Question

Given the system of equations, what is the solution?

x + 3y = 5
x - 3y = -1

Answer 1

`\begin{cases}x+3y=5\\x-3y=-1 \end{cases}\\---------(+)\\\\x+x+3y-3y=5+(-1)\\\\2x=4\qquad|:2\\\\\boxed{x=2}\\\\\\\\x+3y=5\\\\2+3y=5\qquad|-2\\\\2+3y-2=5-2\\\\3y=3\qquad|:3\\\\\boxed{y=1}`

The solution is (2; 1)

Answer 2

`\left \{ {{x + 3y = 5} \atop {x - 3y = -1}} \right.`

Immediately, you can tell that 3y and -3y will cancel out using elimination. That would be the easiest solving method to use here. Set up the problems like this, add them, and solve for the remaining variable:

x + 3y = 5
x - 3y = -1
---------------
2x = 4
x = 2

Now that you know what x is, you can substitute this value for x in one of the equations:

x + 3y = 5

2 + 3y = 5

3y = 3

y = 1

The solution to this system would be (2, 1).

You can perform a check for each equation to verify that this is the correct solution:

2 + 3(1) = 5

2 + 3 = 5

5 = 5
Correct

2 - 3(1) = -1

2 - 3 = -1

-1 = -1
Correct