Given y=x+4 and y=x^2 +4x, find x and y.

y=x+4 ...............(1)

...........(2)

plugging value of y from (1) in (2)

from here we get two values of x as

x-1=0, x=1 and x+4=0, x=-4

plugging x=1 in equation (1)

y=x+4 =1+4=5

plugging x=-4 in equation (1) , we get

y=x+4 =4-4 =0

so for x=1, y=5 and x=-4, y=0

x=1, y=5 and x=-4, y=5 points are (1,5) and (-4,5)

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places