Suppose a, b, c, and d are real numbers which satisfy the system of equations
a + 2b + 3c + 4d = 10
4a + b + 2c + 3d = 4
3a + 4b + c + 2d = -10
2a + 3b + 4c + d = −4.
Find a + b + c + d.
+526 We have the following system of equations -
\(a+2b+3c+4d=10\)
\(4a+b+2c+3d=4\)
\(3a+4b+c+2d=-10\)
\(2a+3b+4c+d=-4\)
Adding all of them,
⇒\(10a+10b+10c+10d=10-10+4-4\)
⇒\(10(a+b+c+d)=0\)
⇒\(a+b+c+d=0\)
∴ The value of a+b+c+d is 0.
P.S. This maybe due to the reason that either pair of a,b,c,d amount to 0 due to summing of negative and positive integers.
+526 We have the following system of equations -
\(a+2b+3c+4d=10\)
\(4a+b+2c+3d=4\)
\(3a+4b+c+2d=-10\)
\(2a+3b+4c+d=-4\)
Adding all of them,
⇒\(10a+10b+10c+10d=10-10+4-4\)
⇒\(10(a+b+c+d)=0\)
⇒\(a+b+c+d=0\)
∴ The value of a+b+c+d is 0.
P.S. This maybe due to the reason that either pair of a,b,c,d amount to 0 due to summing of negative and positive integers.
