Go to the website symbolab.com
or Mathpap.com and that should help give you your answer
The correct question is
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?
we have that
e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1
then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10
the answer is x=1.10
First, using elimination method for eq. 1 and 2 to eliminate one of the variables.
2x - y + 3z = 8
-2x + 3y - 2z = 10
------------------------------
2y + z = 18
z = 18 - 2y
Now eq. 1 and 3, but multiple eq.1 with 2.
2x + 4y - 2z = 8
-2x + 3y - 2z = 10
----------------------------
7y - 4z = 18
Now you have two new equations 7y -4z = 18 and z = 18 - 2y. Solve for y and z, using substitute method.
7y - 4z = 18
7y - 4(18 - 2y) = 18
7y - 72 + 8y = 18
15y = 18+72
y = 90/15
y = 6
Find z using one of the two new equations.
z = 18 - 2y
z = 18 - 2(6)
z = 18 - 12
z = 6
Now you got y and z, find x using any one of the main three equations.
x + 2y - z = 4
X + 2(6) - 6 = 4
x + 12 - 6 = 4
x = 4 - 6
x = -2
Solutions:
x = -2
y = 6
z = 6
