144k views
1 vote
Substitution of quadratic 10x-5y=3 and 6x+30y=81

User Blotto
by
8.7k points

1 Answer

7 votes
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
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories