Solve the system of equations:
First:x+3y=5
Second:3x-y=5
multiply the second equation by 3 then add both equations(I know I have it wrong somewhere I just guessed at the solution and it worked. I got (2,1)
Correct. After performing the addition step you mentioned, you get 10x = 20
Therefore x = 2
To solve the system of equations:
First, let's write the equations down:
First equation: x + 3y = 5
Second equation: 3x - y = 5
To eliminate one of the variables, we can multiply the second equation by a suitable constant so that the coefficients of one variable will cancel out when we add the two equations.
Let's multiply the second equation by 3:
3(3x - y) = 3(5)
which simplifies to:
9x - 3y = 15
Now we have:
First equation: x + 3y = 5
Modified second equation: 9x - 3y = 15
To eliminate the y variable, we can add the two equations together.
(x + 3y) + (9x - 3y) = 5 + 15
Combining like terms, we get:
10x + 0y = 20
which simplifies to:
10x = 20
To solve for x, we divide both sides of the equation by 10:
x = 20 / 10
x = 2
Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use the first equation:
x + 3y = 5
2 + 3y = 5
Subtract 2 from both sides to isolate the y variable:
3y = 5 - 2
3y = 3
Divide both sides by 3 to solve for y:
y = 1
So the solution to the system of equations is (x, y) = (2, 1).