Hello from MrBillDoesMath!
Answer: b
Discussion:
7b + 3 - 4b = 3 - 3(b+4) =>
7b + 3 - 4b = 3 - 3b -12 =>
(7b - 4b) + 3 = -3b - 9 =>
3b + 3 = -3b -9 => Add 9 to both sides
3b + 12 = -3b -9 + 9 =>
3b + 12 = -3b => ( add 3b to both sides)
6b + 12 = 0 => (subtract 12 from both sides)
6b = -12 =>
b = -12/6 = -2
Thank you,
MrB
Answer:
[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]
Step-by-step explanation:
x³+216y³+8z³-36xyz
x³+(6y)³+(2z)³-3×6×2×xyz
As we know
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
Let a=x
b=6y
c=2z
Now.
[x+6y+2z][(x²+(6y)²+(2z)²-x×6y-6y×2z-x×2z]
[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]
Answer:
You are missing some of the equation my dude
Step-by-step explanation:
Step-by-step explanation:
To solve a system of equations, we can add the two equations and solve for one of the remaining variables -- let's try to eliminate the 
variable when we add the two equations together.
Right now, there's a 
term in the first equation, and a 
term in the second equation, so if we add those together, we'll be able to eliminate the variable altogether and solve for 
.
However, when we also have a 
term in the first equation and 
term in the second equation, so adding these together will also eliminate the term, leaving a 
on the left-hand side of the equation.
If we add the two numbers on the right side of the equation, we get 
, which does not equal , meaning there are no solutions to this system of equations.
