The one on the left is 6 (1/2x3x4) and the one on the right is 72 (6x12)
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1)
I:y=3x-4
II:9x-3y=14
substitute y into II:
9x-3*(3x-4)=14
9x-9x+12=14
12=14
this is obviously not equal so there is no solution, the lines are parallel
2)
I:y=4x+6
II:5x-y=6
substitute y into II:
5x-(4x+6)=6
5x-4x-6=6
x=12
substiute x into II:
5*12-y=6
-y=6-60
-y=-54
y=54
the solution is (12,54)
Answer:
x = 1 , y = 1 and z = 0
Step-by-step explanation:
-2x+2y+3z=0 ------(1)
-2x-y+z=-3 --------(2)
2x+3y+3z=5 ---------(3)
<u>To find the values of x,y and z</u>
(3) - (2)⇒ 2y + 4z = 2
y + 2z = 1 --------(4)
(1) - (2)⇒ 3y +2z =3 ---(5)
(5) - (4)⇒ 2y = 2
y = 1
Substituting the value of y in (4) we get z =0
Substituting the value of y and z in (1) we get x = 1
Therefore x = 1 , y = 1 and z = 0
Answer:
12 ??
Step-by-step explanation:
