Answer:
The solution to the system of equations is x=1/5 y= 0
Step-by-step explanation:
solving this system of equation using substitution method
5x-y=1 ….. equation (1)
-10x + 2y =-2 ….. equation (2)
Solving for value of x from equation 2
-10x =-2-2y
so x= -2-2y/-10 = - 2(1+y)/-10=
so x = 1+y/5......... equation 3
The second equation is solved for x, so we can substitute the expression 1+y/5 in for x in the first equation (1)
-10(1+y)/5+2y=-2
-2(1+y) + 2y = -2
-2-2y= -2
-2y= -2+2
y=0
Putting value of y in equation 3
x = 1+y/5.
x= 1+ 0/5 = 1/5
The solution to the system of equations is x=1/5 y= 0
We can check our work by plugging these numbers back into the original equations. Let's try 5x-y=1
5(1/5)-0=1
1 -0= 1
1=1
Yes, our solution checks out.
Answer:
Step-by-step explanation:
Radicals and imaginary numbers ALWAYS come in pairs when it comes to factors of polynomials. This is the called the conjugate theorem. If we are given a solution/root/zero that is
x = 3 + √5, then its conjugate is x = 3 - √5. Going backwards from the solution to the factor, we utilize the Zero Product Property and get
(x - (3 - √5)) which simplifies to (x - 3 + √5). if you are looking for the conjugate of the given zero, the choice you want is the second one down.
Answer:
X=6
Step-by-step explanation:
5x6=30+15
The equation for the line is x= -4 because your line is going through the x axis
