The unbalanced equation is

#"C"_2"H"_2 + "O"_2 → "CO"_2 + "H"_2"O"#

Let's start with the most complicated formula, #"C"_2"H"_2#. We put a 1 in front of it.

#color(red)(1)"C"_2"H"_2 + "O"_2 → "CO"_2+ "H"_2"O"#

We have fixed 2 #"C"# atoms on the left, so we need 2 #"C"# atoms on the right. We put a 2 in front of #"CO"_2#.

#color(red)(1)"C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + "H"_2"O"#

We have also fixed 2 #"H"# atoms in the #"C"_2"H"_2#, so we need 2 #"H"# atoms on the right. We put a 1 in front of the #"H"_2"O"#.

#color(red)(1)"C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + color(blue)(1)"H"_2"O"#

Now we have fixed 5 #"O"# atoms on the right, so we need 5 #"O"# atoms on the left.

Oops! We would need to use 2½ molecules of #"O"_2#.

To avoid fractions, we multiply all coefficients by 2 and get

#color(red)(2)"C"_2"H"_2 + "O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O"#

We now have 10 #"O"# atoms on the right, so we need 10 #"O"# atoms on the left. We can now put a 5 in front of the #"O"_2#.

#color(red)(2)"C"_2"H"_2 + color(orange)(5)"O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O"#

The equation should now be balanced. Let's check.

#"Atom"color(white)(m) "On the left"color(white)(m) "On the right""#

#stackrel(—————————————)(color(white)(m)"C"color(white)(mmmmm)4color(white)(mmmmmm) 4)#

#color(white)(m)"H"color(white)(mmmmm) 4 color(white)(mmmmmm)4#

#color(white)(m)"O"color(white)(mmmml) 10color(white)(mmmmmll) 10#

The equation is balanced!