Answer:
Cedric system of equation has same solution as teacher solution
Explanation:
In this question we will find the solution of the system of the teacher and then we will check that whether it satisfies the other equations or not
The system of equation of the Teacher is
8x - 16y = 14 ..............(i)
-x + 5y = -3 .................(ii)
Now taking equation (ii)
-x + 5y = -3
Subtracting 5y from both sides of the equation
-x + 5y - 5y = -3 -5y
-x = -3 - 5y
Multiplying both sides of the equation with (-1)
it becomes
x = 3 + 5y ................(iii)
Putting this value of x in equation (i)
it becomes
8(3 + 5y) - 16y = 14
24 + 40 y -16 y = 14
24 + 24 y = 14
Subtracting 24 from both sides
24 - 24 + 24 y = 14 -24
24 y = -10
Dividing both sides by 24
Putting this value in equation (iii)
x = 3 + 5
x = 3 +
Taking LCM and solving on RHS
x =
x =
Check:
Katrina system of equation:
3(x)-15(y) = 9
put value of x and y
3(
) - 15 (
) = 9
(
) +(
) = 9
Adding on LHS
(
) = 9
(
) = 9
9 = 9 so first equation satisfies
Second equation
8(x)-16(y) = -7
put value of x and y
8(
) - 16 (
) = -7
(
) +(
) = -7
Adding on LHS
(
) = -7
(
) = -7
14.83 ≠ -7 so second equation does not satisfies
So System of Katrina is not same as Teacher
Cedric system of equation:
-(x)+5(y) = -3
put value of x and y
(
) +5 (
) = -3
(
) -(
) = -3
Adding on LHS
(
) = -3
(
) = -3
-3 = -3 so first equation satisfies
Second equation
-4(x)+8(y) = -7
put value of x and y
-4(
) +8 (
) = -7
(
) -(
) = -7
Adding on LHS
(
) = -7
(
) = -7
-7= -7 so second equation satisfies too
So System of Cedric is same as Teacher