I have a question I have this systems to solve
2x-y = 1
3x-3=y
How do I go about solving this?
Please give me some direction
sub the 2nd into the 1st
2x - y = 1
2x - (3x-3) = 1
2x - 3x + 3 = 1
-x = -2
x = 2
back into the 2nd
3(2)-3 = y
y = 3
Thank you
To solve this system of equations, you can use the method of substitution or the method of elimination. I will explain both methods, and you can choose the one that you feel most comfortable with.
Method 1: Substitution
1. Solve one of the equations for one variable in terms of the other variable. Let's solve the second equation for y:
3x - 3 = y (rearrange the equation)
2. Substitute the value of y from the second equation into the first equation:
2x - (3x - 3) = 1 (substitute y with 3x - 3)
3. Simplify and solve for x:
2x - 3x + 3 = 1 (distribute the negative sign)
-x + 3 = 1 (combine like terms)
-x = -2 (subtract 3 from both sides)
x = 2 (divide both sides by -1)
4. Substitute the value of x back into one of the original equations to solve for y. Let's use the second equation:
3(2) - 3 = y (substitute x with 2)
6 - 3 = y (simplify)
3 = y
Therefore, the solution to the system of equations is x = 2 and y = 3.
Method 2: Elimination
1. Multiply each equation by a constant to make the coefficients of one variable cancel out when adding or subtracting the equations. In this case, let's multiply the first equation by 3 and the second equation by 1 to eliminate y:
Revised equations:
6x - 3y = 3
3x - 3y = 3
2. Subtract the revised equations to eliminate y:
(6x - 3y) - (3x - 3y) = 3 - 3 (subtract the equations)
6x - 3y - 3x + 3y = 0 (simplify)
3x = 0
3. Solve for x:
3x = 0 (divide both sides by 3)
x = 0
4. Substitute the value of x back into one of the original equations to solve for y. Let's use the second equation:
3(0) - 3 = y (substitute x with 0)
-3 = y
Therefore, the solution to the system of equations is x = 0 and y = -3.
Both methods should give you the same result, but the choice of method may depend on personal preference or the complexity of the equations.