2ax+a^2[/tex]
⇒We can write the equation as:
[tex]x^2+2 \cdot 3x+3^2=0[/tex]
then;
[tex](x+3)^2 = 0[/tex]
⇒[tex]x+3 = 0[/tex]
Subtract 3 from both sides we have;
x = -3
Substitute value of x in [1] we have;
[tex]y = -2(-3)+4[/tex]
⇒[tex]y =6+4=10[/tex]
Solution for the given system of equation = (-3, 10)
Therefore, the y-coordinate of the solution is, 10
