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Substitution of quadratic 10x-5y=3 and 6x+30y=81

User Blotto
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To solve the system of equations:

10x - 5y = 3 (Equation 1)
6x + 30y = 81 (Equation 2)

We can use the method of substitution.

From Equation 1, we can isolate x:

10x = 5y + 3
x = (5y + 3) / 10
x = (y + 3/5) / 2 (Equation 3)

Now substitute Equation 3 into Equation 2:

6((y + 3/5) / 2) + 30y = 81

Simplify the equation:

3(y + 3/5) + 30y = 81
3y + 9/5 + 30y = 81
33y + 9/5 = 81
33y = 81 - 9/5
33y = 405/5 - 9/5
33y = 396/5
y = (396/5) / 33
y = 12/5

Substitute the value of y back into Equation 3 to find x:

x = (12/5 + 3/5) / 2
x = 15/5 / 2
x = 3/2

Therefore, the solution to the system of equations is x = 3/2 and y = 12/5.
User Quadfinity
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