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Given the same solution of the system of equations x + y = 3, ax + by = 4 and ax-2by = 22, X-Y = 7, find the value of 3a-16 Quadratic equation of two variables, done tonight
Add x + y = 3 and X-Y = 7 to get x + y + X-Y = 3 + 7, and substitute x = 5 into x + y = 3
2x=10 5+y=3
x=5 y=-2
Substituting x = 5, y = - 2 into ax + by = 4, ax-2by = 22, 5a-2b = 4, 5A + 4B = 22
5a-2b = 4, 5A + 4B = 22 subtracting to get 5a-2b-5a-4b = 4-22
-6b=-18
b=3
Substituting B = 3 into 5a-2b = 4 yields 5a-2 * 3 = 4
5a=10
a=2
Substituting a = 2 into 3a-16 = 6-16 = - 10
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