Three times the 1st number plus the 2nd number plus twice the 3rd is 5 is the same as 3x+y+2z=5. If three times the 2nd number is subtracted from the sum of the 1st and three times the 3rd number, the result is 2 is just x+3z-3y=2. And if the 3rd number is subtracted from two times the 1st number and three times the 2nd, giving a result of 1 means 2x+3y-z=1. Then you use substition on these equations to get a equation where one variable equals 2 others, like using the first to get y=5-2z-3x and then this can be substituted into the other two to get x+3z-3(5-2z-3x)=2 and 2x+3(5-2z-3x)-z=1 we can then simplify and subtract the equations. After simplification we have 10x+9z=17 and 7z+7x=16 which can be turned into 70x+63z=119 and 70x+70z=160 which can be then subtracted to get that 7z=41 and z=41/7. Now we backtrack to a two variable equation like 7z+7x=16 and plug in to find x. So after plugging in we get 41+7x=16 and 7x=-25 so x=-25/7. Now we choose a 3 variable equation and plug in. So taking y=5-2z-3x we plug in 41/7 for z and -25/7 for x to get y=5-82/7+75/7 and y=5-7/7 and y=4. Therefore x = -25/7 y = 4 and z = 41/7.
Answer:
No solution
Step-by-step explanation:
Note how "2x" shows up in both equations. This suggests doing a substitution to solve the system.
Focus first on the first equation. Solving 2x - y = 7 for 2x, we get:
2x = y + 7.
Next, we substitute y + 7 for 2x in the second equation:
y = (y + 7) + 3.
Simplifying this produces:
0 = 10
This is not true and can never be true. Thus, this system has no solution.
Im not sure use a caculator with it
Answer:
f(3) = 68
Step-by-step explanation:
→ f(1) = 2 ⇒ f(2) = f²(2 − 1) + 4
= f²(1) + 4
= 2² + 4
= 8
………………………………………………
→ f(2) = 8 ⇒ f(3) = f²(3 − 1) + 4
= f²(2) + 4
= 8² + 4
= 64 + 4
= 68
