If y= x+9 ______3-xfind the value of y when x=5
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I will go about solving this using the elimination method.
First, convert the equations.
10x + y = -20
4x + y = -12
Second, find the easiest variable to get rid of and get rid of it! (In this case, y) We will subtract to get rid of y.
6x = -8
Third, you want to solve the equation.
6x = -8 (divide by 6)
x =
Fourth, solve for y by inserting the answer for x into one of the equations.
10() + y = -20
+ y = -20 (subtract )
y =
The solution for this system of equations is (, ).
First, convert the equations.
10x + y = -20
4x + y = -12
Second, find the easiest variable to get rid of and get rid of it! (In this case, y) We will subtract to get rid of y.
6x = -8
Third, you want to solve the equation.
6x = -8 (divide by 6)
x =
Fourth, solve for y by inserting the answer for x into one of the equations.
10(
y =
The solution for this system of equations is (
That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
namely if sec(s) = cos(s)
then solving for "s"
thus