Solving Linear Equations: 2x-3(2y+1)=15 and 3(x+1)+3y=2y-2
In this article, we will solve two linear equations: 2x-3(2y+1)=15 and 3(x+1)+3y=2y-2. We will use the principles of algebra to isolate the variables and find the solutions.
Equation 1: 2x-3(2y+1)=15
Let's start by simplifying the left-hand side of the equation:
2x - 3(2y+1) = 2x - 6y - 3 = 15
Next, we can add 3 to both sides of the equation to get:
2x - 6y = 18
Now, we can divide both sides of the equation by 2 to get:
x - 3y = 9
Equation 2: 3(x+1)+3y=2y-2
Let's simplify the left-hand side of the equation:
3(x+1) + 3y = 3x + 3 + 3y = 2y - 2
Next, we can subtract 3x from both sides of the equation to get:
3 + 3y = -3x + 2y - 2
Now, we can subtract 3 from both sides of the equation to get:
3y = -3x + 2y - 5
Solving the System of Equations
We have two linear equations with two variables x and y. We can use the method of substitution or elimination to solve the system of equations.
Let's use the elimination method. We can multiply equation (1) by 3 and equation (2) by 2 to get:
6x - 18y = 54 6x - 6y = -10
Now, we can subtract equation (4) from equation (3) to eliminate x:
-12y = 64
Next, we can divide both sides of the equation by -12 to get:
y = -64/12 y = -16/3
Now, we can substitute y = -16/3 into equation (1):
x - 3(-16/3) = 9 x + 16 = 9 x = -7
Therefore, the solution to the system of equations is x = -7 and y = -16/3.
Conclusion
In this article, we solved two linear equations: 2x-3(2y+1)=15 and 3(x+1)+3y=2y-2. We used the principles of algebra to simplify the equations and solve the system of equations using the elimination method. The solution to the system of equations is x = -7 and y = -16/3.
