solve the following pair of simultaneous equation y=x+2 y=x^2?

Answer 1

You have them both in terms of y, so just equate them:
y = x²
y = x + 2

So:
x² = x + 2

Now put everything on one side:
x² - x - 2 = 0

Factor:
(x - 2)(x + 1) = 0

Therefore x = -1 or x = 2

This leads to y values of 1 and 4 respectively:
(-1, 1) and (2, 4)

Double-check:
-1 + 2 =? (-1)²
1 = 1 <-- check

2 + 2 =? 2²
4 = 4 <-- check

Another way to confirm the answer is to graph the lines and see you have two intersection points at (-1, 1) and (2, 4)

Answer 2

2^(x +2) = 4*2^x so we've 4*2^x + 5 = 6/2^x, enable 2^x = y then we get 4y^2 + 5y - 6 = 0 (4y - 3)(y +2) = 0 so y = 3/4 or - 2 so as that 2^x = 3/4,considering 2^x is often effective we discard 2^x = - 2 so we've in basic terms 2^x = 3/4 so x = (log(3/4))/ log 2.

Answer 3

Since both equations have y =, you can set them equal to each other:

x^2 = x + 2

x^2 - x - 2 = 0

(x - 2)(x+1) = 0

x = 2 or x = -1

when x = 2, y = 4
when x = -1, y = 1

Two solutions: (2,4) and (-1,1)

that's it!

Answer 4

so substitute (x+2) for (y) in the second equation to get:

x+2 = x^2

now just solve this quadratic equation:

x^2-x-2=0

(x-2)(x+1)=0

x = 2 or x = -1

so (-1,1) and (2,4) are the solutions.