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How do you solve the system of equation problem in algebra 8x-10y equals 2 added with 7x-5y equals 13?
8x - 10y = 2
7x - 5y = 13
First, we need to eliminate one of the variables. For this, we would like to have them with the same coefficient but different sign. We can do this if we multiply by -2 the second equation.
8x - 10y = 2
-2(7x - 5y) = -2(13)
8x - 10y = 2
-14x +10y = -26
Now, we can add both equations, and solve for x.
(8x - 10y) + (-14x + 10y) = (2) + (-26)
(8x - 14x) +(-10y + 10y) = (2 - 26)
-6x = - 24
-6x/-6 = - 24/-6
x = 4
Substitute 4 for x in the first equation:
8x - 10y = 2
8(4) - 10y = 2
32 - 10y = 2
32 - 32 - 10y = 2 - 32
-10y = -30
-10y/-10 = -30/-10
y = 3
Thus, the solution of the system is (4, 3).
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Does this system of equations have one solution no solutions or an infinite number of solutions x - 2y equals -6 and x-2y equals 2?
x - 2y = -6 x - 2y = 2 subtract the 2nd equation from the 1st equation 0 = -8 false Therefore, the system of the equations has no solution.
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